{"id":1653,"date":"2016-06-22T13:22:11","date_gmt":"2016-06-22T13:22:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=1653"},"modified":"2023-11-08T13:20:23","modified_gmt":"2023-11-08T13:20:23","slug":"read-or-watch-multiplying-and-dividing-radical-expressions","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/read-or-watch-multiplying-and-dividing-radical-expressions\/","title":{"raw":"9.4 - Algebraic Operations with Radical Expressions","rendered":"9.4 &#8211; Algebraic Operations with Radical Expressions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(9.4.1) - Multiply and divide radical expressions\r\n<ul>\r\n \t<li>Use the product raised to a power rule to multiply radical expressions<\/li>\r\n \t<li>Use the quotient raised to a power rule to divide radical expressions<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>(9.4.2) - Add and subtract radical expressions<\/li>\r\n \t<li>(9.4.3) - Multiply radicals with multiple terms<\/li>\r\n \t<li>(9.4.4) - Rationalize a denominator containing a radical expression\r\n<ul>\r\n \t<li>Rationalize denominators with one term<\/li>\r\n \t<li>Rationalize denominators with higher roots<\/li>\r\n \t<li>Rationalize denominators with multiple terms<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\nWhen you learned how to solve linear equations, you probably learned about like terms first. \u00a0We can only combine terms that are alike, otherwise the terms will lose their meaning. \u00a0In this section, when you learn how to perform algebraic operations on radical expressions you will use the concept of like terms in a new way.\r\n\r\nYou will also use the distributive property, rules for exponents, and methods for multiplying binomials to perform algebraic operations on radical expressions.\r\n<h1>(9.4.1) - Multiply and Divide Radical Expressions<\/h1>\r\nYou can do more than just simplify <strong>radical expressions<\/strong>. You can multiply and divide them, too. The product raised to a power rule that we discussed previously will help us find products of radical expressions. Recall the rule:\r\n<h4>Multiply Radical Expressions<\/h4>\r\n<div class=\"textbox shaded\">\r\n<h3>A Product Raised to a Power Rule<\/h3>\r\nFor any numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex]: [latex] {{(ab)}^{n}}={{a}^{n}}\\cdot {{b}^{n}}[\/latex]\r\n\r\nFor any numbers [latex]a[\/latex] and [latex]b[\/latex] and any positive integer [latex]n[\/latex]: [latex]\\large {{(ab)}^{\\frac{1}{n}}}={{a}^{\\frac{1}{n}}}\\cdot {{b}^{\\frac{1}{n}}}[\/latex]\r\n\r\nFor any numbers [latex]a[\/latex] and [latex]b[\/latex] and any positive integer [latex]n[\/latex]: [latex] \\sqrt[n]{ab}=\\sqrt[n]{a}\\cdot \\sqrt[n]{b}[\/latex]\r\n\r\n<\/div>\r\nThe Product Raised to a Power Rule is important because you can use it to multiply radical expressions. Note that you can\u2019t multiply a square root and a cube root using this rule. The indices of the radicals must match in order to multiply them. In our first example we will work with integers, then we will move on to expressions with variable radicands.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt{18}\\cdot \\sqrt{16}[\/latex]\r\n\r\n[reveal-answer q=\"888021\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"888021\"]Use the rule [latex] \\sqrt[n]{a}\\cdot \\sqrt[n]{b}=\\sqrt[n]{ab}[\/latex] to multiply the radicands.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\sqrt{18\\cdot 16}\\\\\\sqrt{288}\\end{array}[\/latex]<\/p>\r\nLook for perfect squares in the radicand, and rewrite the radicand as the product of two factors.\r\n<p style=\"text-align: center\">[latex] \\sqrt{144\\cdot 2}[\/latex]<\/p>\r\nIdentify perfect squares.\r\n<p style=\"text-align: center\">[latex] \\sqrt{{{(12)}^{2}}\\cdot 2}[\/latex]<\/p>\r\nRewrite as the product of two radicals.\r\n<p style=\"text-align: center\">[latex] \\sqrt{{{(12)}^{2}}}\\cdot \\sqrt{2}[\/latex]<\/p>\r\nSimplify, using [latex] \\sqrt{{{x}^{2}}}=\\left| x \\right|[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\left| 12 \\right|\\cdot \\sqrt{2}\\\\12\\cdot \\sqrt{2}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{18}\\cdot \\sqrt{16}=12\\sqrt{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou may have also noticed that both [latex] \\sqrt{18}[\/latex] and [latex] \\sqrt{16}[\/latex] can be written as products involving perfect square factors. How would the expression change if you simplified each radical first, <i>before<\/i> multiplying? In the next example we will\u00a0use the same product from above to show that you can simplify before multiplying and get the same result.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt{18}\\cdot \\sqrt{16}[\/latex]\r\n\r\n[reveal-answer q=\"479810\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"479810\"]Look for perfect squares in each radicand, and rewrite as the product of two factors.\r\n<p style=\"text-align: center\">[latex] \\begin{array}{r}\\sqrt{9\\cdot 2}\\cdot \\sqrt{4\\cdot 4}\\\\\\sqrt{3\\cdot 3\\cdot 2}\\cdot \\sqrt{4\\cdot 4}\\end{array}[\/latex]<\/p>\r\nIdentify perfect squares.\r\n<p style=\"text-align: center\">[latex] \\sqrt{{{(3)}^{2}}\\cdot 2}\\cdot \\sqrt{{{(4)}^{2}}}[\/latex]<\/p>\r\nRewrite as the product of radicals.\r\n<p style=\"text-align: center\">[latex] \\sqrt{{{(3)}^{2}}}\\cdot \\sqrt{2}\\cdot \\sqrt{{{(4)}^{2}}}[\/latex]<\/p>\r\nSimplify, using [latex] \\sqrt{{{x}^{2}}}=\\left| x \\right|[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\left|3\\right|\\cdot\\sqrt{2}\\cdot\\left|4\\right|\\\\3\\cdot\\sqrt{2}\\cdot4\\end{array}[\/latex]<\/p>\r\nMultiply.\r\n<p style=\"text-align: center\">[latex] 12\\sqrt{2}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{18}\\cdot \\sqrt{16}=12\\sqrt{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn both cases, you arrive at the same product, [latex] 12\\sqrt{2}[\/latex]. It does not matter whether you multiply the radicands or simplify each radical first.\r\n\r\nYou multiply radical expressions that contain variables in the same manner. As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Look at the two examples that follow. In both problems, the Product Raised to a Power Rule is used right away and then the expression is simplified. Note that we specify that the variable is non-negative, [latex] x\\ge 0[\/latex], thus allowing us to avoid the need for absolute value.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt{12{{x}^{4}}}\\cdot \\sqrt{3x^2}[\/latex], [latex] x\\ge 0[\/latex]\r\n\r\n[reveal-answer q=\"843487\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843487\"]Use the rule [latex] \\sqrt[n]{a}\\cdot \\sqrt[n]{b}=\\sqrt[n]{ab}[\/latex] to multiply the radicands.\r\n<p style=\"text-align: center\">[latex] \\sqrt{12{{x}^{4}}\\cdot 3x^2} = \\sqrt{12\\cdot 3\\cdot {{x}^{4}}\\cdot x^2}[\/latex]<\/p>\r\nRecall that [latex] {{x}^{4}}\\cdot x^2={{x}^{4+2}}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\sqrt{36\\cdot {{x}^{4+2}}}\\\\\\sqrt{36\\cdot {{x}^{6}}}\\end{array}[\/latex]<\/p>\r\nLook for perfect squares in the radicand.\r\n<p style=\"text-align: center\">[latex] \\sqrt{{{(6)}^{2}}\\cdot {{({{x}^{3}})}^{2}}}[\/latex]<\/p>\r\nRewrite as the product of radicals.\r\n<p style=\"text-align: center\">[latex] \\begin{array}{c}\\sqrt{{{(6)}^{2}}}\\cdot \\sqrt{{{({{x}^{3}})}^{2}}}\\\\6\\cdot {{x}^{3}}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{12{{x}^{4}}}\\cdot \\sqrt{3x^2}=6{{x}^{3}}[\/latex]\r\n<h4>Analysis of the solution<\/h4>\r\nEven though our answer contained a variable with an odd exponent that was simplified from an even indexed root, we don't need to write our answer with absolute value because we specified before we simplified that\u00a0[latex] x\\ge 0[\/latex]. It is important to read the problem very well when you are doing math. \u00a0Even the smallest statement like\u00a0[latex] x\\ge 0[\/latex] can influence the way you write your answer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example we will multiply two cube roots.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt[3]{{{x}^{5}}{{y}^{2}}}\\cdot 5\\sqrt[3]{8{{x}^{2}}{{y}^{4}}}[\/latex]\r\n[reveal-answer q=\"399955\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"399955\"]Notice that <i>both<\/i> radicals are cube roots, so you can use the rule [latex] [\/latex] to multiply the radicands.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}5\\sqrt[3]{{{x}^{5}}{{y}^{2}}\\cdot 8{{x}^{2}}{{y}^{4}}}\\\\5\\sqrt[3]{8\\cdot {{x}^{5}}\\cdot {{x}^{2}}\\cdot {{y}^{2}}\\cdot {{y}^{4}}}\\\\5\\sqrt[3]{8\\cdot {{x}^{5+2}}\\cdot {{y}^{2+4}}}\\\\5\\sqrt[3]{8\\cdot {{x}^{7}}\\cdot {{y}^{6}}}\\end{array}[\/latex]<\/p>\r\nLook for perfect cubes in the radicand. Since [latex] {{x}^{7}}[\/latex] is not a perfect cube, it has to be rewritten as [latex] {{x}^{6+1}}={{({{x}^{2}})}^{3}}\\cdot x[\/latex].\r\n<p style=\"text-align: center\">[latex] 5\\sqrt[3]{{{(2)}^{3}}\\cdot {{({{x}^{2}})}^{3}}\\cdot x\\cdot {{({{y}^{2}})}^{3}}}[\/latex]<\/p>\r\nRewrite as the product of radicals.\r\n<p style=\"text-align: center\">[latex] \\begin{array}{r}5\\sqrt[3]{{{(2)}^{3}}}\\cdot \\sqrt[3]{{{({{x}^{2}})}^{3}}}\\cdot \\sqrt[3]{{{({{y}^{2}})}^{3}}}\\cdot \\sqrt[3]{x}\\\\5\\cdot 2\\cdot {{x}^{2}}\\cdot {{y}^{2}}\\cdot \\sqrt[3]{x}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt[3]{{{x}^{5}}{{y}^{2}}}\\cdot 5\\sqrt[3]{8{{x}^{2}}{{y}^{4}}}=10{{x}^{2}}{{y}^{2}}\\sqrt[3]{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we present more examples of how to multiply radical expressions.\r\n\r\nhttps:\/\/youtu.be\/PQs10_rFrSM\r\n\r\nThis next example is slightly more complicated because there are more than two radicals being multiplied. In this case, notice how the radicals are simplified before multiplication takes place. (Remember that the order you choose to use is up to you\u2014you will find that sometimes it is easier to multiply before simplifying, and other times it is easier to simplify before multiplying. With some practice, you may be able to tell which is which before you approach the problem, but either order will work for all problems.)\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] 2\\sqrt[4]{16{{x}^{9}}}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\sqrt[4]{81{{x}^{3}}y}[\/latex], [latex] x\\ge 0[\/latex], [latex] y\\ge 0[\/latex]\r\n\r\n[reveal-answer q=\"257458\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"257458\"]Notice this expression is multiplying three radicals with the same (fourth) root. Simplify each radical, if possible, before multiplying. Be looking for powers of 4 in each radicand.\r\n<p style=\"text-align: center\">[latex] 2\\sqrt[4]{{{(2)}^{4}}\\cdot {{({{x}^{2}})}^{4}}\\cdot x}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\sqrt[4]{{{(3)}^{4}}\\cdot {{x}^{3}}y}[\/latex]<\/p>\r\nRewrite as the product of radicals.\r\n<p style=\"text-align: center\">[latex] 2\\sqrt[4]{{{(2)}^{4}}}\\cdot \\sqrt[4]{{{({{x}^{2}})}^{4}}}\\cdot \\sqrt[4]{x}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\sqrt[4]{{{(3)}^{4}}}\\cdot \\sqrt[4]{{{x}^{3}}y}[\/latex]<\/p>\r\nIdentify and pull out powers of 4, using the fact that [latex] \\sqrt[4]{{{x}^{4}}}=\\left| x \\right|[\/latex].\r\n<p style=\"text-align: center\">[latex] \\begin{array}{r}2\\cdot \\left| 2 \\right|\\cdot \\left| {{x}^{2}} \\right|\\cdot \\sqrt[4]{x}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\left| 3 \\right|\\cdot \\sqrt[4]{{{x}^{3}}y}\\\\2\\cdot 2\\cdot {{x}^{2}}\\cdot \\sqrt[4]{x}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot 3\\cdot \\sqrt[4]{{{x}^{3}}y}\\end{array}[\/latex]<\/p>\r\nSince all the radicals are fourth roots, you can use the rule [latex] \\sqrt[x]{ab}=\\sqrt[x]{a}\\cdot \\sqrt[x]{b}[\/latex] to multiply the radicands.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2\\cdot 2\\cdot 3\\cdot {{x}^{2}}\\cdot \\sqrt[4]{x\\cdot {{y}^{3}}\\cdot {{x}^{3}}y}\\\\12{{x}^{2}}\\sqrt[4]{{{x}^{1+3}}\\cdot {{y}^{3+1}}}\\end{array}[\/latex]<\/p>\r\nNow that the radicands have been multiplied, look again for powers of 4, and pull them out. We can drop the absolute value signs in our final answer because at the start of the problem we were told [latex] x\\ge 0[\/latex], [latex] y\\ge 0[\/latex].\r\n<p style=\"text-align: center\">[latex] \\begin{array}{l}12{{x}^{2}}\\sqrt[4]{{{x}^{4}}\\cdot {{y}^{4}}}\\\\12{{x}^{2}}\\sqrt[4]{{{x}^{4}}}\\cdot \\sqrt[4]{{{y}^{4}}}\\\\12{{x}^{2}}\\cdot \\left| x \\right|\\cdot \\left| y \\right|\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 2\\sqrt[4]{16{{x}^{9}}}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\sqrt[4]{81{{x}^{3}}y}=12{{x}^{3}}y,\\,\\,x\\ge 0,\\,\\,y\\ge 0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show more examples of multiplying cube roots.\r\n\r\nhttps:\/\/youtu.be\/cxRXofdelIM\r\n<h4>Dividing Radical Expressions<\/h4>\r\nYou can use the same ideas to help you figure out how to simplify and divide radical expressions. Recall that the Product Raised to a Power Rule states that [latex] \\sqrt[n]{ab}=\\sqrt[n]{a}\\cdot \\sqrt[n]{b}[\/latex]. Well, what if you are dealing with a quotient instead of a product?\r\n\r\nThere is a rule for that, too. The <strong>Quotient Raised to a Power Rule<\/strong> states that [latex]\\displaystyle {{\\left( \\frac{a}{b} \\right)}^{n}}=\\frac{{{a}^{n}}}{{{b}^{n}}}[\/latex]. Again, if you imagine that the exponent is a rational number, then you can make this rule applicable for roots as well: [latex]\\displaystyle {{\\left( \\frac{a}{b} \\right)}^{\\frac{1}{n}}}=\\frac{{{a}^{\\frac{1}{n}}}}{{{b}^{\\frac{1}{n}}}}[\/latex], so [latex]\\displaystyle \\sqrt[n]{\\frac{a}{b}}=\\frac{\\sqrt[n]{a}}{\\sqrt[n]{b}}[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>A Quotient Raised to a Power Rule<\/h3>\r\nFor any real numbers [latex]a[\/latex] and [latex]b[\/latex] ([latex]b \\ne 0[\/latex]) and any positive integer [latex]n[\/latex]: [latex]\\large {{\\left( \\frac{a}{b} \\right)}^{\\frac{1}{n}}}=\\frac{{{a}^{\\frac{1}{n}}}}{{{b}^{\\frac{1}{n}}}}[\/latex]\r\n\r\nFor any real numbers [latex]a[\/latex] and [latex]b[\/latex] ([latex]b \\ne 0[\/latex])\u00a0and any positive integer [latex]n[\/latex]: [latex]\\large \\sqrt[n]{\\frac{a}{b}}=\\frac{\\sqrt[n]{a}}{\\sqrt[n]{b}}[\/latex]\r\n\r\n<\/div>\r\nAs you did with multiplication, you will start with some examples featuring integers before moving on to more complex expressions like [latex]\\displaystyle \\frac{\\sqrt[3]{24x{{y}^{4}}}}{\\sqrt[3]{8y}}[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex]\\displaystyle \\sqrt{\\frac{48}{25}}[\/latex]\r\n\r\n[reveal-answer q=\"883744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"883744\"]Use the rule [latex]\\displaystyle \\sqrt[n]{\\frac{a}{b}}=\\frac{\\sqrt[n]{a}}{\\sqrt[n]{b}}[\/latex] to create two radicals; one in the numerator and one in the denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{48}}{\\sqrt{25}}[\/latex]<\/p>\r\nSimplify each radical. Look for perfect square factors in the radicand, and rewrite the radicand as a product of factors.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\displaystyle \\frac{\\sqrt{16\\cdot 3}}{\\sqrt{25}}\\\\ \\text{or}\\\\ \\displaystyle \\frac{\\sqrt{4\\cdot 4\\cdot 3}}{\\sqrt{5\\cdot 5}}\\end{array}[\/latex]<\/p>\r\nIdentify and pull out perfect squares.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}\\frac{\\sqrt{{{(4)}^{2}}\\cdot 3}}{\\sqrt{{{(5)}^{2}}}}\\\\\\\\\\frac{\\sqrt{{{(4)}^{2}}}\\cdot \\sqrt{3}}{\\sqrt{{{(5)}^{2}}}}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{4\\cdot \\sqrt{3}}{5}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle \\sqrt{\\frac{48}{25}}=\\frac{4\\sqrt{3}}{5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex]\\displaystyle \\sqrt[3]{\\frac{640}{40}}[\/latex]\r\n\r\n[reveal-answer q=\"725564\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"725564\"]Rewrite using the Quotient Raised to a Power Rule.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt[3]{640}}{\\sqrt[3]{40}}[\/latex]<\/p>\r\nSimplify each radical. Look for perfect cubes in the radicand, and rewrite the radicand as a product of factors.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt[3]{64\\cdot 10}}{\\sqrt[3]{8\\cdot 5}}[\/latex]<\/p>\r\nIdentify and pull out perfect cubes.\r\n<p style=\"text-align: center\">[latex]\\Large \\begin{array}{r}\\frac{\\sqrt[3]{{{(4)}^{3}}\\cdot 10}}{\\sqrt[3]{{{(2)}^{3}}\\cdot 5}}\\\\\\\\\\frac{\\sqrt[3]{{{(4)}^{3}}}\\cdot \\sqrt[3]{10}}{\\sqrt[3]{{{(2)}^{3}}}\\cdot \\sqrt[3]{5}}\\\\\\\\\\frac{4\\cdot \\sqrt[3]{10}}{2\\cdot \\sqrt[3]{5}}\\end{array}[\/latex]<\/p>\r\nYou can simplify this expression even further by looking for common factors in the numerator and denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{4\\sqrt[3]{10}}{2\\sqrt[3]{5}}[\/latex]<\/p>\r\nRewrite the numerator as a product of factors.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2\\cdot 2\\sqrt[3]{5}\\cdot \\sqrt[3]{2}}{2\\sqrt[3]{5}}[\/latex]<\/p>\r\nIdentify factors of 1, and simplify.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}2\\cdot \\frac{2\\sqrt[3]{5}}{2\\sqrt[3]{5}}\\cdot \\sqrt[3]{2}\\\\\\\\2\\cdot 1\\cdot \\sqrt[3]{2}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle \\sqrt[3]{\\frac{640}{40}}=2\\sqrt[3]{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThat was a lot of effort, but you were able to simplify using the Quotient Raised to a Power Rule. What if you found the quotient of this expression by dividing within the radical first, and then took the cube root of the quotient?\r\n\r\nLet\u2019s take another look at that problem.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex]\\displaystyle \\frac{\\sqrt[3]{640}}{\\sqrt[3]{40}}[\/latex]\r\n\r\n[reveal-answer q=\"403134\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"403134\"]Since both radicals are cube roots, you can use the rule [latex]\\displaystyle \\frac{\\sqrt[x]{a}}{\\sqrt[x]{b}}=\\sqrt[x]{\\frac{a}{b}}[\/latex] to create a single rational expression underneath the radical.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\sqrt[3]{\\frac{640}{40}}[\/latex]<\/p>\r\nWithin the radical, divide [latex]640[\/latex] by [latex]40[\/latex].\r\n<p style=\"text-align: center\">[latex] \\begin{array}{r}640\\div 40=16\\\\\\sqrt[3]{16}\\end{array}[\/latex]<\/p>\r\nLook for perfect cubes in the radicand, and rewrite the radicand as a product of factors.\r\n<p style=\"text-align: center\">[latex]\\sqrt[3]{8\\cdot2}[\/latex]<\/p>\r\nIdentify perfect cubes and pull them out.\r\n<p style=\"text-align: center\">[latex] \\begin{array}{r}\\sqrt[3]{{{(2)}^{3}}\\cdot 2}\\\\\\sqrt[3]{2}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center\">[latex]2\\cdot\\sqrt[3]{2}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle\\frac{\\sqrt[3]{640}}{\\sqrt[3]{40}}=2\\sqrt[3]{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThat was a more straightforward approach, wasn\u2019t it?\r\n\r\nIn the next video we show more examples of simplifying a radical that contains a quotient.\r\n\r\nhttps:\/\/youtu.be\/SxImTm9GVNo\r\n\r\nAs with multiplication, the main idea here is that sometimes it makes sense to divide and then simplify, and other times it makes sense to simplify and then divide. Whichever order you choose, though, you should arrive at the same final expression.\r\n\r\nNow let\u2019s turn to some radical expressions containing division. Notice that the process for dividing these is the same as it is for dividing integers.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex]\\displaystyle \\frac{\\sqrt{30x}}{\\sqrt{10x}},x&gt;0[\/latex]\r\n\r\n[reveal-answer q=\"236188\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"236188\"]Use the Quotient Raised to a Power Rule to rewrite this expression.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\sqrt{\\frac{30x}{10x}}[\/latex]<\/p>\r\nSimplify [latex]\\displaystyle \\sqrt{\\frac{30x}{10x}}[\/latex] by identifying similar factors in the numerator and denominator and then identifying factors of 1.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}\\sqrt{\\frac{3\\cdot10x}{10x}}\\\\\\\\\\sqrt{3\\cdot\\frac{10x}{10x}}\\\\\\\\\\sqrt{3\\cdot1}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle \\frac{\\sqrt{30x}}{\\sqrt{10x}}=\\sqrt{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex]\\displaystyle \\frac{\\sqrt[3]{24x{{y}^{4}}}}{\\sqrt[3]{8y}},\\,\\,y\\ne 0[\/latex]\r\n\r\n[reveal-answer q=\"95343\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"95343\"]Use the Quotient Raised to a Power Rule to rewrite this expression.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\sqrt[3]{\\frac{24x{{y}^{4}}}{8y}}[\/latex]<\/p>\r\nSimplify [latex]\\displaystyle \\sqrt[3]{\\frac{24x{{y}^{4}}}{8y}}[\/latex] by identifying similar factors in the numerator and denominator and then identifying factors of 1.\r\n<p style=\"text-align: center\">[latex]\\Large \\begin{array}{l}\\sqrt[3]{\\frac{8\\cdot 3\\cdot x\\cdot {{y}^{3}}\\cdot y}{8\\cdot y}}\\\\\\\\\\sqrt[3]{\\frac{3\\cdot x\\cdot {{y}^{3}}}{1}\\cdot \\frac{8y}{8y}}\\\\\\\\\\sqrt[3]{\\frac{3\\cdot x\\cdot {{y}^{3}}}{1}\\cdot 1}\\end{array}[\/latex]<\/p>\r\nIdentify perfect cubes and pull them out of the radical.\r\n<p style=\"text-align: center\">[latex] \\sqrt[3]{3x{{y}^{3}}} = \\sqrt[3]{{{(y)}^{3}}\\cdot \\,3x}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center\">[latex] \\sqrt[3]{{{(y)}^{3}}}\\cdot \\,\\sqrt[3]{3x}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\frac{\\sqrt[3]{24x{{y}^{4}}}}{\\sqrt[3]{8y}}=y\\,\\sqrt[3]{3x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last video we show more examples of simplifying radicals that contain quotients with variables.\r\n\r\nhttps:\/\/youtu.be\/04X-hMgb0tA\r\n\r\nAs you become more familiar with dividing and simplifying radical expressions, make sure you continue to pay attention to the roots of the radicals that you are dividing. For example, while you can think of [latex]\\displaystyle \\frac{\\sqrt{8{{y}^{2}}}}{\\sqrt{225{{y}^{4}}}}[\/latex] as equivalent to [latex]\\displaystyle \\sqrt{\\frac{8{{y}^{2}}}{225{{y}^{4}}}}[\/latex] since both the numerator and the denominator are square roots, notice that you cannot express [latex]\\displaystyle \\frac{\\sqrt{8{{y}^{2}}}}{\\sqrt[4]{225{{y}^{4}}}}[\/latex] as [latex]\\displaystyle \\sqrt[4]{\\frac{8{{y}^{2}}}{225{{y}^{4}}}}[\/latex]. In this second case, the numerator is a square root and the denominator is a fourth root.\r\n<h1>(9.4.2) - Add and Subtract Radical Expressions<\/h1>\r\nThere are two keys to combining radicals by addition or subtraction: look at the <strong>index<\/strong>, and look at the <strong>radicand<\/strong>. If these are the same, then addition and subtraction are possible. If not, then you cannot combine the two radicals. In the graphic below, the index of the\u00a0expression [latex]12\\sqrt[3]{xy}[\/latex] is 3 and the radicand is [latex]xy[\/latex].\r\n\r\n<img class=\" wp-image-3200 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/29230521\/Screen-Shot-2016-07-29-at-4.04.52-PM-300x141.png\" alt=\"Screen Shot 2016-07-29 at 4.04.52 PM\" width=\"511\" height=\"240\" \/>\r\n\r\nMaking sense of a string of radicals may be difficult. One helpful tip is to think of radicals as variables, and treat them the same way. When you add and subtract variables, you look for like terms, which is the same thing you will do when you add and subtract radicals.\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Definition<\/h3>\r\nRadical expressions with the same indices and the same radicands are called <strong>like radicals.<\/strong>\r\n\r\n<\/div>\r\n<h4>Adding Radicals<\/h4>\r\nIn this first example, both radicals have the same radicand and index.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nAdd. [latex] 3\\sqrt{11}+7\\sqrt{11}[\/latex]\r\n[reveal-answer q=\"971281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"971281\"]The two radicals are the same, [latex] [\/latex]. This means you can combine them as you would combine the terms [latex] 3a+7a[\/latex].\r\n<p style=\"text-align: center\">[latex] \\text{3}\\sqrt{11}\\text{ + 7}\\sqrt{11}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 3\\sqrt{11}+7\\sqrt{11}=10\\sqrt{11}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThis next example contains more addends, or terms that are being added together. Notice how you can combine <i>like<\/i> terms (radicals that have the same root and index) but you cannot combine <i>unlike<\/i> terms.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nAdd. [latex] 5\\sqrt{2}+\\sqrt{3}+4\\sqrt{3}+2\\sqrt{2}[\/latex]\r\n\r\n[reveal-answer q=\"687881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"687881\"]Rearrange terms so that like radicals are next to each other. Then add.\r\n<p style=\"text-align: center\">[latex] 5\\sqrt{2}+2\\sqrt{2}+\\sqrt{3}+4\\sqrt{3}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 5\\sqrt{2}+\\sqrt{3}+4\\sqrt{3}+2\\sqrt{2}=7\\sqrt{2}+5\\sqrt{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that the expression in the previous example is simplified even though it has two terms: [latex] 7\\sqrt{2}[\/latex] and [latex] 5\\sqrt{3}[\/latex]. It would be a mistake to try to combine them further! (Some people make the mistake that [latex] 7\\sqrt{2}+5\\sqrt{3}=12\\sqrt{5}[\/latex]. This is incorrect because[latex] \\sqrt{2}[\/latex] and [latex]\\sqrt{3}[\/latex] are not like radicals so they cannot be added.)\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nAdd. [latex] 3\\sqrt{x}+12\\sqrt[3]{xy}+\\sqrt{x}[\/latex]\r\n\r\n[reveal-answer q=\"885242\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"885242\"]Rearrange terms so that like radicals are next to each other. Then add.\r\n<p style=\"text-align: center\">[latex] 3\\sqrt{x}+\\sqrt{x}+12\\sqrt[3]{xy}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 3\\sqrt{x}+12\\sqrt[3]{xy}+\\sqrt{x}=4\\sqrt{x}+12\\sqrt[3]{xy}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show more examples of how to identify and add like radicals.\r\n\r\nhttps:\/\/youtu.be\/ihcZhgm3yBg\r\n\r\nSometimes you may need to add <i>and<\/i> simplify the radical. If the radicals are different, try simplifying first\u2014you may end up being able to combine the radicals at the end, as shown in these next two examples.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nAdd and simplify. [latex] 2\\sqrt[3]{40}+\\sqrt[3]{135}[\/latex]\r\n\r\n[reveal-answer q=\"638886\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"638886\"]Simplify each radical by identifying perfect cubes.\r\n<p style=\"text-align: center\">[latex] \\begin{array}{r}2\\sqrt[3]{8\\cdot 5}+\\sqrt[3]{27\\cdot 5} &amp;=&amp; 2\\sqrt[3]{{{(2)}^{3}}\\cdot 5}+\\sqrt[3]{{{(3)}^{3}}\\cdot 5}\\\\ &amp;=&amp; 2\\sqrt[3]{{{(2)}^{3}}}\\cdot \\sqrt[3]{5}+\\sqrt[3]{{{(3)}^{3}}}\\cdot \\sqrt[3]{5}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center\">[latex] 2\\cdot 2\\cdot \\sqrt[3]{5}+3\\cdot \\sqrt[3]{5}[\/latex]<\/p>\r\nAdd.\r\n<p style=\"text-align: center\">[latex]4\\sqrt[3]{5}+3\\sqrt[3]{5}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 2\\sqrt[3]{40}+\\sqrt[3]{135}=7\\sqrt[3]{5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nAdd and simplify. [latex] x\\sqrt[3]{x{{y}^{4}}}+y\\sqrt[3]{{{x}^{4}}y}[\/latex]\r\n\r\n[reveal-answer q=\"95976\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"95976\"]Simplify each radical by identifying perfect cubes.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x\\sqrt[3]{x\\cdot {{y}^{3}}\\cdot y}+y\\sqrt[3]{{{x}^{3}}\\cdot x\\cdot y} &amp;=&amp; x\\sqrt[3]{{{y}^{3}}}\\cdot \\sqrt[3]{xy}+y\\sqrt[3]{{{x}^{3}}}\\cdot \\sqrt[3]{xy}\\\\ &amp;=&amp; xy\\cdot \\sqrt[3]{xy}+xy\\cdot \\sqrt[3]{xy}\\end{array}[\/latex]<\/p>\r\nAdd like radicals.\r\n<p style=\"text-align: center\">[latex] xy\\sqrt[3]{xy}+xy\\sqrt[3]{xy}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] x\\sqrt[3]{x{{y}^{4}}}+y\\sqrt[3]{{{x}^{4}}y}=2xy\\sqrt[3]{xy}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows more examples of adding radicals that require simplification.\r\n\r\nhttps:\/\/youtu.be\/S3fGUeALy7E\r\n<h4>Subtracting Radicals<\/h4>\r\nSubtraction of radicals follows the same set of rules and approaches as addition\u2014the radicands and the indices \u00a0must be the same for two (or more) radicals to be subtracted. In the three examples that follow, subtraction has been rewritten as addition of the opposite.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSubtract. [latex] 5\\sqrt{13}-3\\sqrt{13}[\/latex]\r\n\r\n[reveal-answer q=\"107411\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"107411\"]The radicands and indices are the same, so these two radicals can be combined.\r\n<p style=\"text-align: center\">[latex] 5\\sqrt{13}-3\\sqrt{13}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 5\\sqrt{13}-3\\sqrt{13}=2\\sqrt{13}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSubtract. [latex] 4\\sqrt[3]{5a}-\\sqrt[3]{3a}-2\\sqrt[3]{5a}[\/latex]\r\n\r\n[reveal-answer q=\"491962\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"491962\"]Two of the radicals have the same index and radicand, so they can be combined. Rewrite the expression so that like radicals are next to each other.\r\n<p style=\"text-align: center\">[latex] 4\\sqrt[3]{5a}+(-\\sqrt[3]{3a})+(-2\\sqrt[3]{5a}) = 4\\sqrt[3]{5a}+(-2\\sqrt[3]{5a})+(-\\sqrt[3]{3a})[\/latex]<\/p>\r\nCombine. Although the indices of [latex] 2\\sqrt[3]{5a}[\/latex] and [latex] -\\sqrt[3]{3a}[\/latex] are the same, the radicands are not\u2014so they cannot be combined.\r\n<p style=\"text-align: center\">[latex] 2\\sqrt[3]{5a}+(-\\sqrt[3]{3a})[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 4\\sqrt[3]{5a}-\\sqrt[3]{3a}-2\\sqrt[3]{5a}=2\\sqrt[3]{5a}-\\sqrt[3]{3a}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show more examples of subtracting radical expressions when no simplifying is required.\r\n\r\nhttps:\/\/youtu.be\/77TR9HsPZ6M\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSubtract and simplify. [latex] 5\\sqrt[4]{{{a}^{5}}b}-a\\sqrt[4]{16ab}[\/latex], where [latex]a\\ge 0[\/latex] and [latex]b\\ge 0[\/latex]\r\n\r\n[reveal-answer q=\"802638\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"802638\"]Simplify each radical by identifying and pulling out powers of 4.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}5\\sqrt[4]{{{a}^{4}}\\cdot a\\cdot b}-a\\sqrt[4]{{{(2)}^{4}}\\cdot a\\cdot b} &amp;=&amp; 5\\cdot a\\sqrt[4]{a\\cdot b}-a\\cdot 2\\sqrt[4]{a\\cdot b}\\\\&amp;=&amp; 5a\\sqrt[4]{ab}-2a\\sqrt[4]{ab}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 5\\sqrt[4]{{{a}^{5}}b}-a\\sqrt[4]{16ab}=3a\\sqrt[4]{ab}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last video we show more examples of subtracting radicals that require simplifying.\r\n\r\nhttps:\/\/youtu.be\/6MogonN1PRQ\r\n<h1>(9.4.3) - Multiply radicals with multiple terms<\/h1>\r\nWhen multiplying multiple term radical expressions it is important to follow the <strong>Distributive Property of Multiplication<\/strong>, as when you are multiplying regular, non-radical expressions.\r\n\r\nRadicals follow the same mathematical rules that other real numbers do. So, although the expression [latex] \\sqrt{x}(3\\sqrt{x}-5)[\/latex] may look different than [latex] a(3a-5)[\/latex], you can treat them the same way.\r\n\r\nLet\u2019s have a look at how to apply the Distributive Property. First let\u2019s do a problem with the variable [latex]a[\/latex], and then solve the same problem replacing [latex]a[\/latex] with [latex] \\sqrt{x}[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] a(3a-5)[\/latex]\r\n\r\n[reveal-answer q=\"653719\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"653719\"]Use the Distributive Property of Multiplication over Subtraction.\r\n<p style=\"text-align: center\">[latex]a(3a)-a(5) =3a^2-5a[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] a(3a-5)=3{{a}^{2}}-5a[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt{x}(3\\sqrt{x}-5)[\/latex]\r\n\r\n[reveal-answer q=\"886472\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"886472\"]Use the Distributive Property of Multiplication over Subtraction.\r\n<p style=\"text-align: center\">[latex] \\sqrt{x}(3\\sqrt{x})-\\sqrt{x}(5)[\/latex]<\/p>\r\nApply the rules of multiplying radicals: [latex] \\sqrt{a}\\cdot \\sqrt{b}=\\sqrt{ab}[\/latex] to multiply [latex] \\sqrt{x}(3\\sqrt{x})[\/latex].\r\n<p style=\"text-align: center\">[latex] 3\\sqrt{{{x}^{2}}}-5\\sqrt{x}[\/latex]<\/p>\r\nBe sure to simplify radicals when you can: [latex] \\sqrt{{{x}^{2}}}=\\left| x \\right|[\/latex], so [latex] 3\\sqrt{{{x}^{2}}}=3\\left| x \\right|[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt{x}(3\\sqrt{x}-5)=3\\left| x \\right|-5\\sqrt{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe answers to the previous two problems should look similar to you. The only difference is that in the second problem, [latex] \\sqrt{x}[\/latex] has replaced the variable [latex]a[\/latex]\u00a0(and so [latex] \\left| x \\right|[\/latex] has replaced [latex]a^2[\/latex]). The process of multiplying is very much the same in both problems.\r\n\r\nIn these next two problems, each term contains a radical.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] 7\\sqrt{x}\\left( 2\\sqrt{xy}+\\sqrt{y} \\right)[\/latex]\r\n\r\n[reveal-answer q=\"732671\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"732671\"]Use the Distributive Property of Multiplication over Addition to multiply each term within parentheses by [latex] 7\\sqrt{x}[\/latex].\r\n<p style=\"text-align: center\">[latex] 7\\sqrt{x}\\left( 2\\sqrt{xy} \\right)+7\\sqrt{x}\\left( \\sqrt{y} \\right)[\/latex]<\/p>\r\nApply the rules of multiplying radicals.\r\n<p style=\"text-align: center\">[latex] 7\\cdot 2\\sqrt{{{x}^{2}}y}+7\\sqrt{xy}[\/latex]<\/p>\r\n[latex] \\sqrt{{{x}^{2}}}=\\left| x \\right|[\/latex], so [latex] \\left| x \\right|[\/latex] can be pulled out of the radical.\r\n<p style=\"text-align: center\">[latex] 14|x|\\sqrt{y}+7\\sqrt{xy}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] 7\\sqrt{x}\\left( 2\\sqrt{xy}+\\sqrt{y} \\right)=14\\left| x \\right|\\sqrt{y}+7\\sqrt{xy}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex] \\sqrt[3]{a}\\left( 2\\sqrt[3]{{{a}^{2}}}-4\\sqrt[3]{{{a}^{5}}}+8\\sqrt[3]{{{a}^{8}}} \\right)[\/latex]\r\n\r\n[reveal-answer q=\"100802\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"100802\"]Use the Distributive Property.\r\n<p style=\"text-align: center\">[latex] \\sqrt[3]{a}\\left( 2\\sqrt[3]{{{a}^{2}}} \\right)-\\sqrt[3]{a}\\left( 4\\sqrt[3]{{{a}^{5}}} \\right)+\\sqrt[3]{a}\\left( 8\\sqrt[3]{{{a}^{8}}} \\right)[\/latex]<\/p>\r\nApply the rules of multiplying radicals.\r\n<p style=\"text-align: center\">[latex]2\\sqrt[3]{a\\cdot {{a}^{2}}}-4\\sqrt[3]{a\\cdot {{a}^{5}}}+8\\sqrt[3]{a\\cdot {{a}^{8}}} = 2\\sqrt[3]{{{a}^{3}}}-4\\sqrt[3]{{{a}^{6}}}+8\\sqrt[3]{{{a}^{9}}}[\/latex]<\/p>\r\nIdentify cubes in each of the radicals.\r\n<p style=\"text-align: center\">[latex] 2\\sqrt[3]{{{a}^{3}}}-4\\sqrt[3]{{{\\left( {{a}^{2}} \\right)}^{3}}}+8\\sqrt[3]{{{\\left( {{a}^{3}} \\right)}^{3}}}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\sqrt[3]{a}\\left( 2\\sqrt[3]{{{a}^{2}}}-4\\sqrt[3]{{{a}^{5}}}+8\\sqrt[3]{{{a}^{8}}} \\right)=2a-4{{a}^{2}}+8{{a}^{3}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show more examples of how to multiply radical expressions using distribution.\r\n\r\nhttps:\/\/youtu.be\/hizqmgBjW0k\r\n\r\nIn all of these examples, multiplication of radicals has been shown following the pattern [latex] \\sqrt{a}\\cdot \\sqrt{b}=\\sqrt{ab}[\/latex]. Then, only after multiplying, some radicals have been simplified\u2014like in the last problem. After you have worked with radical expressions a bit more, you may feel more comfortable identifying quantities such as [latex] \\sqrt{x}\\cdot \\sqrt{x}=x[\/latex] without going through the intermediate step of finding that [latex] \\sqrt{x}\\cdot \\sqrt{x}=\\sqrt{{{x}^{2}}}[\/latex]. In the rest of the examples that follow, though, each step is shown.\r\n<h4>Multiply Binomial Expressions That Contain Radicals<\/h4>\r\nYou can use the same technique for multiplying binomials to multiply binomial\u00a0expressions with radicals.\r\n\r\nAs a refresher, here is the process for multiplying two binomials. If you like using the expression \u201cFOIL\u201d (First, Outside, Inside, Last) to help you figure out the order in which the terms should be multiplied, you can use it here, too.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nMultiply. [latex] \\left( 2x+5 \\right)\\left( 3x-2 \\right)[\/latex]\r\n\r\n[reveal-answer q=\"577475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"577475\"]Use the Distributive Property.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{First}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x\\cdot 3x=6{{x}^{2}}\\\\\\text{Outside}:\\,\\,\\,2x\\cdot \\left( -2 \\right)=-4x\\\\\\text{Inside}:\\,\\,\\,\\,\\,\\,\\,\\,5\\cdot 3x=15x\\\\\\text{Last}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\cdot \\left( -2 \\right)=-10\\end{array}[\/latex]<\/p>\r\nRecord the terms, and then combine like terms.\r\n<p style=\"text-align: center\">[latex] 6{{x}^{2}}-4x+15x-10[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\left( 2x+5 \\right)\\left( 3x-2 \\right)=6{{x}^{2}}+11x-10[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is the same problem, with [latex] \\sqrt{b}[\/latex] replacing the variable [latex]x[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nMultiply. [latex] \\left( 2\\sqrt{b}+5 \\right)\\left( 3\\sqrt{b}-2 \\right),\\,\\,b\\ge 0[\/latex]\r\n\r\n[reveal-answer q=\"674608\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"674608\"]Use the Distributive Property to multiply. Simplify using [latex] \\sqrt{x}\\cdot \\sqrt{x}=x[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{First}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\sqrt{b}\\cdot 3\\sqrt{b}=2\\cdot 3\\cdot \\sqrt{b}\\cdot \\sqrt{b}=6b\\\\\\text{Outside}:\\,\\,\\,2\\sqrt{b}\\cdot \\left( -2 \\right)=-4\\sqrt{b}\\\\\\text{Inside}:\\,\\,\\,\\,\\,\\,\\,\\,5\\cdot 3\\sqrt{b}=15\\sqrt{b}\\\\\\text{Last}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\cdot \\left( -2 \\right)=-10\\end{array}[\/latex]<\/p>\r\nRecord the terms, and then combine like terms.\r\n<p style=\"text-align: center\">[latex] 6b-4\\sqrt{b}+15\\sqrt{b}-10[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\left( 2\\sqrt{b}+5 \\right)\\left( 3\\sqrt{b}-2 \\right)=6b+11\\sqrt{b}-10[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe multiplication works the same way in both problems; you just have to pay attention to the index of the radical (that is, whether the roots are square roots, cube roots, etc.) when multiplying radical expressions.\r\n<div class=\"textbox shaded\">\r\n<h3>Multiplying Two-Term\u00a0Radical Expressions<\/h3>\r\nTo multiply radical expressions, use the same method as used to multiply polynomials.\r\n<ul>\r\n \t<li>Use the Distributive Property (or, if you prefer, the shortcut FOIL method);<\/li>\r\n \t<li>Remember that [latex] \\sqrt{a}\\cdot \\sqrt{b}=\\sqrt{ab}[\/latex]; and<\/li>\r\n \t<li>Combine like terms.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nMultiply. [latex] \\left( 4{{x}^{2}}+\\sqrt[3]{x} \\right)\\left( \\sqrt[3]{{{x}^{2}}}+2 \\right)[\/latex]\r\n\r\n[reveal-answer q=\"865344\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"865344\"]Use FOIL to multiply.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{First}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4x^{2}\\cdot\\sqrt[3]{x^{2}}=4x^{2}\\sqrt[3]{x^{2}}\\\\\\text{Outside}:\\,\\,\\,4x^{2}\\cdot 2=8x^{2}\\\\\\text{Inside}:\\,\\,\\,\\,\\,\\,\\,\\,\\sqrt[3]{x}\\cdot\\sqrt[3]{x^{2}}=\\sqrt[3]{x^{2}\\cdot x}=\\sqrt[3]{x^{3}}=x\\\\\\text{Last}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\sqrt[3]{x}\\cdot 2=2\\sqrt[3]{x}\\end{array}[\/latex]<\/p>\r\nRecord the terms, and then combine like terms (if possible). Here, there are no like terms to combine.\r\n<p style=\"text-align: center\">[latex] 4{{x}^{2}}\\sqrt[3]{{{x}^{2}}}+8{{x}^{2}}+x+2\\sqrt[3]{x}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] \\left( 4{{x}^{2}}+\\sqrt[3]{x} \\right)\\left( \\sqrt[3]{{{x}^{2}}}+2 \\right)=4{{x}^{2}}\\sqrt[3]{{{x}^{2}}}+8{{x}^{2}}+x+2\\sqrt[3]{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show more examples of how to multiply two binomials that contain radicals.\r\n\r\nhttps:\/\/youtu.be\/VUWIBk3ga5I\r\n<h1>(9.4.4) - Rationalize Denominators<\/h1>\r\nAlthough radicals follow the same rules that integers do, it is often difficult to figure out the value of an expression containing radicals. For example, you probably have a good sense of how much [latex]\\displaystyle \\frac{4}{8},\\ 0.75[\/latex] and [latex]\\displaystyle \\frac{6}{9}[\/latex] are, but what about the quantities [latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex] and [latex]\\displaystyle \\frac{1}{\\sqrt{5}}[\/latex]? These are much harder to visualize.\r\n\r\nThat said, sometimes you have to work with expressions that contain many radicals. Often the value of these expressions is not immediately clear. In cases where you have a fraction with a radical in the denominator, you can use a technique called <strong>rationalizing a denominator<\/strong> to eliminate the radical. The point of rationalizing a denominator is to make it easier to understand what the quantity really is by removing radicals from the denominators.\r\n\r\nThe idea of rationalizing a denominator makes a bit more sense if you consider the definition of \u201crationalize.\u201d Recall that the numbers 5, [latex]\\displaystyle \\frac{1}{2}[\/latex], and [latex] 0.75[\/latex] are all known as rational numbers\u2014they can each be expressed as a ratio of two integers ([latex]\\displaystyle \\frac{5}{1},\\frac{1}{2}[\/latex]<i>,<\/i> and [latex]\\displaystyle \\frac{3}{4}[\/latex] respectively). Some radicals are irrational numbers because they cannot be represented as a ratio of two integers. As a result, the point of rationalizing a denominator is to change the expression so that the denominator becomes a rational number.\r\n\r\nHere are some examples of irrational and rational denominators.\r\n<table style=\"width: 20%\">\r\n<thead>\r\n<tr>\r\n<th>\r\n<p style=\"text-align: center\">Irrational<\/p>\r\n<\/th>\r\n<th><\/th>\r\n<th>\r\n<p style=\"text-align: center\">Rational<\/p>\r\n<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex]<\/p>\r\n<\/td>\r\n<td>\r\n<p style=\"text-align: center\">=<\/p>\r\n<\/td>\r\n<td>\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{2}}{2}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/p>\r\n<\/td>\r\n<td>\r\n<p style=\"text-align: center\">=<\/p>\r\n<\/td>\r\n<td>\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2\\sqrt{3}+3}{3}[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow let\u2019s examine how to get from irrational to rational denominators.\r\n<h3>Rationalizing Denominators with One Term<\/h3>\r\nLet\u2019s start with the fraction [latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex]. Its denominator is [latex] \\sqrt{2}[\/latex], an irrational number. This makes it difficult to figure out what the value of [latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex] is.\r\n\r\nYou can rename this fraction without changing its value, if you multiply it by 1. In this case, set 1 equal to [latex]\\displaystyle \\frac{\\sqrt{2}}{\\sqrt{2}}[\/latex]. Watch what happens.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{1}{\\sqrt{2}}\\cdot 1=\\frac{1}{\\sqrt{2}}\\cdot \\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{\\sqrt{2\\cdot 2}}=\\frac{\\sqrt{2}}{\\sqrt{4}}=\\frac{\\sqrt{2}}{2}[\/latex]<\/p>\r\nThe denominator of the new fraction is no longer a radical (notice, however, that the numerator is).\r\n\r\nSo why choose to multiply [latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex] by [latex]\\displaystyle \\frac{\\sqrt{2}}{\\sqrt{2}}[\/latex]? You knew that the square root of a number times itself will be a whole number. In algebraic terms, this idea is represented by [latex] \\sqrt{x}\\cdot \\sqrt{x}=x[\/latex]. Look back to the denominators in the multiplication of [latex]\\displaystyle \\frac{1}{\\sqrt{2}}\\cdot 1[\/latex]. Do you see where [latex] \\sqrt{2}\\cdot \\sqrt{2}=\\sqrt{4}=2[\/latex]?\r\n\r\nIn the following videos we show examples of rationalizing the denominator of a radical expression that contains integer radicands.\r\n\r\nhttps:\/\/youtu.be\/K7NdhPLVl7g\r\n\r\nHere are some more examples. Notice how the value of the fraction is not changed at all\u2014it is simply being multiplied by another name for 1.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRationalize the denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/p>\r\n[reveal-answer q=\"551606\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"551606\"]The denominator of this fraction is [latex] \\sqrt{3}[\/latex]. To make it into a rational number, multiply it by [latex] \\sqrt{3}[\/latex], since [latex] \\sqrt{3}\\cdot \\sqrt{3}=3[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/p>\r\nMultiply the entire fraction by another name for 1, [latex]\\displaystyle \\frac{\\sqrt{3}}{\\sqrt{3}}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}\\cdot \\frac{\\sqrt{3}}{\\sqrt{3}} = \\frac{\\sqrt{3}(2+\\sqrt{3})}{\\sqrt{3}\\cdot \\sqrt{3}}[\/latex]<\/p>\r\nUse the Distributive Property to multiply [latex] \\sqrt{3}(2+\\sqrt{3})[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2\\sqrt{3}+\\sqrt{3}\\cdot \\sqrt{3}}{\\sqrt{9}}[\/latex]<\/p>\r\nSimplify the radicals, where possible. [latex] \\sqrt{9}=3[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2\\sqrt{3}+\\sqrt{9}}{\\sqrt{9}}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}=\\frac{2\\sqrt{3}+3}{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou can use the same method to rationalize denominators to simplify fractions with radicals that contain a variable. As long as you multiply the original expression by another name for 1, you can eliminate a radical in the denominator without changing the value of the expression itself.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRationalize the denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}+\\sqrt{y}}{\\sqrt{x}},\\text{ where }x\\ne \\text{0}[\/latex]<\/p>\r\n[reveal-answer q=\"642546\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"642546\"]\r\n\r\nThe denominator is [latex] \\sqrt{x}[\/latex], so the entire expression can be multiplied by [latex]\\displaystyle \\frac{\\sqrt{x}}{\\sqrt{x}}[\/latex] to get rid of the radical in the denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}+\\sqrt{y}}{\\sqrt{x}}\\cdot \\frac{\\sqrt{x}}{\\sqrt{x}} = \\frac{\\sqrt{x}(\\sqrt{x}+\\sqrt{y})}{\\sqrt{x}\\cdot \\sqrt{x}}[\/latex]<\/p>\r\nUse the Distributive Property. Simplify the radicals, where possible. Remember that [latex] \\sqrt{x}\\cdot \\sqrt{x}=x[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}\\cdot \\sqrt{x}+\\sqrt{x}\\cdot \\sqrt{y}}{\\sqrt{x}\\cdot \\sqrt{x}}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle \\frac{\\sqrt{x}+\\sqrt{y}}{\\sqrt{x}}=\\frac{x+\\sqrt{xy}}{x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRationalize the denominator and simplify.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\sqrt{\\frac{100x}{11y}},\\text{ where }y\\ne \\text{0}[\/latex]<\/p>\r\n[reveal-answer q=\"197340\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"197340\"]Rewrite [latex]\\displaystyle \\sqrt{\\frac{a}{b}}[\/latex] as [latex]\\displaystyle \\frac{\\sqrt{a}}{\\sqrt{b}}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{100x}}{\\sqrt{11y}}[\/latex]<\/p>\r\nThe denominator is [latex] \\sqrt{11y}[\/latex], so multiplying the entire expression by [latex]\\displaystyle \\frac{\\sqrt{11y}}{\\sqrt{11y}}[\/latex] will rationalize the denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{100x\\cdot11y}}{\\sqrt{11y}\\cdot\\sqrt{11y}}[\/latex]<\/p>\r\nMultiply and simplify the radicals, where possible.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{100\\cdot 11xy}}{\\sqrt{11y}\\cdot \\sqrt{11y}}[\/latex]<\/p>\r\n100 is a perfect square.\u00a0Remember that[latex] \\sqrt{100}=10[\/latex]\u00a0and [latex] \\sqrt{x}\\cdot \\sqrt{x}=x[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{100}\\cdot \\sqrt{11xy}}{\\sqrt{11y}\\cdot \\sqrt{11y}}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle \\sqrt{\\frac{100x}{11y}}=\\frac{10\\sqrt{11xy}}{11y}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Rationalizing the Denominator with Higher Roots<\/h3>\r\n<p id=\"fs-id1169148883936\">When we rationalized a square root, we multiplied the numerator and denominator by a square root that would give us a perfect square under the radical in the denominator. When we took the square root, the denominator no longer had a radical.<\/p>\r\n<p id=\"fs-id1169149026920\">We will follow a similar process to rationalize higher roots. To rationalize a denominator with a higher index radical, we multiply the numerator and denominator by a radical that would give us a radicand that is a perfect power of the index. When we simplify the new radical, the denominator will no longer have a radical.<\/p>\r\n<p id=\"fs-id1169149100260\">For example,<\/p>\r\n\r\n<h2><img class=\"wp-image-5049 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2016\/06\/31021356\/CNX_IntAlg_Figure_08_05_004_img-300x113.jpg\" alt=\"\" width=\"823\" height=\"310\" \/><\/h2>\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nRationalize the denominator and simplify.\u00a0[latex]\\displaystyle \\frac{1}{\\sqrt[3]{6}}[\/latex]\r\n\r\n[reveal-answer q=\"672181\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"672181\"]\r\n\r\nThe radical in the denominator has one factor of 6. We multiply both the numerator and the denominator by [latex]\\sqrt[3]{6^2}[\/latex], which gives us 2 more factors of 6.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{1 \\cdot (\\sqrt[3]{6^2})}{\\sqrt[3]{6} \\cdot (\\sqrt[3]{6^2})}[\/latex]<\/p>\r\nMultiply. Notice the radicand in the denominator has 3 powers of 6.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt[3]{6^2}}{\\sqrt[3]{6^3}}[\/latex]<\/p>\r\nFinally, simplify the cube root in the denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt[3]{36}}{6}[\/latex]<\/p>\r\n<strong>Answer<\/strong>\r\n\r\n[latex]\\displaystyle \\frac{\\sqrt[3]{36}}{6}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nRationalize the denominator and simplify.\u00a0[latex]\\displaystyle \\frac{3}{\\sqrt[3]{4x}}[\/latex]\r\n\r\n[reveal-answer q=\"604990\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"604990\"]\r\n\r\nRewrite the radicand to show the factors.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{3}{\\sqrt[3]{2^2 \\cdot x}}[\/latex]<\/p>\r\nMultiply the numerator and denominator by [latex]\\sqrt[3]{2 \\cdot x^2}[\/latex]. This will get us 3 factors of 2 and 3 factors of [latex]x[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{3 \\cdot (\\sqrt[3]{2 \\cdot x^2})}{\\sqrt[3]{2^2 \\cdot x} \\cdot (\\sqrt[3]{2 \\cdot x^2})}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{3\\sqrt[3]{2x^2}}{\\sqrt[3]{2^3x^3}}[\/latex]<\/p>\r\nSimplify the radical in the denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{3\\sqrt[3]{2x^2}}{2x}[\/latex]<\/p>\r\n<strong>Answer<\/strong>\r\n\r\n[latex]\\displaystyle \\frac{3\\sqrt[3]{2x^2}}{2x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]19018[\/ohm_question]\r\n\r\n<\/div>\r\n<h3>Rationalizing Denominators with Two Terms<\/h3>\r\nDenominators do not always contain just one term, as shown in the previous examples. Sometimes, you will see expressions like [latex]\\displaystyle \\frac{3}{\\sqrt{2}+3}[\/latex] where the denominator is composed of two terms, [latex] \\sqrt{2}[\/latex] and [latex]+3[\/latex].\r\n\r\nUnfortunately, you cannot rationalize these denominators the same way you rationalize single-term denominators. If you multiply [latex] \\sqrt{2}+3[\/latex] by [latex] \\sqrt{2}[\/latex], you get [latex] 2+3\\sqrt{2}[\/latex]. The original [latex] \\sqrt{2}[\/latex] is gone, but now the quantity [latex] 3\\sqrt{2}[\/latex] has appeared...this is no better!\r\n\r\nIn order to rationalize this denominator, you want to square the radical term and somehow prevent the integer term from being multiplied by a radical. Is this possible?\r\n\r\nIt is possible\u2014and you have already seen how to do it!\r\n\r\nRecall that when binomials of the form [latex] (a+b)(a-b)[\/latex] are multiplied, the product is [latex] [\/latex]. So, for example, [latex] (x+3)(x-3)={{x}^{2}}-3x+3x-9={{x}^{2}}-9[\/latex]; notice that the terms [latex]\u22123x[\/latex] and [latex]+3x[\/latex] combine to 0. Now for the connection to rationalizing denominators: what if you replaced <i>x<\/i> with [latex] \\sqrt{2}[\/latex]?\r\n\r\nLook at the side by side examples below. Just as [latex] -3x+3x[\/latex] combines to 0 on the left, [latex] -3\\sqrt{2}+3\\sqrt{2}[\/latex] combines to 0 on the right.\r\n<table style=\"border-collapse: collapse;width: 47.6861%;height: 116px\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 50%\">[latex](x+3)(x-3)[\/latex]<\/td>\r\n<td style=\"width: 50%\">[latex]\\left( \\sqrt{2}+3 \\right)\\left( \\sqrt{2}-3 \\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">[latex]={{x}^{2}}-3x+3x-9[\/latex]<\/td>\r\n<td style=\"width: 50%\">[latex]={{\\left( \\sqrt{2} \\right)}^{2}}-3\\sqrt{2}+3\\sqrt{2}-9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\">[latex]={{x}^{2}}-9[\/latex]<\/td>\r\n<td style=\"width: 50%\">[latex]={{\\left( \\sqrt{2} \\right)}^{2}}-9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\"><\/td>\r\n<td style=\"width: 50%\">[latex]=2-9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%\"><\/td>\r\n<td style=\"width: 50%\">[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere you have it! Multiplying [latex] \\sqrt{2}+3[\/latex] by [latex] \\sqrt{2}-3[\/latex] removed one radical without adding another.\r\n\r\nIn this example, [latex] \\sqrt{2}-3[\/latex] is known as a <strong>conjugate<\/strong>, and [latex] \\sqrt{2}+3[\/latex] and [latex] \\sqrt{2}-3[\/latex] are known as a <strong>conjugate pair<\/strong>. To find the conjugate of a binomial that includes radicals, change the sign of the second term to its opposite as shown in the table below.\r\n<table style=\"width: 50%\">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\">Term<\/th>\r\n<th style=\"text-align: center\">Conjugate<\/th>\r\n<th style=\"text-align: center\">Product<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex] \\sqrt{2}+3[\/latex]<\/td>\r\n<td>[latex] \\sqrt{2}-3[\/latex]<\/td>\r\n<td>[latex] \\left( \\sqrt{2}+3 \\right)\\left( \\sqrt{2}-3 \\right)={{\\left( \\sqrt{2} \\right)}^{2}}-{{\\left( 3 \\right)}^{2}}=2-9=-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\sqrt{x}-5[\/latex]<\/td>\r\n<td>[latex] \\sqrt{x}+5[\/latex]<\/td>\r\n<td>[latex] \\left( \\sqrt{x}-5 \\right)\\left( \\sqrt{x}+5 \\right)={{\\left( \\sqrt{x} \\right)}^{2}}-{{\\left( 5 \\right)}^{2}}=x-25[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] 8-2\\sqrt{x}[\/latex]<\/td>\r\n<td>[latex] 8+2\\sqrt{x}[\/latex]<\/td>\r\n<td>[latex] \\left( 8-2\\sqrt{x} \\right)\\left( 8+2\\sqrt{x} \\right)={{\\left( 8 \\right)}^{2}}-{{\\left( 2\\sqrt{x} \\right)}^{2}}=64-4x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] 1+\\sqrt{xy}[\/latex]<\/td>\r\n<td>[latex] 1-\\sqrt{xy}[\/latex]<\/td>\r\n<td>[latex] \\left( 1+\\sqrt{xy} \\right)\\left( 1-\\sqrt{xy} \\right)={{\\left( 1 \\right)}^{2}}-{{\\left( \\sqrt{xy} \\right)}^{2}}=1-xy[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRationalize the denominator and simplify.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{5-\\sqrt{7}}{3+\\sqrt{5}}[\/latex]<\/p>\r\n[reveal-answer q=\"808741\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"808741\"]Find the conjugate of [latex] 3+\\sqrt{5}[\/latex]. Then multiply the entire expression by\u00a0[latex]\\displaystyle \\frac{3-\\sqrt{5}}{3-\\sqrt{5}}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{5-\\sqrt{7}}{3+\\sqrt{5}}\\cdot \\frac{3-\\sqrt{5}}{3-\\sqrt{5}} = \\frac{\\left( 5-\\sqrt{7} \\right)\\left( 3-\\sqrt{5} \\right)}{\\left( 3+\\sqrt{5} \\right)\\left( 3-\\sqrt{5} \\right)}[\/latex]<\/p>\r\nUse the Distributive Property to multiply the binomials in the numerator and denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{5\\cdot 3-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{7}\\cdot \\sqrt{5}}{3\\cdot 3-3\\sqrt{5}+3\\sqrt{5}-\\sqrt{5}\\cdot \\sqrt{5}}[\/latex]<\/p>\r\nSince you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to 0.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{15-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{35}}{9-3\\sqrt{5}+3\\sqrt{5}-\\sqrt{25}}[\/latex]<\/p>\r\nSimplify radicals where possible.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{15-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{35}}{9-\\sqrt{25}} = \\frac{15-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{35}}{9-5}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle \\frac{5-\\sqrt{7}}{3+\\sqrt{5}}=\\frac{15-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{35}}{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRationalize the denominator and simplify.\u00a0[latex]\\displaystyle \\frac{\\sqrt{x}}{\\sqrt{x}+2}[\/latex]\r\n\r\n[reveal-answer q=\"695658\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"695658\"]Find the conjugate of [latex] \\sqrt{x}+2[\/latex]. Then multiply the numerator and denominator by [latex]\\displaystyle \\frac{\\sqrt{x}-2}{\\sqrt{x}-2}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}}{\\sqrt{x}+2}\\cdot \\frac{\\sqrt{x}-2}{\\sqrt{x}-2} = \\frac{\\sqrt{x}\\left( \\sqrt{x}-2 \\right)}{\\left( \\sqrt{x}+2 \\right)\\left( \\sqrt{x}-2 \\right)}[\/latex]<\/p>\r\nUse the Distributive Property to multiply the binomials in the numerator and denominator.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}\\cdot \\sqrt{x}-2\\sqrt{x}}{\\sqrt{x}\\cdot \\sqrt{x}-2\\sqrt{x}+2\\sqrt{x}-2\\cdot 2}[\/latex]<\/p>\r\nSimplify. Remember that [latex] \\sqrt{x}\\cdot \\sqrt{x}=x[\/latex].\u00a0Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to 0.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}\\cdot \\sqrt{x}-2\\sqrt{x}}{\\sqrt{x}\\cdot \\sqrt{x}-2\\sqrt{x}+2\\sqrt{x}-4}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle \\frac{\\sqrt{x}}{\\sqrt{x}+2}=\\frac{x-2\\sqrt{x}}{x-4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOne word of caution: this method will work for binomials that include a square root, but not for binomials with roots greater than 2. This is because squaring a root that has an index greater than 2 does not remove the root, as shown below.\r\n<p style=\"text-align: center\">[latex] \\begin{array}{l}\\left( \\sqrt[3]{10}+5 \\right)\\left( \\sqrt[3]{10}-5 \\right)\\\\={{\\left( \\sqrt[3]{10} \\right)}^{2}}-5\\sqrt[3]{10}+5\\sqrt[3]{10}-25\\\\={{\\left( \\sqrt[3]{10} \\right)}^{2}}-25\\\\=\\sqrt[3]{100}-25\\end{array}[\/latex]<\/p>\r\n[latex] \\sqrt[3]{100}[\/latex] cannot be simplified any further, since its prime factors are [latex] 2\\cdot 2\\cdot 5\\cdot 5[\/latex]. There are no cubed numbers to pull out! Multiplying [latex] \\sqrt[3]{10}+5[\/latex] by its conjugate does not result in a radical-free expression.\r\n\r\nIn the following video we show more examples of how to rationalize a denominator using the conjugate.\r\n\r\nhttps:\/\/youtu.be\/vINRIRgeKqU\r\n\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(9.4.1) &#8211; Multiply and divide radical expressions\n<ul>\n<li>Use the product raised to a power rule to multiply radical expressions<\/li>\n<li>Use the quotient raised to a power rule to divide radical expressions<\/li>\n<\/ul>\n<\/li>\n<li>(9.4.2) &#8211; Add and subtract radical expressions<\/li>\n<li>(9.4.3) &#8211; Multiply radicals with multiple terms<\/li>\n<li>(9.4.4) &#8211; Rationalize a denominator containing a radical expression\n<ul>\n<li>Rationalize denominators with one term<\/li>\n<li>Rationalize denominators with higher roots<\/li>\n<li>Rationalize denominators with multiple terms<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>When you learned how to solve linear equations, you probably learned about like terms first. \u00a0We can only combine terms that are alike, otherwise the terms will lose their meaning. \u00a0In this section, when you learn how to perform algebraic operations on radical expressions you will use the concept of like terms in a new way.<\/p>\n<p>You will also use the distributive property, rules for exponents, and methods for multiplying binomials to perform algebraic operations on radical expressions.<\/p>\n<h1>(9.4.1) &#8211; Multiply and Divide Radical Expressions<\/h1>\n<p>You can do more than just simplify <strong>radical expressions<\/strong>. You can multiply and divide them, too. The product raised to a power rule that we discussed previously will help us find products of radical expressions. Recall the rule:<\/p>\n<h4>Multiply Radical Expressions<\/h4>\n<div class=\"textbox shaded\">\n<h3>A Product Raised to a Power Rule<\/h3>\n<p>For any numbers [latex]a[\/latex] and [latex]b[\/latex] and any integer [latex]n[\/latex]: [latex]{{(ab)}^{n}}={{a}^{n}}\\cdot {{b}^{n}}[\/latex]<\/p>\n<p>For any numbers [latex]a[\/latex] and [latex]b[\/latex] and any positive integer [latex]n[\/latex]: [latex]\\large {{(ab)}^{\\frac{1}{n}}}={{a}^{\\frac{1}{n}}}\\cdot {{b}^{\\frac{1}{n}}}[\/latex]<\/p>\n<p>For any numbers [latex]a[\/latex] and [latex]b[\/latex] and any positive integer [latex]n[\/latex]: [latex]\\sqrt[n]{ab}=\\sqrt[n]{a}\\cdot \\sqrt[n]{b}[\/latex]<\/p>\n<\/div>\n<p>The Product Raised to a Power Rule is important because you can use it to multiply radical expressions. Note that you can\u2019t multiply a square root and a cube root using this rule. The indices of the radicals must match in order to multiply them. In our first example we will work with integers, then we will move on to expressions with variable radicands.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt{18}\\cdot \\sqrt{16}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q888021\">Show Solution<\/span><\/p>\n<div id=\"q888021\" class=\"hidden-answer\" style=\"display: none\">Use the rule [latex]\\sqrt[n]{a}\\cdot \\sqrt[n]{b}=\\sqrt[n]{ab}[\/latex] to multiply the radicands.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\sqrt{18\\cdot 16}\\\\\\sqrt{288}\\end{array}[\/latex]<\/p>\n<p>Look for perfect squares in the radicand, and rewrite the radicand as the product of two factors.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt{144\\cdot 2}[\/latex]<\/p>\n<p>Identify perfect squares.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt{{{(12)}^{2}}\\cdot 2}[\/latex]<\/p>\n<p>Rewrite as the product of two radicals.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt{{{(12)}^{2}}}\\cdot \\sqrt{2}[\/latex]<\/p>\n<p>Simplify, using [latex]\\sqrt{{{x}^{2}}}=\\left| x \\right|[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\left| 12 \\right|\\cdot \\sqrt{2}\\\\12\\cdot \\sqrt{2}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{18}\\cdot \\sqrt{16}=12\\sqrt{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You may have also noticed that both [latex]\\sqrt{18}[\/latex] and [latex]\\sqrt{16}[\/latex] can be written as products involving perfect square factors. How would the expression change if you simplified each radical first, <i>before<\/i> multiplying? In the next example we will\u00a0use the same product from above to show that you can simplify before multiplying and get the same result.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt{18}\\cdot \\sqrt{16}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q479810\">Show Solution<\/span><\/p>\n<div id=\"q479810\" class=\"hidden-answer\" style=\"display: none\">Look for perfect squares in each radicand, and rewrite as the product of two factors.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\sqrt{9\\cdot 2}\\cdot \\sqrt{4\\cdot 4}\\\\\\sqrt{3\\cdot 3\\cdot 2}\\cdot \\sqrt{4\\cdot 4}\\end{array}[\/latex]<\/p>\n<p>Identify perfect squares.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt{{{(3)}^{2}}\\cdot 2}\\cdot \\sqrt{{{(4)}^{2}}}[\/latex]<\/p>\n<p>Rewrite as the product of radicals.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt{{{(3)}^{2}}}\\cdot \\sqrt{2}\\cdot \\sqrt{{{(4)}^{2}}}[\/latex]<\/p>\n<p>Simplify, using [latex]\\sqrt{{{x}^{2}}}=\\left| x \\right|[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\left|3\\right|\\cdot\\sqrt{2}\\cdot\\left|4\\right|\\\\3\\cdot\\sqrt{2}\\cdot4\\end{array}[\/latex]<\/p>\n<p>Multiply.<\/p>\n<p style=\"text-align: center\">[latex]12\\sqrt{2}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{18}\\cdot \\sqrt{16}=12\\sqrt{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In both cases, you arrive at the same product, [latex]12\\sqrt{2}[\/latex]. It does not matter whether you multiply the radicands or simplify each radical first.<\/p>\n<p>You multiply radical expressions that contain variables in the same manner. As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Look at the two examples that follow. In both problems, the Product Raised to a Power Rule is used right away and then the expression is simplified. Note that we specify that the variable is non-negative, [latex]x\\ge 0[\/latex], thus allowing us to avoid the need for absolute value.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt{12{{x}^{4}}}\\cdot \\sqrt{3x^2}[\/latex], [latex]x\\ge 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843487\">Show Solution<\/span><\/p>\n<div id=\"q843487\" class=\"hidden-answer\" style=\"display: none\">Use the rule [latex]\\sqrt[n]{a}\\cdot \\sqrt[n]{b}=\\sqrt[n]{ab}[\/latex] to multiply the radicands.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt{12{{x}^{4}}\\cdot 3x^2} = \\sqrt{12\\cdot 3\\cdot {{x}^{4}}\\cdot x^2}[\/latex]<\/p>\n<p>Recall that [latex]{{x}^{4}}\\cdot x^2={{x}^{4+2}}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\sqrt{36\\cdot {{x}^{4+2}}}\\\\\\sqrt{36\\cdot {{x}^{6}}}\\end{array}[\/latex]<\/p>\n<p>Look for perfect squares in the radicand.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt{{{(6)}^{2}}\\cdot {{({{x}^{3}})}^{2}}}[\/latex]<\/p>\n<p>Rewrite as the product of radicals.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\sqrt{{{(6)}^{2}}}\\cdot \\sqrt{{{({{x}^{3}})}^{2}}}\\\\6\\cdot {{x}^{3}}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{12{{x}^{4}}}\\cdot \\sqrt{3x^2}=6{{x}^{3}}[\/latex]<\/p>\n<h4>Analysis of the solution<\/h4>\n<p>Even though our answer contained a variable with an odd exponent that was simplified from an even indexed root, we don&#8217;t need to write our answer with absolute value because we specified before we simplified that\u00a0[latex]x\\ge 0[\/latex]. It is important to read the problem very well when you are doing math. \u00a0Even the smallest statement like\u00a0[latex]x\\ge 0[\/latex] can influence the way you write your answer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example we will multiply two cube roots.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt[3]{{{x}^{5}}{{y}^{2}}}\\cdot 5\\sqrt[3]{8{{x}^{2}}{{y}^{4}}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q399955\">Show Solution<\/span><\/p>\n<div id=\"q399955\" class=\"hidden-answer\" style=\"display: none\">Notice that <i>both<\/i> radicals are cube roots, so you can use the rule [latex][\/latex] to multiply the radicands.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}5\\sqrt[3]{{{x}^{5}}{{y}^{2}}\\cdot 8{{x}^{2}}{{y}^{4}}}\\\\5\\sqrt[3]{8\\cdot {{x}^{5}}\\cdot {{x}^{2}}\\cdot {{y}^{2}}\\cdot {{y}^{4}}}\\\\5\\sqrt[3]{8\\cdot {{x}^{5+2}}\\cdot {{y}^{2+4}}}\\\\5\\sqrt[3]{8\\cdot {{x}^{7}}\\cdot {{y}^{6}}}\\end{array}[\/latex]<\/p>\n<p>Look for perfect cubes in the radicand. Since [latex]{{x}^{7}}[\/latex] is not a perfect cube, it has to be rewritten as [latex]{{x}^{6+1}}={{({{x}^{2}})}^{3}}\\cdot x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]5\\sqrt[3]{{{(2)}^{3}}\\cdot {{({{x}^{2}})}^{3}}\\cdot x\\cdot {{({{y}^{2}})}^{3}}}[\/latex]<\/p>\n<p>Rewrite as the product of radicals.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}5\\sqrt[3]{{{(2)}^{3}}}\\cdot \\sqrt[3]{{{({{x}^{2}})}^{3}}}\\cdot \\sqrt[3]{{{({{y}^{2}})}^{3}}}\\cdot \\sqrt[3]{x}\\\\5\\cdot 2\\cdot {{x}^{2}}\\cdot {{y}^{2}}\\cdot \\sqrt[3]{x}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt[3]{{{x}^{5}}{{y}^{2}}}\\cdot 5\\sqrt[3]{8{{x}^{2}}{{y}^{4}}}=10{{x}^{2}}{{y}^{2}}\\sqrt[3]{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we present more examples of how to multiply radical expressions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Multiply Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/PQs10_rFrSM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>This next example is slightly more complicated because there are more than two radicals being multiplied. In this case, notice how the radicals are simplified before multiplication takes place. (Remember that the order you choose to use is up to you\u2014you will find that sometimes it is easier to multiply before simplifying, and other times it is easier to simplify before multiplying. With some practice, you may be able to tell which is which before you approach the problem, but either order will work for all problems.)<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]2\\sqrt[4]{16{{x}^{9}}}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\sqrt[4]{81{{x}^{3}}y}[\/latex], [latex]x\\ge 0[\/latex], [latex]y\\ge 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q257458\">Show Solution<\/span><\/p>\n<div id=\"q257458\" class=\"hidden-answer\" style=\"display: none\">Notice this expression is multiplying three radicals with the same (fourth) root. Simplify each radical, if possible, before multiplying. Be looking for powers of 4 in each radicand.<\/p>\n<p style=\"text-align: center\">[latex]2\\sqrt[4]{{{(2)}^{4}}\\cdot {{({{x}^{2}})}^{4}}\\cdot x}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\sqrt[4]{{{(3)}^{4}}\\cdot {{x}^{3}}y}[\/latex]<\/p>\n<p>Rewrite as the product of radicals.<\/p>\n<p style=\"text-align: center\">[latex]2\\sqrt[4]{{{(2)}^{4}}}\\cdot \\sqrt[4]{{{({{x}^{2}})}^{4}}}\\cdot \\sqrt[4]{x}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\sqrt[4]{{{(3)}^{4}}}\\cdot \\sqrt[4]{{{x}^{3}}y}[\/latex]<\/p>\n<p>Identify and pull out powers of 4, using the fact that [latex]\\sqrt[4]{{{x}^{4}}}=\\left| x \\right|[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2\\cdot \\left| 2 \\right|\\cdot \\left| {{x}^{2}} \\right|\\cdot \\sqrt[4]{x}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\left| 3 \\right|\\cdot \\sqrt[4]{{{x}^{3}}y}\\\\2\\cdot 2\\cdot {{x}^{2}}\\cdot \\sqrt[4]{x}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot 3\\cdot \\sqrt[4]{{{x}^{3}}y}\\end{array}[\/latex]<\/p>\n<p>Since all the radicals are fourth roots, you can use the rule [latex]\\sqrt[x]{ab}=\\sqrt[x]{a}\\cdot \\sqrt[x]{b}[\/latex] to multiply the radicands.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2\\cdot 2\\cdot 3\\cdot {{x}^{2}}\\cdot \\sqrt[4]{x\\cdot {{y}^{3}}\\cdot {{x}^{3}}y}\\\\12{{x}^{2}}\\sqrt[4]{{{x}^{1+3}}\\cdot {{y}^{3+1}}}\\end{array}[\/latex]<\/p>\n<p>Now that the radicands have been multiplied, look again for powers of 4, and pull them out. We can drop the absolute value signs in our final answer because at the start of the problem we were told [latex]x\\ge 0[\/latex], [latex]y\\ge 0[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}12{{x}^{2}}\\sqrt[4]{{{x}^{4}}\\cdot {{y}^{4}}}\\\\12{{x}^{2}}\\sqrt[4]{{{x}^{4}}}\\cdot \\sqrt[4]{{{y}^{4}}}\\\\12{{x}^{2}}\\cdot \\left| x \\right|\\cdot \\left| y \\right|\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]2\\sqrt[4]{16{{x}^{9}}}\\cdot \\sqrt[4]{{{y}^{3}}}\\cdot \\sqrt[4]{81{{x}^{3}}y}=12{{x}^{3}}y,\\,\\,x\\ge 0,\\,\\,y\\ge 0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show more examples of multiplying cube roots.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Multiply Cube Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/cxRXofdelIM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h4>Dividing Radical Expressions<\/h4>\n<p>You can use the same ideas to help you figure out how to simplify and divide radical expressions. Recall that the Product Raised to a Power Rule states that [latex]\\sqrt[n]{ab}=\\sqrt[n]{a}\\cdot \\sqrt[n]{b}[\/latex]. Well, what if you are dealing with a quotient instead of a product?<\/p>\n<p>There is a rule for that, too. The <strong>Quotient Raised to a Power Rule<\/strong> states that [latex]\\displaystyle {{\\left( \\frac{a}{b} \\right)}^{n}}=\\frac{{{a}^{n}}}{{{b}^{n}}}[\/latex]. Again, if you imagine that the exponent is a rational number, then you can make this rule applicable for roots as well: [latex]\\displaystyle {{\\left( \\frac{a}{b} \\right)}^{\\frac{1}{n}}}=\\frac{{{a}^{\\frac{1}{n}}}}{{{b}^{\\frac{1}{n}}}}[\/latex], so [latex]\\displaystyle \\sqrt[n]{\\frac{a}{b}}=\\frac{\\sqrt[n]{a}}{\\sqrt[n]{b}}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>A Quotient Raised to a Power Rule<\/h3>\n<p>For any real numbers [latex]a[\/latex] and [latex]b[\/latex] ([latex]b \\ne 0[\/latex]) and any positive integer [latex]n[\/latex]: [latex]\\large {{\\left( \\frac{a}{b} \\right)}^{\\frac{1}{n}}}=\\frac{{{a}^{\\frac{1}{n}}}}{{{b}^{\\frac{1}{n}}}}[\/latex]<\/p>\n<p>For any real numbers [latex]a[\/latex] and [latex]b[\/latex] ([latex]b \\ne 0[\/latex])\u00a0and any positive integer [latex]n[\/latex]: [latex]\\large \\sqrt[n]{\\frac{a}{b}}=\\frac{\\sqrt[n]{a}}{\\sqrt[n]{b}}[\/latex]<\/p>\n<\/div>\n<p>As you did with multiplication, you will start with some examples featuring integers before moving on to more complex expressions like [latex]\\displaystyle \\frac{\\sqrt[3]{24x{{y}^{4}}}}{\\sqrt[3]{8y}}[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\displaystyle \\sqrt{\\frac{48}{25}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q883744\">Show Solution<\/span><\/p>\n<div id=\"q883744\" class=\"hidden-answer\" style=\"display: none\">Use the rule [latex]\\displaystyle \\sqrt[n]{\\frac{a}{b}}=\\frac{\\sqrt[n]{a}}{\\sqrt[n]{b}}[\/latex] to create two radicals; one in the numerator and one in the denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{48}}{\\sqrt{25}}[\/latex]<\/p>\n<p>Simplify each radical. Look for perfect square factors in the radicand, and rewrite the radicand as a product of factors.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\displaystyle \\frac{\\sqrt{16\\cdot 3}}{\\sqrt{25}}\\\\ \\text{or}\\\\ \\displaystyle \\frac{\\sqrt{4\\cdot 4\\cdot 3}}{\\sqrt{5\\cdot 5}}\\end{array}[\/latex]<\/p>\n<p>Identify and pull out perfect squares.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}\\frac{\\sqrt{{{(4)}^{2}}\\cdot 3}}{\\sqrt{{{(5)}^{2}}}}\\\\\\\\\\frac{\\sqrt{{{(4)}^{2}}}\\cdot \\sqrt{3}}{\\sqrt{{{(5)}^{2}}}}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{4\\cdot \\sqrt{3}}{5}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\sqrt{\\frac{48}{25}}=\\frac{4\\sqrt{3}}{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\displaystyle \\sqrt[3]{\\frac{640}{40}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q725564\">Show Solution<\/span><\/p>\n<div id=\"q725564\" class=\"hidden-answer\" style=\"display: none\">Rewrite using the Quotient Raised to a Power Rule.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt[3]{640}}{\\sqrt[3]{40}}[\/latex]<\/p>\n<p>Simplify each radical. Look for perfect cubes in the radicand, and rewrite the radicand as a product of factors.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt[3]{64\\cdot 10}}{\\sqrt[3]{8\\cdot 5}}[\/latex]<\/p>\n<p>Identify and pull out perfect cubes.<\/p>\n<p style=\"text-align: center\">[latex]\\Large \\begin{array}{r}\\frac{\\sqrt[3]{{{(4)}^{3}}\\cdot 10}}{\\sqrt[3]{{{(2)}^{3}}\\cdot 5}}\\\\\\\\\\frac{\\sqrt[3]{{{(4)}^{3}}}\\cdot \\sqrt[3]{10}}{\\sqrt[3]{{{(2)}^{3}}}\\cdot \\sqrt[3]{5}}\\\\\\\\\\frac{4\\cdot \\sqrt[3]{10}}{2\\cdot \\sqrt[3]{5}}\\end{array}[\/latex]<\/p>\n<p>You can simplify this expression even further by looking for common factors in the numerator and denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{4\\sqrt[3]{10}}{2\\sqrt[3]{5}}[\/latex]<\/p>\n<p>Rewrite the numerator as a product of factors.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2\\cdot 2\\sqrt[3]{5}\\cdot \\sqrt[3]{2}}{2\\sqrt[3]{5}}[\/latex]<\/p>\n<p>Identify factors of 1, and simplify.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}2\\cdot \\frac{2\\sqrt[3]{5}}{2\\sqrt[3]{5}}\\cdot \\sqrt[3]{2}\\\\\\\\2\\cdot 1\\cdot \\sqrt[3]{2}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\sqrt[3]{\\frac{640}{40}}=2\\sqrt[3]{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>That was a lot of effort, but you were able to simplify using the Quotient Raised to a Power Rule. What if you found the quotient of this expression by dividing within the radical first, and then took the cube root of the quotient?<\/p>\n<p>Let\u2019s take another look at that problem.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\displaystyle \\frac{\\sqrt[3]{640}}{\\sqrt[3]{40}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q403134\">Show Solution<\/span><\/p>\n<div id=\"q403134\" class=\"hidden-answer\" style=\"display: none\">Since both radicals are cube roots, you can use the rule [latex]\\displaystyle \\frac{\\sqrt[x]{a}}{\\sqrt[x]{b}}=\\sqrt[x]{\\frac{a}{b}}[\/latex] to create a single rational expression underneath the radical.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\sqrt[3]{\\frac{640}{40}}[\/latex]<\/p>\n<p>Within the radical, divide [latex]640[\/latex] by [latex]40[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}640\\div 40=16\\\\\\sqrt[3]{16}\\end{array}[\/latex]<\/p>\n<p>Look for perfect cubes in the radicand, and rewrite the radicand as a product of factors.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt[3]{8\\cdot2}[\/latex]<\/p>\n<p>Identify perfect cubes and pull them out.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\sqrt[3]{{{(2)}^{3}}\\cdot 2}\\\\\\sqrt[3]{2}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center\">[latex]2\\cdot\\sqrt[3]{2}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle\\frac{\\sqrt[3]{640}}{\\sqrt[3]{40}}=2\\sqrt[3]{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>That was a more straightforward approach, wasn\u2019t it?<\/p>\n<p>In the next video we show more examples of simplifying a radical that contains a quotient.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Dividing Radicals without Variables (Basic with no rationalizing)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/SxImTm9GVNo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>As with multiplication, the main idea here is that sometimes it makes sense to divide and then simplify, and other times it makes sense to simplify and then divide. Whichever order you choose, though, you should arrive at the same final expression.<\/p>\n<p>Now let\u2019s turn to some radical expressions containing division. Notice that the process for dividing these is the same as it is for dividing integers.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\displaystyle \\frac{\\sqrt{30x}}{\\sqrt{10x}},x>0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q236188\">Show Solution<\/span><\/p>\n<div id=\"q236188\" class=\"hidden-answer\" style=\"display: none\">Use the Quotient Raised to a Power Rule to rewrite this expression.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\sqrt{\\frac{30x}{10x}}[\/latex]<\/p>\n<p>Simplify [latex]\\displaystyle \\sqrt{\\frac{30x}{10x}}[\/latex] by identifying similar factors in the numerator and denominator and then identifying factors of 1.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}\\sqrt{\\frac{3\\cdot10x}{10x}}\\\\\\\\\\sqrt{3\\cdot\\frac{10x}{10x}}\\\\\\\\\\sqrt{3\\cdot1}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{\\sqrt{30x}}{\\sqrt{10x}}=\\sqrt{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\displaystyle \\frac{\\sqrt[3]{24x{{y}^{4}}}}{\\sqrt[3]{8y}},\\,\\,y\\ne 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q95343\">Show Solution<\/span><\/p>\n<div id=\"q95343\" class=\"hidden-answer\" style=\"display: none\">Use the Quotient Raised to a Power Rule to rewrite this expression.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\sqrt[3]{\\frac{24x{{y}^{4}}}{8y}}[\/latex]<\/p>\n<p>Simplify [latex]\\displaystyle \\sqrt[3]{\\frac{24x{{y}^{4}}}{8y}}[\/latex] by identifying similar factors in the numerator and denominator and then identifying factors of 1.<\/p>\n<p style=\"text-align: center\">[latex]\\Large \\begin{array}{l}\\sqrt[3]{\\frac{8\\cdot 3\\cdot x\\cdot {{y}^{3}}\\cdot y}{8\\cdot y}}\\\\\\\\\\sqrt[3]{\\frac{3\\cdot x\\cdot {{y}^{3}}}{1}\\cdot \\frac{8y}{8y}}\\\\\\\\\\sqrt[3]{\\frac{3\\cdot x\\cdot {{y}^{3}}}{1}\\cdot 1}\\end{array}[\/latex]<\/p>\n<p>Identify perfect cubes and pull them out of the radical.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt[3]{3x{{y}^{3}}} = \\sqrt[3]{{{(y)}^{3}}\\cdot \\,3x}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt[3]{{{(y)}^{3}}}\\cdot \\,\\sqrt[3]{3x}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\frac{\\sqrt[3]{24x{{y}^{4}}}}{\\sqrt[3]{8y}}=y\\,\\sqrt[3]{3x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last video we show more examples of simplifying radicals that contain quotients with variables.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Dividing Radicals with Variables (Basic with no rationalizing)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/04X-hMgb0tA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>As you become more familiar with dividing and simplifying radical expressions, make sure you continue to pay attention to the roots of the radicals that you are dividing. For example, while you can think of [latex]\\displaystyle \\frac{\\sqrt{8{{y}^{2}}}}{\\sqrt{225{{y}^{4}}}}[\/latex] as equivalent to [latex]\\displaystyle \\sqrt{\\frac{8{{y}^{2}}}{225{{y}^{4}}}}[\/latex] since both the numerator and the denominator are square roots, notice that you cannot express [latex]\\displaystyle \\frac{\\sqrt{8{{y}^{2}}}}{\\sqrt[4]{225{{y}^{4}}}}[\/latex] as [latex]\\displaystyle \\sqrt[4]{\\frac{8{{y}^{2}}}{225{{y}^{4}}}}[\/latex]. In this second case, the numerator is a square root and the denominator is a fourth root.<\/p>\n<h1>(9.4.2) &#8211; Add and Subtract Radical Expressions<\/h1>\n<p>There are two keys to combining radicals by addition or subtraction: look at the <strong>index<\/strong>, and look at the <strong>radicand<\/strong>. If these are the same, then addition and subtraction are possible. If not, then you cannot combine the two radicals. In the graphic below, the index of the\u00a0expression [latex]12\\sqrt[3]{xy}[\/latex] is 3 and the radicand is [latex]xy[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3200 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/29230521\/Screen-Shot-2016-07-29-at-4.04.52-PM-300x141.png\" alt=\"Screen Shot 2016-07-29 at 4.04.52 PM\" width=\"511\" height=\"240\" \/><\/p>\n<p>Making sense of a string of radicals may be difficult. One helpful tip is to think of radicals as variables, and treat them the same way. When you add and subtract variables, you look for like terms, which is the same thing you will do when you add and subtract radicals.<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Definition<\/h3>\n<p>Radical expressions with the same indices and the same radicands are called <strong>like radicals.<\/strong><\/p>\n<\/div>\n<h4>Adding Radicals<\/h4>\n<p>In this first example, both radicals have the same radicand and index.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Add. [latex]3\\sqrt{11}+7\\sqrt{11}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q971281\">Show Solution<\/span><\/p>\n<div id=\"q971281\" class=\"hidden-answer\" style=\"display: none\">The two radicals are the same, [latex][\/latex]. This means you can combine them as you would combine the terms [latex]3a+7a[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\text{3}\\sqrt{11}\\text{ + 7}\\sqrt{11}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]3\\sqrt{11}+7\\sqrt{11}=10\\sqrt{11}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>This next example contains more addends, or terms that are being added together. Notice how you can combine <i>like<\/i> terms (radicals that have the same root and index) but you cannot combine <i>unlike<\/i> terms.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Add. [latex]5\\sqrt{2}+\\sqrt{3}+4\\sqrt{3}+2\\sqrt{2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q687881\">Show Solution<\/span><\/p>\n<div id=\"q687881\" class=\"hidden-answer\" style=\"display: none\">Rearrange terms so that like radicals are next to each other. Then add.<\/p>\n<p style=\"text-align: center\">[latex]5\\sqrt{2}+2\\sqrt{2}+\\sqrt{3}+4\\sqrt{3}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]5\\sqrt{2}+\\sqrt{3}+4\\sqrt{3}+2\\sqrt{2}=7\\sqrt{2}+5\\sqrt{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice that the expression in the previous example is simplified even though it has two terms: [latex]7\\sqrt{2}[\/latex] and [latex]5\\sqrt{3}[\/latex]. It would be a mistake to try to combine them further! (Some people make the mistake that [latex]7\\sqrt{2}+5\\sqrt{3}=12\\sqrt{5}[\/latex]. This is incorrect because[latex]\\sqrt{2}[\/latex] and [latex]\\sqrt{3}[\/latex] are not like radicals so they cannot be added.)<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Add. [latex]3\\sqrt{x}+12\\sqrt[3]{xy}+\\sqrt{x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q885242\">Show Solution<\/span><\/p>\n<div id=\"q885242\" class=\"hidden-answer\" style=\"display: none\">Rearrange terms so that like radicals are next to each other. Then add.<\/p>\n<p style=\"text-align: center\">[latex]3\\sqrt{x}+\\sqrt{x}+12\\sqrt[3]{xy}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]3\\sqrt{x}+12\\sqrt[3]{xy}+\\sqrt{x}=4\\sqrt{x}+12\\sqrt[3]{xy}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show more examples of how to identify and add like radicals.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Adding Radicals (Basic With No Simplifying)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ihcZhgm3yBg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes you may need to add <i>and<\/i> simplify the radical. If the radicals are different, try simplifying first\u2014you may end up being able to combine the radicals at the end, as shown in these next two examples.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Add and simplify. [latex]2\\sqrt[3]{40}+\\sqrt[3]{135}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q638886\">Show Solution<\/span><\/p>\n<div id=\"q638886\" class=\"hidden-answer\" style=\"display: none\">Simplify each radical by identifying perfect cubes.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}2\\sqrt[3]{8\\cdot 5}+\\sqrt[3]{27\\cdot 5} &=& 2\\sqrt[3]{{{(2)}^{3}}\\cdot 5}+\\sqrt[3]{{{(3)}^{3}}\\cdot 5}\\\\ &=& 2\\sqrt[3]{{{(2)}^{3}}}\\cdot \\sqrt[3]{5}+\\sqrt[3]{{{(3)}^{3}}}\\cdot \\sqrt[3]{5}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center\">[latex]2\\cdot 2\\cdot \\sqrt[3]{5}+3\\cdot \\sqrt[3]{5}[\/latex]<\/p>\n<p>Add.<\/p>\n<p style=\"text-align: center\">[latex]4\\sqrt[3]{5}+3\\sqrt[3]{5}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]2\\sqrt[3]{40}+\\sqrt[3]{135}=7\\sqrt[3]{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Add and simplify. [latex]x\\sqrt[3]{x{{y}^{4}}}+y\\sqrt[3]{{{x}^{4}}y}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q95976\">Show Solution<\/span><\/p>\n<div id=\"q95976\" class=\"hidden-answer\" style=\"display: none\">Simplify each radical by identifying perfect cubes.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x\\sqrt[3]{x\\cdot {{y}^{3}}\\cdot y}+y\\sqrt[3]{{{x}^{3}}\\cdot x\\cdot y} &=& x\\sqrt[3]{{{y}^{3}}}\\cdot \\sqrt[3]{xy}+y\\sqrt[3]{{{x}^{3}}}\\cdot \\sqrt[3]{xy}\\\\ &=& xy\\cdot \\sqrt[3]{xy}+xy\\cdot \\sqrt[3]{xy}\\end{array}[\/latex]<\/p>\n<p>Add like radicals.<\/p>\n<p style=\"text-align: center\">[latex]xy\\sqrt[3]{xy}+xy\\sqrt[3]{xy}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x\\sqrt[3]{x{{y}^{4}}}+y\\sqrt[3]{{{x}^{4}}y}=2xy\\sqrt[3]{xy}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows more examples of adding radicals that require simplification.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Adding Radicals That Requires Simplifying\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/S3fGUeALy7E?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h4>Subtracting Radicals<\/h4>\n<p>Subtraction of radicals follows the same set of rules and approaches as addition\u2014the radicands and the indices \u00a0must be the same for two (or more) radicals to be subtracted. In the three examples that follow, subtraction has been rewritten as addition of the opposite.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Subtract. [latex]5\\sqrt{13}-3\\sqrt{13}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q107411\">Show Solution<\/span><\/p>\n<div id=\"q107411\" class=\"hidden-answer\" style=\"display: none\">The radicands and indices are the same, so these two radicals can be combined.<\/p>\n<p style=\"text-align: center\">[latex]5\\sqrt{13}-3\\sqrt{13}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]5\\sqrt{13}-3\\sqrt{13}=2\\sqrt{13}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Subtract. [latex]4\\sqrt[3]{5a}-\\sqrt[3]{3a}-2\\sqrt[3]{5a}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q491962\">Show Solution<\/span><\/p>\n<div id=\"q491962\" class=\"hidden-answer\" style=\"display: none\">Two of the radicals have the same index and radicand, so they can be combined. Rewrite the expression so that like radicals are next to each other.<\/p>\n<p style=\"text-align: center\">[latex]4\\sqrt[3]{5a}+(-\\sqrt[3]{3a})+(-2\\sqrt[3]{5a}) = 4\\sqrt[3]{5a}+(-2\\sqrt[3]{5a})+(-\\sqrt[3]{3a})[\/latex]<\/p>\n<p>Combine. Although the indices of [latex]2\\sqrt[3]{5a}[\/latex] and [latex]-\\sqrt[3]{3a}[\/latex] are the same, the radicands are not\u2014so they cannot be combined.<\/p>\n<p style=\"text-align: center\">[latex]2\\sqrt[3]{5a}+(-\\sqrt[3]{3a})[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]4\\sqrt[3]{5a}-\\sqrt[3]{3a}-2\\sqrt[3]{5a}=2\\sqrt[3]{5a}-\\sqrt[3]{3a}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show more examples of subtracting radical expressions when no simplifying is required.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Subtracting Radicals (Basic With No Simplifying)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/77TR9HsPZ6M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Subtract and simplify. [latex]5\\sqrt[4]{{{a}^{5}}b}-a\\sqrt[4]{16ab}[\/latex], where [latex]a\\ge 0[\/latex] and [latex]b\\ge 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q802638\">Show Solution<\/span><\/p>\n<div id=\"q802638\" class=\"hidden-answer\" style=\"display: none\">Simplify each radical by identifying and pulling out powers of 4.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}5\\sqrt[4]{{{a}^{4}}\\cdot a\\cdot b}-a\\sqrt[4]{{{(2)}^{4}}\\cdot a\\cdot b} &=& 5\\cdot a\\sqrt[4]{a\\cdot b}-a\\cdot 2\\sqrt[4]{a\\cdot b}\\\\&=& 5a\\sqrt[4]{ab}-2a\\sqrt[4]{ab}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]5\\sqrt[4]{{{a}^{5}}b}-a\\sqrt[4]{16ab}=3a\\sqrt[4]{ab}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last video we show more examples of subtracting radicals that require simplifying.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Subtracting Radicals That Requires Simplifying\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/6MogonN1PRQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(9.4.3) &#8211; Multiply radicals with multiple terms<\/h1>\n<p>When multiplying multiple term radical expressions it is important to follow the <strong>Distributive Property of Multiplication<\/strong>, as when you are multiplying regular, non-radical expressions.<\/p>\n<p>Radicals follow the same mathematical rules that other real numbers do. So, although the expression [latex]\\sqrt{x}(3\\sqrt{x}-5)[\/latex] may look different than [latex]a(3a-5)[\/latex], you can treat them the same way.<\/p>\n<p>Let\u2019s have a look at how to apply the Distributive Property. First let\u2019s do a problem with the variable [latex]a[\/latex], and then solve the same problem replacing [latex]a[\/latex] with [latex]\\sqrt{x}[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]a(3a-5)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q653719\">Show Solution<\/span><\/p>\n<div id=\"q653719\" class=\"hidden-answer\" style=\"display: none\">Use the Distributive Property of Multiplication over Subtraction.<\/p>\n<p style=\"text-align: center\">[latex]a(3a)-a(5) =3a^2-5a[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]a(3a-5)=3{{a}^{2}}-5a[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt{x}(3\\sqrt{x}-5)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q886472\">Show Solution<\/span><\/p>\n<div id=\"q886472\" class=\"hidden-answer\" style=\"display: none\">Use the Distributive Property of Multiplication over Subtraction.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt{x}(3\\sqrt{x})-\\sqrt{x}(5)[\/latex]<\/p>\n<p>Apply the rules of multiplying radicals: [latex]\\sqrt{a}\\cdot \\sqrt{b}=\\sqrt{ab}[\/latex] to multiply [latex]\\sqrt{x}(3\\sqrt{x})[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]3\\sqrt{{{x}^{2}}}-5\\sqrt{x}[\/latex]<\/p>\n<p>Be sure to simplify radicals when you can: [latex]\\sqrt{{{x}^{2}}}=\\left| x \\right|[\/latex], so [latex]3\\sqrt{{{x}^{2}}}=3\\left| x \\right|[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt{x}(3\\sqrt{x}-5)=3\\left| x \\right|-5\\sqrt{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The answers to the previous two problems should look similar to you. The only difference is that in the second problem, [latex]\\sqrt{x}[\/latex] has replaced the variable [latex]a[\/latex]\u00a0(and so [latex]\\left| x \\right|[\/latex] has replaced [latex]a^2[\/latex]). The process of multiplying is very much the same in both problems.<\/p>\n<p>In these next two problems, each term contains a radical.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]7\\sqrt{x}\\left( 2\\sqrt{xy}+\\sqrt{y} \\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q732671\">Show Solution<\/span><\/p>\n<div id=\"q732671\" class=\"hidden-answer\" style=\"display: none\">Use the Distributive Property of Multiplication over Addition to multiply each term within parentheses by [latex]7\\sqrt{x}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]7\\sqrt{x}\\left( 2\\sqrt{xy} \\right)+7\\sqrt{x}\\left( \\sqrt{y} \\right)[\/latex]<\/p>\n<p>Apply the rules of multiplying radicals.<\/p>\n<p style=\"text-align: center\">[latex]7\\cdot 2\\sqrt{{{x}^{2}}y}+7\\sqrt{xy}[\/latex]<\/p>\n<p>[latex]\\sqrt{{{x}^{2}}}=\\left| x \\right|[\/latex], so [latex]\\left| x \\right|[\/latex] can be pulled out of the radical.<\/p>\n<p style=\"text-align: center\">[latex]14|x|\\sqrt{y}+7\\sqrt{xy}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]7\\sqrt{x}\\left( 2\\sqrt{xy}+\\sqrt{y} \\right)=14\\left| x \\right|\\sqrt{y}+7\\sqrt{xy}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\sqrt[3]{a}\\left( 2\\sqrt[3]{{{a}^{2}}}-4\\sqrt[3]{{{a}^{5}}}+8\\sqrt[3]{{{a}^{8}}} \\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q100802\">Show Solution<\/span><\/p>\n<div id=\"q100802\" class=\"hidden-answer\" style=\"display: none\">Use the Distributive Property.<\/p>\n<p style=\"text-align: center\">[latex]\\sqrt[3]{a}\\left( 2\\sqrt[3]{{{a}^{2}}} \\right)-\\sqrt[3]{a}\\left( 4\\sqrt[3]{{{a}^{5}}} \\right)+\\sqrt[3]{a}\\left( 8\\sqrt[3]{{{a}^{8}}} \\right)[\/latex]<\/p>\n<p>Apply the rules of multiplying radicals.<\/p>\n<p style=\"text-align: center\">[latex]2\\sqrt[3]{a\\cdot {{a}^{2}}}-4\\sqrt[3]{a\\cdot {{a}^{5}}}+8\\sqrt[3]{a\\cdot {{a}^{8}}} = 2\\sqrt[3]{{{a}^{3}}}-4\\sqrt[3]{{{a}^{6}}}+8\\sqrt[3]{{{a}^{9}}}[\/latex]<\/p>\n<p>Identify cubes in each of the radicals.<\/p>\n<p style=\"text-align: center\">[latex]2\\sqrt[3]{{{a}^{3}}}-4\\sqrt[3]{{{\\left( {{a}^{2}} \\right)}^{3}}}+8\\sqrt[3]{{{\\left( {{a}^{3}} \\right)}^{3}}}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\sqrt[3]{a}\\left( 2\\sqrt[3]{{{a}^{2}}}-4\\sqrt[3]{{{a}^{5}}}+8\\sqrt[3]{{{a}^{8}}} \\right)=2a-4{{a}^{2}}+8{{a}^{3}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show more examples of how to multiply radical expressions using distribution.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Multiplying Radical Expressions with Variables  Using Distribution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/hizqmgBjW0k?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In all of these examples, multiplication of radicals has been shown following the pattern [latex]\\sqrt{a}\\cdot \\sqrt{b}=\\sqrt{ab}[\/latex]. Then, only after multiplying, some radicals have been simplified\u2014like in the last problem. After you have worked with radical expressions a bit more, you may feel more comfortable identifying quantities such as [latex]\\sqrt{x}\\cdot \\sqrt{x}=x[\/latex] without going through the intermediate step of finding that [latex]\\sqrt{x}\\cdot \\sqrt{x}=\\sqrt{{{x}^{2}}}[\/latex]. In the rest of the examples that follow, though, each step is shown.<\/p>\n<h4>Multiply Binomial Expressions That Contain Radicals<\/h4>\n<p>You can use the same technique for multiplying binomials to multiply binomial\u00a0expressions with radicals.<\/p>\n<p>As a refresher, here is the process for multiplying two binomials. If you like using the expression \u201cFOIL\u201d (First, Outside, Inside, Last) to help you figure out the order in which the terms should be multiplied, you can use it here, too.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Multiply. [latex]\\left( 2x+5 \\right)\\left( 3x-2 \\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q577475\">Show Solution<\/span><\/p>\n<div id=\"q577475\" class=\"hidden-answer\" style=\"display: none\">Use the Distributive Property.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{First}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x\\cdot 3x=6{{x}^{2}}\\\\\\text{Outside}:\\,\\,\\,2x\\cdot \\left( -2 \\right)=-4x\\\\\\text{Inside}:\\,\\,\\,\\,\\,\\,\\,\\,5\\cdot 3x=15x\\\\\\text{Last}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\cdot \\left( -2 \\right)=-10\\end{array}[\/latex]<\/p>\n<p>Record the terms, and then combine like terms.<\/p>\n<p style=\"text-align: center\">[latex]6{{x}^{2}}-4x+15x-10[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left( 2x+5 \\right)\\left( 3x-2 \\right)=6{{x}^{2}}+11x-10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is the same problem, with [latex]\\sqrt{b}[\/latex] replacing the variable [latex]x[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Multiply. [latex]\\left( 2\\sqrt{b}+5 \\right)\\left( 3\\sqrt{b}-2 \\right),\\,\\,b\\ge 0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q674608\">Show Solution<\/span><\/p>\n<div id=\"q674608\" class=\"hidden-answer\" style=\"display: none\">Use the Distributive Property to multiply. Simplify using [latex]\\sqrt{x}\\cdot \\sqrt{x}=x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{First}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\sqrt{b}\\cdot 3\\sqrt{b}=2\\cdot 3\\cdot \\sqrt{b}\\cdot \\sqrt{b}=6b\\\\\\text{Outside}:\\,\\,\\,2\\sqrt{b}\\cdot \\left( -2 \\right)=-4\\sqrt{b}\\\\\\text{Inside}:\\,\\,\\,\\,\\,\\,\\,\\,5\\cdot 3\\sqrt{b}=15\\sqrt{b}\\\\\\text{Last}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5\\cdot \\left( -2 \\right)=-10\\end{array}[\/latex]<\/p>\n<p>Record the terms, and then combine like terms.<\/p>\n<p style=\"text-align: center\">[latex]6b-4\\sqrt{b}+15\\sqrt{b}-10[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left( 2\\sqrt{b}+5 \\right)\\left( 3\\sqrt{b}-2 \\right)=6b+11\\sqrt{b}-10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The multiplication works the same way in both problems; you just have to pay attention to the index of the radical (that is, whether the roots are square roots, cube roots, etc.) when multiplying radical expressions.<\/p>\n<div class=\"textbox shaded\">\n<h3>Multiplying Two-Term\u00a0Radical Expressions<\/h3>\n<p>To multiply radical expressions, use the same method as used to multiply polynomials.<\/p>\n<ul>\n<li>Use the Distributive Property (or, if you prefer, the shortcut FOIL method);<\/li>\n<li>Remember that [latex]\\sqrt{a}\\cdot \\sqrt{b}=\\sqrt{ab}[\/latex]; and<\/li>\n<li>Combine like terms.<\/li>\n<\/ul>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Multiply. [latex]\\left( 4{{x}^{2}}+\\sqrt[3]{x} \\right)\\left( \\sqrt[3]{{{x}^{2}}}+2 \\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q865344\">Show Solution<\/span><\/p>\n<div id=\"q865344\" class=\"hidden-answer\" style=\"display: none\">Use FOIL to multiply.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{First}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4x^{2}\\cdot\\sqrt[3]{x^{2}}=4x^{2}\\sqrt[3]{x^{2}}\\\\\\text{Outside}:\\,\\,\\,4x^{2}\\cdot 2=8x^{2}\\\\\\text{Inside}:\\,\\,\\,\\,\\,\\,\\,\\,\\sqrt[3]{x}\\cdot\\sqrt[3]{x^{2}}=\\sqrt[3]{x^{2}\\cdot x}=\\sqrt[3]{x^{3}}=x\\\\\\text{Last}:\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\sqrt[3]{x}\\cdot 2=2\\sqrt[3]{x}\\end{array}[\/latex]<\/p>\n<p>Record the terms, and then combine like terms (if possible). Here, there are no like terms to combine.<\/p>\n<p style=\"text-align: center\">[latex]4{{x}^{2}}\\sqrt[3]{{{x}^{2}}}+8{{x}^{2}}+x+2\\sqrt[3]{x}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left( 4{{x}^{2}}+\\sqrt[3]{x} \\right)\\left( \\sqrt[3]{{{x}^{2}}}+2 \\right)=4{{x}^{2}}\\sqrt[3]{{{x}^{2}}}+8{{x}^{2}}+x+2\\sqrt[3]{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show more examples of how to multiply two binomials that contain radicals.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-10\" title=\"Multiplying Binomial Radical Expressions with Variables\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/VUWIBk3ga5I?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(9.4.4) &#8211; Rationalize Denominators<\/h1>\n<p>Although radicals follow the same rules that integers do, it is often difficult to figure out the value of an expression containing radicals. For example, you probably have a good sense of how much [latex]\\displaystyle \\frac{4}{8},\\ 0.75[\/latex] and [latex]\\displaystyle \\frac{6}{9}[\/latex] are, but what about the quantities [latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex] and [latex]\\displaystyle \\frac{1}{\\sqrt{5}}[\/latex]? These are much harder to visualize.<\/p>\n<p>That said, sometimes you have to work with expressions that contain many radicals. Often the value of these expressions is not immediately clear. In cases where you have a fraction with a radical in the denominator, you can use a technique called <strong>rationalizing a denominator<\/strong> to eliminate the radical. The point of rationalizing a denominator is to make it easier to understand what the quantity really is by removing radicals from the denominators.<\/p>\n<p>The idea of rationalizing a denominator makes a bit more sense if you consider the definition of \u201crationalize.\u201d Recall that the numbers 5, [latex]\\displaystyle \\frac{1}{2}[\/latex], and [latex]0.75[\/latex] are all known as rational numbers\u2014they can each be expressed as a ratio of two integers ([latex]\\displaystyle \\frac{5}{1},\\frac{1}{2}[\/latex]<i>,<\/i> and [latex]\\displaystyle \\frac{3}{4}[\/latex] respectively). Some radicals are irrational numbers because they cannot be represented as a ratio of two integers. As a result, the point of rationalizing a denominator is to change the expression so that the denominator becomes a rational number.<\/p>\n<p>Here are some examples of irrational and rational denominators.<\/p>\n<table style=\"width: 20%\">\n<thead>\n<tr>\n<th>\n<p style=\"text-align: center\">Irrational<\/p>\n<\/th>\n<th><\/th>\n<th>\n<p style=\"text-align: center\">Rational<\/p>\n<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex]<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center\">=<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{2}}{2}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center\">=<\/p>\n<\/td>\n<td>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2\\sqrt{3}+3}{3}[\/latex]<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now let\u2019s examine how to get from irrational to rational denominators.<\/p>\n<h3>Rationalizing Denominators with One Term<\/h3>\n<p>Let\u2019s start with the fraction [latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex]. Its denominator is [latex]\\sqrt{2}[\/latex], an irrational number. This makes it difficult to figure out what the value of [latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex] is.<\/p>\n<p>You can rename this fraction without changing its value, if you multiply it by 1. In this case, set 1 equal to [latex]\\displaystyle \\frac{\\sqrt{2}}{\\sqrt{2}}[\/latex]. Watch what happens.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{1}{\\sqrt{2}}\\cdot 1=\\frac{1}{\\sqrt{2}}\\cdot \\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{\\sqrt{2\\cdot 2}}=\\frac{\\sqrt{2}}{\\sqrt{4}}=\\frac{\\sqrt{2}}{2}[\/latex]<\/p>\n<p>The denominator of the new fraction is no longer a radical (notice, however, that the numerator is).<\/p>\n<p>So why choose to multiply [latex]\\displaystyle \\frac{1}{\\sqrt{2}}[\/latex] by [latex]\\displaystyle \\frac{\\sqrt{2}}{\\sqrt{2}}[\/latex]? You knew that the square root of a number times itself will be a whole number. In algebraic terms, this idea is represented by [latex]\\sqrt{x}\\cdot \\sqrt{x}=x[\/latex]. Look back to the denominators in the multiplication of [latex]\\displaystyle \\frac{1}{\\sqrt{2}}\\cdot 1[\/latex]. Do you see where [latex]\\sqrt{2}\\cdot \\sqrt{2}=\\sqrt{4}=2[\/latex]?<\/p>\n<p>In the following videos we show examples of rationalizing the denominator of a radical expression that contains integer radicands.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-11\" title=\"Ex 1:  Rationalize the Denominator of a Radical Expression\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/K7NdhPLVl7g?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Here are some more examples. Notice how the value of the fraction is not changed at all\u2014it is simply being multiplied by another name for 1.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rationalize the denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q551606\">Show Solution<\/span><\/p>\n<div id=\"q551606\" class=\"hidden-answer\" style=\"display: none\">The denominator of this fraction is [latex]\\sqrt{3}[\/latex]. To make it into a rational number, multiply it by [latex]\\sqrt{3}[\/latex], since [latex]\\sqrt{3}\\cdot \\sqrt{3}=3[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}[\/latex]<\/p>\n<p>Multiply the entire fraction by another name for 1, [latex]\\displaystyle \\frac{\\sqrt{3}}{\\sqrt{3}}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}\\cdot \\frac{\\sqrt{3}}{\\sqrt{3}} = \\frac{\\sqrt{3}(2+\\sqrt{3})}{\\sqrt{3}\\cdot \\sqrt{3}}[\/latex]<\/p>\n<p>Use the Distributive Property to multiply [latex]\\sqrt{3}(2+\\sqrt{3})[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2\\sqrt{3}+\\sqrt{3}\\cdot \\sqrt{3}}{\\sqrt{9}}[\/latex]<\/p>\n<p>Simplify the radicals, where possible. [latex]\\sqrt{9}=3[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{2\\sqrt{3}+\\sqrt{9}}{\\sqrt{9}}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{2+\\sqrt{3}}{\\sqrt{3}}=\\frac{2\\sqrt{3}+3}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You can use the same method to rationalize denominators to simplify fractions with radicals that contain a variable. As long as you multiply the original expression by another name for 1, you can eliminate a radical in the denominator without changing the value of the expression itself.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rationalize the denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}+\\sqrt{y}}{\\sqrt{x}},\\text{ where }x\\ne \\text{0}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q642546\">Show Solution<\/span><\/p>\n<div id=\"q642546\" class=\"hidden-answer\" style=\"display: none\">\n<p>The denominator is [latex]\\sqrt{x}[\/latex], so the entire expression can be multiplied by [latex]\\displaystyle \\frac{\\sqrt{x}}{\\sqrt{x}}[\/latex] to get rid of the radical in the denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}+\\sqrt{y}}{\\sqrt{x}}\\cdot \\frac{\\sqrt{x}}{\\sqrt{x}} = \\frac{\\sqrt{x}(\\sqrt{x}+\\sqrt{y})}{\\sqrt{x}\\cdot \\sqrt{x}}[\/latex]<\/p>\n<p>Use the Distributive Property. Simplify the radicals, where possible. Remember that [latex]\\sqrt{x}\\cdot \\sqrt{x}=x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}\\cdot \\sqrt{x}+\\sqrt{x}\\cdot \\sqrt{y}}{\\sqrt{x}\\cdot \\sqrt{x}}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{\\sqrt{x}+\\sqrt{y}}{\\sqrt{x}}=\\frac{x+\\sqrt{xy}}{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rationalize the denominator and simplify.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\sqrt{\\frac{100x}{11y}},\\text{ where }y\\ne \\text{0}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q197340\">Show Solution<\/span><\/p>\n<div id=\"q197340\" class=\"hidden-answer\" style=\"display: none\">Rewrite [latex]\\displaystyle \\sqrt{\\frac{a}{b}}[\/latex] as [latex]\\displaystyle \\frac{\\sqrt{a}}{\\sqrt{b}}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{100x}}{\\sqrt{11y}}[\/latex]<\/p>\n<p>The denominator is [latex]\\sqrt{11y}[\/latex], so multiplying the entire expression by [latex]\\displaystyle \\frac{\\sqrt{11y}}{\\sqrt{11y}}[\/latex] will rationalize the denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{100x\\cdot11y}}{\\sqrt{11y}\\cdot\\sqrt{11y}}[\/latex]<\/p>\n<p>Multiply and simplify the radicals, where possible.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{100\\cdot 11xy}}{\\sqrt{11y}\\cdot \\sqrt{11y}}[\/latex]<\/p>\n<p>100 is a perfect square.\u00a0Remember that[latex]\\sqrt{100}=10[\/latex]\u00a0and [latex]\\sqrt{x}\\cdot \\sqrt{x}=x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{100}\\cdot \\sqrt{11xy}}{\\sqrt{11y}\\cdot \\sqrt{11y}}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\sqrt{\\frac{100x}{11y}}=\\frac{10\\sqrt{11xy}}{11y}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Rationalizing the Denominator with Higher Roots<\/h3>\n<p id=\"fs-id1169148883936\">When we rationalized a square root, we multiplied the numerator and denominator by a square root that would give us a perfect square under the radical in the denominator. When we took the square root, the denominator no longer had a radical.<\/p>\n<p id=\"fs-id1169149026920\">We will follow a similar process to rationalize higher roots. To rationalize a denominator with a higher index radical, we multiply the numerator and denominator by a radical that would give us a radicand that is a perfect power of the index. When we simplify the new radical, the denominator will no longer have a radical.<\/p>\n<p id=\"fs-id1169149100260\">For example,<\/p>\n<h2><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-5049 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2016\/06\/31021356\/CNX_IntAlg_Figure_08_05_004_img-300x113.jpg\" alt=\"\" width=\"823\" height=\"310\" \/><\/h2>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Rationalize the denominator and simplify.\u00a0[latex]\\displaystyle \\frac{1}{\\sqrt[3]{6}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q672181\">Show Answer<\/span><\/p>\n<div id=\"q672181\" class=\"hidden-answer\" style=\"display: none\">\n<p>The radical in the denominator has one factor of 6. We multiply both the numerator and the denominator by [latex]\\sqrt[3]{6^2}[\/latex], which gives us 2 more factors of 6.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{1 \\cdot (\\sqrt[3]{6^2})}{\\sqrt[3]{6} \\cdot (\\sqrt[3]{6^2})}[\/latex]<\/p>\n<p>Multiply. Notice the radicand in the denominator has 3 powers of 6.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt[3]{6^2}}{\\sqrt[3]{6^3}}[\/latex]<\/p>\n<p>Finally, simplify the cube root in the denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt[3]{36}}{6}[\/latex]<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>[latex]\\displaystyle \\frac{\\sqrt[3]{36}}{6}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Rationalize the denominator and simplify.\u00a0[latex]\\displaystyle \\frac{3}{\\sqrt[3]{4x}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604990\">Show Answer<\/span><\/p>\n<div id=\"q604990\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rewrite the radicand to show the factors.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{3}{\\sqrt[3]{2^2 \\cdot x}}[\/latex]<\/p>\n<p>Multiply the numerator and denominator by [latex]\\sqrt[3]{2 \\cdot x^2}[\/latex]. This will get us 3 factors of 2 and 3 factors of [latex]x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{3 \\cdot (\\sqrt[3]{2 \\cdot x^2})}{\\sqrt[3]{2^2 \\cdot x} \\cdot (\\sqrt[3]{2 \\cdot x^2})}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{3\\sqrt[3]{2x^2}}{\\sqrt[3]{2^3x^3}}[\/latex]<\/p>\n<p>Simplify the radical in the denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{3\\sqrt[3]{2x^2}}{2x}[\/latex]<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>[latex]\\displaystyle \\frac{3\\sqrt[3]{2x^2}}{2x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm19018\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19018&theme=oea&iframe_resize_id=ohm19018&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3>Rationalizing Denominators with Two Terms<\/h3>\n<p>Denominators do not always contain just one term, as shown in the previous examples. Sometimes, you will see expressions like [latex]\\displaystyle \\frac{3}{\\sqrt{2}+3}[\/latex] where the denominator is composed of two terms, [latex]\\sqrt{2}[\/latex] and [latex]+3[\/latex].<\/p>\n<p>Unfortunately, you cannot rationalize these denominators the same way you rationalize single-term denominators. If you multiply [latex]\\sqrt{2}+3[\/latex] by [latex]\\sqrt{2}[\/latex], you get [latex]2+3\\sqrt{2}[\/latex]. The original [latex]\\sqrt{2}[\/latex] is gone, but now the quantity [latex]3\\sqrt{2}[\/latex] has appeared&#8230;this is no better!<\/p>\n<p>In order to rationalize this denominator, you want to square the radical term and somehow prevent the integer term from being multiplied by a radical. Is this possible?<\/p>\n<p>It is possible\u2014and you have already seen how to do it!<\/p>\n<p>Recall that when binomials of the form [latex](a+b)(a-b)[\/latex] are multiplied, the product is [latex][\/latex]. So, for example, [latex](x+3)(x-3)={{x}^{2}}-3x+3x-9={{x}^{2}}-9[\/latex]; notice that the terms [latex]\u22123x[\/latex] and [latex]+3x[\/latex] combine to 0. Now for the connection to rationalizing denominators: what if you replaced <i>x<\/i> with [latex]\\sqrt{2}[\/latex]?<\/p>\n<p>Look at the side by side examples below. Just as [latex]-3x+3x[\/latex] combines to 0 on the left, [latex]-3\\sqrt{2}+3\\sqrt{2}[\/latex] combines to 0 on the right.<\/p>\n<table style=\"border-collapse: collapse;width: 47.6861%;height: 116px\">\n<tbody>\n<tr>\n<td style=\"width: 50%\">[latex](x+3)(x-3)[\/latex]<\/td>\n<td style=\"width: 50%\">[latex]\\left( \\sqrt{2}+3 \\right)\\left( \\sqrt{2}-3 \\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">[latex]={{x}^{2}}-3x+3x-9[\/latex]<\/td>\n<td style=\"width: 50%\">[latex]={{\\left( \\sqrt{2} \\right)}^{2}}-3\\sqrt{2}+3\\sqrt{2}-9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\">[latex]={{x}^{2}}-9[\/latex]<\/td>\n<td style=\"width: 50%\">[latex]={{\\left( \\sqrt{2} \\right)}^{2}}-9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\"><\/td>\n<td style=\"width: 50%\">[latex]=2-9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%\"><\/td>\n<td style=\"width: 50%\">[latex]-7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>There you have it! Multiplying [latex]\\sqrt{2}+3[\/latex] by [latex]\\sqrt{2}-3[\/latex] removed one radical without adding another.<\/p>\n<p>In this example, [latex]\\sqrt{2}-3[\/latex] is known as a <strong>conjugate<\/strong>, and [latex]\\sqrt{2}+3[\/latex] and [latex]\\sqrt{2}-3[\/latex] are known as a <strong>conjugate pair<\/strong>. To find the conjugate of a binomial that includes radicals, change the sign of the second term to its opposite as shown in the table below.<\/p>\n<table style=\"width: 50%\">\n<thead>\n<tr>\n<th style=\"text-align: center\">Term<\/th>\n<th style=\"text-align: center\">Conjugate<\/th>\n<th style=\"text-align: center\">Product<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\sqrt{2}+3[\/latex]<\/td>\n<td>[latex]\\sqrt{2}-3[\/latex]<\/td>\n<td>[latex]\\left( \\sqrt{2}+3 \\right)\\left( \\sqrt{2}-3 \\right)={{\\left( \\sqrt{2} \\right)}^{2}}-{{\\left( 3 \\right)}^{2}}=2-9=-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\sqrt{x}-5[\/latex]<\/td>\n<td>[latex]\\sqrt{x}+5[\/latex]<\/td>\n<td>[latex]\\left( \\sqrt{x}-5 \\right)\\left( \\sqrt{x}+5 \\right)={{\\left( \\sqrt{x} \\right)}^{2}}-{{\\left( 5 \\right)}^{2}}=x-25[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]8-2\\sqrt{x}[\/latex]<\/td>\n<td>[latex]8+2\\sqrt{x}[\/latex]<\/td>\n<td>[latex]\\left( 8-2\\sqrt{x} \\right)\\left( 8+2\\sqrt{x} \\right)={{\\left( 8 \\right)}^{2}}-{{\\left( 2\\sqrt{x} \\right)}^{2}}=64-4x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1+\\sqrt{xy}[\/latex]<\/td>\n<td>[latex]1-\\sqrt{xy}[\/latex]<\/td>\n<td>[latex]\\left( 1+\\sqrt{xy} \\right)\\left( 1-\\sqrt{xy} \\right)={{\\left( 1 \\right)}^{2}}-{{\\left( \\sqrt{xy} \\right)}^{2}}=1-xy[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rationalize the denominator and simplify.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{5-\\sqrt{7}}{3+\\sqrt{5}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q808741\">Show Solution<\/span><\/p>\n<div id=\"q808741\" class=\"hidden-answer\" style=\"display: none\">Find the conjugate of [latex]3+\\sqrt{5}[\/latex]. Then multiply the entire expression by\u00a0[latex]\\displaystyle \\frac{3-\\sqrt{5}}{3-\\sqrt{5}}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{5-\\sqrt{7}}{3+\\sqrt{5}}\\cdot \\frac{3-\\sqrt{5}}{3-\\sqrt{5}} = \\frac{\\left( 5-\\sqrt{7} \\right)\\left( 3-\\sqrt{5} \\right)}{\\left( 3+\\sqrt{5} \\right)\\left( 3-\\sqrt{5} \\right)}[\/latex]<\/p>\n<p>Use the Distributive Property to multiply the binomials in the numerator and denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{5\\cdot 3-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{7}\\cdot \\sqrt{5}}{3\\cdot 3-3\\sqrt{5}+3\\sqrt{5}-\\sqrt{5}\\cdot \\sqrt{5}}[\/latex]<\/p>\n<p>Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to 0.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{15-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{35}}{9-3\\sqrt{5}+3\\sqrt{5}-\\sqrt{25}}[\/latex]<\/p>\n<p>Simplify radicals where possible.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{15-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{35}}{9-\\sqrt{25}} = \\frac{15-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{35}}{9-5}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{5-\\sqrt{7}}{3+\\sqrt{5}}=\\frac{15-5\\sqrt{5}-3\\sqrt{7}+\\sqrt{35}}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rationalize the denominator and simplify.\u00a0[latex]\\displaystyle \\frac{\\sqrt{x}}{\\sqrt{x}+2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q695658\">Show Solution<\/span><\/p>\n<div id=\"q695658\" class=\"hidden-answer\" style=\"display: none\">Find the conjugate of [latex]\\sqrt{x}+2[\/latex]. Then multiply the numerator and denominator by [latex]\\displaystyle \\frac{\\sqrt{x}-2}{\\sqrt{x}-2}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}}{\\sqrt{x}+2}\\cdot \\frac{\\sqrt{x}-2}{\\sqrt{x}-2} = \\frac{\\sqrt{x}\\left( \\sqrt{x}-2 \\right)}{\\left( \\sqrt{x}+2 \\right)\\left( \\sqrt{x}-2 \\right)}[\/latex]<\/p>\n<p>Use the Distributive Property to multiply the binomials in the numerator and denominator.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}\\cdot \\sqrt{x}-2\\sqrt{x}}{\\sqrt{x}\\cdot \\sqrt{x}-2\\sqrt{x}+2\\sqrt{x}-2\\cdot 2}[\/latex]<\/p>\n<p>Simplify. Remember that [latex]\\sqrt{x}\\cdot \\sqrt{x}=x[\/latex].\u00a0Since you multiplied by the conjugate of the denominator, the radical terms in the denominator will combine to 0.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{\\sqrt{x}\\cdot \\sqrt{x}-2\\sqrt{x}}{\\sqrt{x}\\cdot \\sqrt{x}-2\\sqrt{x}+2\\sqrt{x}-4}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle \\frac{\\sqrt{x}}{\\sqrt{x}+2}=\\frac{x-2\\sqrt{x}}{x-4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>One word of caution: this method will work for binomials that include a square root, but not for binomials with roots greater than 2. This is because squaring a root that has an index greater than 2 does not remove the root, as shown below.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\left( \\sqrt[3]{10}+5 \\right)\\left( \\sqrt[3]{10}-5 \\right)\\\\={{\\left( \\sqrt[3]{10} \\right)}^{2}}-5\\sqrt[3]{10}+5\\sqrt[3]{10}-25\\\\={{\\left( \\sqrt[3]{10} \\right)}^{2}}-25\\\\=\\sqrt[3]{100}-25\\end{array}[\/latex]<\/p>\n<p>[latex]\\sqrt[3]{100}[\/latex] cannot be simplified any further, since its prime factors are [latex]2\\cdot 2\\cdot 5\\cdot 5[\/latex]. There are no cubed numbers to pull out! Multiplying [latex]\\sqrt[3]{10}+5[\/latex] by its conjugate does not result in a radical-free expression.<\/p>\n<p>In the following video we show more examples of how to rationalize a denominator using the conjugate.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-12\" title=\"Ex:  Rationalize the Denominator of a Radical Expression - Conjugate\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vINRIRgeKqU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1653\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Multiply Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/PQs10_rFrSM\">https:\/\/youtu.be\/PQs10_rFrSM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Multiply Cube Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/cxRXofdelIM\">https:\/\/youtu.be\/cxRXofdelIM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Dividing Radicals without Variables (Basic with no rationalizing). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/SxImTm9GVNo\">https:\/\/youtu.be\/SxImTm9GVNo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Dividing Radicals with Variables (Basic with no rationalizing). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/04X-hMgb0tA\">https:\/\/youtu.be\/04X-hMgb0tA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Adding Radicals (Basic With No Simplifying). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ihcZhgm3yBg\">https:\/\/youtu.be\/ihcZhgm3yBg<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Adding Radicals That Requires Simplifying. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/S3fGUeALy7E\">https:\/\/youtu.be\/S3fGUeALy7E<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Subtracting Radicals (Basic With No Simplifying). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/77TR9HsPZ6M\">https:\/\/youtu.be\/77TR9HsPZ6M<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Subtracting Radicals That Requires Simplifying. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/6MogonN1PRQ\">https:\/\/youtu.be\/6MogonN1PRQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Multiplying Radical Expressions with Variables Using Distribution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hizqmgBjW0k\">https:\/\/youtu.be\/hizqmgBjW0k<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Multiplying Binomial Radical Expressions with Variables. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/VUWIBk3ga5I\">https:\/\/youtu.be\/VUWIBk3ga5I<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Authored by<\/strong>: Abramson, Jay. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Dwonload fro free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/li><li>Ex 1: Rationalize the Denominator of a Radical Expression. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/K7NdhPLVl7g\">https:\/\/youtu.be\/K7NdhPLVl7g<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Rationalize the Denominator of a Radical Expression - Conjugate. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/vINRIRgeKqU\">https:\/\/youtu.be\/vINRIRgeKqU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Intermediate Algebra. <strong>Authored by<\/strong>: Lynn Marecek. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/02776133-d49d-49cb-bfaa-67c7f61b25a1@3.16\">http:\/\/cnx.org\/contents\/02776133-d49d-49cb-bfaa-67c7f61b25a1@3.16<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/02776133-d49d-49cb-bfaa-67c7f61b25a1@3.16<\/li><li>Question ID#19018. <strong>Authored by<\/strong>: Wallace,Tyler, mb Lippman,David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Multiply Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/PQs10_rFrSM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Multiply Cube Roots\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen 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