{"id":175,"date":"2023-11-08T16:10:14","date_gmt":"2023-11-08T16:10:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-multiply-and-divide-rational-expressions\/"},"modified":"2024-09-23T04:15:50","modified_gmt":"2024-09-23T04:15:50","slug":"4-2-multiplying-and-dividing-rational-expressions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/4-2-multiplying-and-dividing-rational-expressions\/","title":{"raw":"4.2 Multiplying and Dividing Rational Expressions","rendered":"4.2 Multiplying and Dividing Rational Expressions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Multiply and simplify rational expressions.<\/li>\r\n \t<li>Divide and simplify rational expressions.<\/li>\r\n<\/ul>\r\n<\/div>\r\nJust as we can multiply and divide fractions, we can multiply and divide <strong>rational expressions<\/strong>. In fact, we use the same processes for multiplying and dividing rational expressions as we use for multiplying and dividing fractions. The process is the same even though the expressions look different!\r\n<h3>Multiply Rational Expressions<\/h3>\r\nRemember that there are two ways to multiply fractions.\r\n\r\nOne way is to multiply the numerators and the denominators separately, and then simplify the product, as shown here. Notice we factor out the <strong>greatest common factor<\/strong>\u00a0(between the numerator and denominator) for efficiency, though we could have worked with the prime factorization as well.\r\n<p style=\"text-align: center;\">[latex] \\require{color} \\displaystyle \\frac{4}{5}\\cdot \\frac{9}{8}= \\frac{4\\cdot 9}{5\\cdot 8} = \\frac{36}{40}=\\frac{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 9}{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 10}=\\frac{9}{10}[\/latex]<\/p>\r\nA second way is to factor and simplify between the fractions' numerators and denominators\u00a0<i>before<\/i> performing the multiplication.\r\n<p style=\"text-align: center;\">[latex]\r\n\\displaystyle \\frac{4}{5}\\cdot\\frac{9}{8}\r\n=\\frac{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 1}{5}\\cdot\\frac{9}{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot2}\r\n=\\frac{1\\cdot9}{5\\cdot2}=\\frac{9}{10}[\/latex]<\/p>\r\nAlthough both methods result in the same product, it usually makes more sense to simplify fractions <em>before<\/em> multiplying.\r\n\r\nThe same two approaches can be applied to rational expressions. Our first two\u00a0examples apply\u00a0both techniques to one expression. After that we will\u00a0let you decide which works best for you. You can factor out <i>any<\/i> common factors, but finding <em>the greatest one<\/em> will take fewer steps.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFor the following expression, multiply, and state the product in simplest form.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{5a^2}{14}\\cdot \\frac{7}{10a^3}[\/latex]<\/p>\r\n[reveal-answer q=\"518862\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"518862\"]\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{5a^2}{14}\\cdot \\frac{7}{10a^3} &amp;= \\frac{5a^2\\cdot 7}{14\\cdot 10a^3} &amp;&amp; {\\color{blue}\\textsf{multiply the numerators and denominators accordingly}}\\\\[5pt]\r\n&amp;= \\frac{35a^2}{140a^3} &amp;&amp; {\\color{blue}\\textsf{simplify the numerator and denominator}}\\\\[5pt]\r\n&amp;= \\frac{35 \\cdot a^2}{35 \\cdot 4 \\cdot a^2\\cdot a} &amp;&amp; {\\color{blue}\\textsf{factor the numerator and denominator; look for the GCF}}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{35}}}\\cdot {\\color{red}\\cancel{\\color{black}{a^2}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{35}}} \\cdot 4\\cdot {\\color{red}\\cancel{\\color{black}{a^2}}}\\cdot a} &amp;&amp; {\\color{blue}\\textsf{divide out the GCF between the numerator and denominator}}\\\\[5pt]\r\n&amp;= \\frac{1}{4a} &amp;&amp; {\\color{blue}\\textsf{simplify}}\\\\[5pt]\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOkay, that worked. This time, let us factor first, then simplify.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFor the following expression, multiply, and state the product in simplest form.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{5a^2}{14}\\cdot \\frac{7}{10a^3}[\/latex]<\/p>\r\n[reveal-answer q=\"724339\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"724339\"]\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{5a^2}{14}\\cdot \\frac{7}{10a^3} &amp;= \\frac{5\\cdot a^2}{7\\cdot 2}\\cdot \\frac{7}{5\\cdot 2\\cdot a^2 \\cdot a} &amp;&amp; {\\color{blue}\\textsf{factor the numerators and denominators; look for the GCFs}}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot {\\color{red}\\cancel{\\color{black}{a^2}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{7}}}\\cdot 2}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{7}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot 2\\cdot {\\color{red}\\cancel{\\color{black}{a^2}}} \\cdot a} &amp;&amp; {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]\r\n&amp;= \\frac{1}{4a} &amp;&amp; {\\color{blue}\\textsf{simplify}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nBoth methods produced the same answer whereas the second method appeared faster and easier.\r\n\r\nRemember that when working with rational expressions, we should get into the habit of identifying the restricted values of the variable, any values that would result in division by\u00a0[latex]0[\/latex]. This will be very important later when we solve rational equations. In the example above, [latex] \\displaystyle \\frac{5{{a}^{2}}}{14}\\cdot \\frac{7}{10{{a}^{3}}}[\/latex], the only restricted value of [latex]a[\/latex] is [latex]0[\/latex]. When [latex]a=0[\/latex], the denominator of the\u00a0rational expression [latex]\\dfrac{7}{10a^3}[\/latex]\u00a0equals\u00a0[latex]0[\/latex], which will make the rational expression undefined.\r\n\r\nRational expressions might contain multi-term polynomials. To multiply such rational expressions, the best approach is to first factor the polynomials and then look for common factors to divide out. Multiplying the numerators and denominators accordingly before factoring will often create complicated polynomials and then we will have to factor these polynomials anyway! For this reason, it is easier to multiply by factoring and then simplifying. Just take it step by step like in the examples below.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFor the following expression, state the restricted values of the variable, multiply, and\u00a0state the product in simplest form.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{a^2-a-2}{5a}\\cdot \\frac{10a}{a+1}[\/latex]<\/p>\r\n[reveal-answer q=\"794041\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"794041\"]\r\n\r\nSince [latex]a=0[\/latex] makes the denominator of the first rational expression equal to [latex]0[\/latex] and [latex]a=-1[\/latex] makes the denominator of the second rational expression equal to [latex]0[\/latex], the restricted values of [latex]a[\/latex] are [latex]0[\/latex] and [latex]-1[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{a^2-a-2}{5a}\\cdot \\frac{10a}{a+1} &amp;= \\frac{(a-2)\\cdot (a+1)}{5\\cdot a}\\cdot \\frac{5\\cdot 2 \\cdot a}{(a+1)} &amp;&amp; {\\color{blue}\\textsf{factor the numerators and denominators}}\\\\[5pt]\r\n&amp;= \\frac{(a-2)\\cdot {\\color{red}\\cancel{\\color{black}{(a+1)}}}}{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot {\\color{red}\\cancel{\\color{black}{a}}}\\cdot 1}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot 2 \\cdot {\\color{red}\\cancel{\\color{black}{a}}}}{{\\color{red}\\cancel{\\color{black}{(a+1)}}}\\cdot 1} &amp;&amp; {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]\r\n&amp;= \\frac{2(a-2)}{1} &amp;&amp; {\\color{blue}\\textsf{multiply}}\\\\[5pt]\r\n&amp;= 2(a-2) &amp;&amp; {\\color{blue}\\textsf{simplify}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The final answer may also be written in expanded (distributed) form\u00a0[latex]2a-4[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that\u00a0the factors containing more than one term were put in parentheses. Make sure to remember to always do that.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFor the following expression, state the restricted values of the variable, multiply, and\u00a0state the product in simplest form.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{t^2+4t+4}{2t^2-t-10}\\cdot\\frac{t+5}{t^2+2t}[\/latex]<\/p>\r\n[reveal-answer q=\"980309\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"980309\"]\r\n\r\nSince [latex]2t^2-t-10=(2t-5)(t+2)[\/latex], using the Zero-Product Property, we acknowledge that [latex]t=\\dfrac{5}{2}[\/latex] and [latex]t=-2[\/latex] make the denominator of the first rational expression equal to [latex]0[\/latex]. Also, [latex]t^2+2t=t(t+2)[\/latex], so [latex]t=0[\/latex] and [latex]t=-2[\/latex] make\u00a0the denominator of the second rational expression equal to [latex]0[\/latex]. Hence, the restricted values of [latex]t[\/latex] are [latex]\\dfrac{5}{2}[\/latex], [latex]0[\/latex], and [latex]-2[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{t^2+4t+4}{2t^2-t-10}\\cdot\\frac{t+5}{t^2+2t} &amp;= \\frac{(t+2)\\cdot (t+2)}{(2t-5)\\cdot (t+2)}\\cdot\\frac{(t+5)}{t\\cdot (t+2)} &amp;&amp; {\\color{blue}\\textsf{factor the numerators and denominators}}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{(t+2)}}}\\cdot {\\color{red}\\cancel{\\color{black}{(t+2)}}}\\cdot1}{(2t-5)\\cdot {\\color{red}\\cancel{\\color{black}{(t+2)}}}}\\cdot\\frac{(t+5)}{t\\cdot {\\color{red}\\cancel{\\color{black}{(t+2)}}}} &amp;&amp; {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]\r\n&amp;= \\frac{t+5}{t(2t-5)} &amp;&amp; {\\color{blue}\\textsf{multiply and simplify}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The final answer may also be written in expanded (distributed) form\u00a0[latex]\\dfrac{t+5}{2t^2-5t}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" style=\"text-align: center;\">\r\n<table style=\"font-size: 110%; width: 280px; background-color: transparent; border-style: none;\" cellspacing=\"0px\" cellpadding=\"0px\">\r\n<tbody style=\"background-color: transparent;\">\r\n<tr style=\"height: 20px; background-color: transparent;\">\r\n<td style=\"width: 80px; background-color: transparent;\"><img class=\"wp-image-2132\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" \/><\/td>\r\n<td style=\"width: 20px; background-color: transparent; vertical-align: middle; font-size: 130%;\"><strong>CAUTION!<\/strong><\/td>\r\n<td style=\"width: 80px; background-color: transparent;\"><img class=\"wp-image-2132\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left;\"><strong><span style=\"color: #ff0000;\">DON'T EVEN THINK ABOUT\r\n<\/span><\/strong><span style=\"color: #000000;\">simplifying the [latex]t[\/latex]s or [latex]5[\/latex]s between the numerator and denominator in the final answer in the example above!<\/span><\/p>\r\n\r\n<\/div>\r\nIn the following video, we present another example of multiplying rational expressions.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=Hj6gF1SNttk&amp;feature=youtu.be\r\n\r\nThe following example involves a difference of cubes.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFor the following expression, state the restricted values of the variable, multiply, and simplify.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{8x^3-1}{x^2}\\cdot\\frac{3x}{4x^2-1}[\/latex]<\/p>\r\n[reveal-answer q=\"944439\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"944439\"]\r\n\r\nSince [latex]x^2=0[\/latex] only if [latex]x=0[\/latex], and [latex]4x^2-1=0[\/latex], equivalent to [latex](2x)^2-1^2=0[\/latex], equivalent to [latex](2x-1)(2x+1)=0[\/latex], results in [latex]x=\\dfrac{1}{2}[\/latex] or [latex]x=-\\dfrac{1}{2}[\/latex]\u00a0by the Zero-Factor Property, the restricted values of [latex]x[\/latex] are [latex]0[\/latex],\u00a0[latex]\\dfrac{1}{2}[\/latex], and [latex]-\\dfrac{1}{2}[\/latex].\r\n\r\nNow, referring to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/3-7-factoring-the-sum-or-difference-of-cubes-and-a-general-approach-to-factoring\/\" target=\"_blank\" rel=\"noopener\">Sec. 3.7<\/a> how to factor a difference of cubes, we have\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{8x^3-1}{x^2}\\cdot\\frac{3x}{4x^2-1}&amp;=\\frac{(2x)^3-1^3}{x\\cdot {\\color{red}\\cancel{\\color{black}{x}}}}\\cdot\\frac{3\\cdot {\\color{red}\\cancel{\\color{black}{x}}}}{(2x)^2-1^2}\\\\[5pt]\r\n&amp;=\\frac{{\\color{red}\\cancel{\\color{black}{(2x-1)}}}\\cdot\\left((2x)^2+2x\\cdot 1+1^2\\right)}{x}\\cdot\\frac{3}{{\\color{red}\\cancel{\\color{black}{(2x-1)}}}\\cdot(2x+1)}\\\\[5pt]\r\n&amp;=\\frac{4x^2+2x+1}{x}\\cdot\\frac{3}{2x+1}\\\\[5pt]\r\n&amp;=\\frac{3\\left(4x^2+2x+1\\right)}{x(2x+1)}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Divide Rational Expressions<\/h2>\r\nYou have seen that we multiply rational expressions as we multiply fractions. It should come as no surprise that we also divide rational expressions the same way we divide fractions. Recall that to divide means to multiply by the reciprocal. To divide rational expressions, keep the first rational expression, change the division sign to multiplication, and then take the reciprocal of the second rational expression.\r\n\r\nLet us begin by recalling division of fractions.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{2}{3}\\div\\frac{5}{9}=\\frac{2}{3}\\cdot\\frac{9}{5}=\\frac{2}{{\\color{red}\\cancel{\\color{black}{3}}}\\cdot 1}\\cdot\\frac{{\\color{red}\\cancel{\\color{black}{3}}}\\cdot 3}{5}=\\frac{6}{5}[\/latex]<\/p>\r\nUse the same process to divide rational expressions. Once division is presented as multiplication by the reciprocal, use what you know about multiplication to simplify.\r\n\r\nWe still need to think about the restricted values of the variable. There is a new consideration this time \u2013 because we divide by a rational expression, we also need to find the values that would make the\u00a0<i>numerator <\/i>of that expression equal zero. Have a look.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFor the following expression, state the restricted values of the variable, divide, and simplify.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{5x^2}{9}\\div\\frac{15x^3}{27}[\/latex]<\/p>\r\n[reveal-answer q=\"688236\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"688236\"]\r\n\r\nFor the restricted values of [latex]x[\/latex], first notice the two denominators, [latex]9[\/latex] and\u00a0[latex]27[\/latex]. These never equal\u00a0zero. Since we divide by\u00a0[latex]\\dfrac{15x^3}{27}[\/latex], the numerator of that rational expression cannot equal zero either. Hence, [latex]x\\ne 0[\/latex] and the\u00a0only restricted value of [latex]x[\/latex] is [latex]0[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{5x^2}{9}\\div\\frac{15x^3}{27} &amp;= \\frac{5x^2}{9}\\cdot\\frac{27}{15x^3} &amp;&amp; {\\color{blue}\\textsf{rewrite division as multiplication by the reciprocal}}\\\\[5pt]\r\n&amp;= \\frac{5\\cdot x^2}{9}\\cdot\\frac{9\\cdot 3}{5\\cdot 3 \\cdot x^2\\cdot x} &amp;&amp; {\\color{blue}\\textsf{factor the numerators and denominators; look for the GCFs}}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot {\\color{red}\\cancel{\\color{black}{x^2}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{9}}}\\cdot 1}\\cdot\\frac{{\\color{red}\\cancel{\\color{black}{9}}}\\cdot {\\color{red}\\cancel{\\color{black}{3}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot {\\color{red}\\cancel{\\color{black}{3}}} \\cdot {\\color{red}\\cancel{\\color{black}{x^2}}}\\cdot x} &amp;&amp; {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]\r\n&amp;= \\frac{1}{x} &amp;&amp; {\\color{blue}\\textsf{multiply and simplify}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFor the following expression, state the restricted values of the variable, divide, and simplify.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{3x^2}{x+2}\\div\\frac{6x^4}{x^2+5x+6}[\/latex]<\/p>\r\n[reveal-answer q=\"53255\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"53255\"]\r\n\r\nFor the restricted values of [latex]x[\/latex], first notice that [latex]x=-2[\/latex] makes the denominator of the first rational expression, [latex]x+2[\/latex], equal to zero. Since [latex]x^2+5x+6=(x+3)(x+2)[\/latex], using the Zero-Product Property, we acknowledge that [latex]x=-3[\/latex] and [latex]x=-2[\/latex] make the denominator of the second rational expression equal to zero. Since we divide by\u00a0[latex]\\dfrac{6x^4}{x^2+5x+6}[\/latex], the numerator of that rational expression, [latex]6x^4[\/latex], cannot equal zero either. Hence, [latex]x\\ne 0[\/latex].\u00a0Summarizing, the restricted values of [latex]x[\/latex] are [latex]-3[\/latex], [latex]-2[\/latex], and [latex]0[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{3x^2}{x+2}\\div\\frac{6x^4}{x^2+5x+6} &amp;= \\frac{3x^2}{x+2}\\cdot\\frac{x^2+5x+6}{6x^4} &amp;&amp; {\\color{blue}\\textsf{rewrite division as multiplication by the reciprocal}}\\\\[5pt]\r\n&amp;= \\frac{3\\cdot x^2}{(x+2)}\\cdot\\frac{(x+3)\\cdot (x+2)}{3\\cdot 2\\cdot x^2\\cdot x^2} &amp;&amp; {\\color{blue}\\textsf{factor the numerators and denominators; look for the GCFs}}\\\\[5pt]\r\n&amp;= \\frac{{\\color{red}\\cancel{\\color{black}{3}}}\\cdot {\\color{red}\\cancel{\\color{black}{x^2}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{(x+2)}}}\\cdot 1}\\cdot\\frac{(x+3)\\cdot {\\color{red}\\cancel{\\color{black}{(x+2)}}}}{{\\color{red}\\cancel{\\color{black}{3}}}\\cdot 2\\cdot {\\color{red}\\cancel{\\color{black}{x^2}}}\\cdot x^2} &amp;&amp; {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]\r\n&amp;= \\frac{x+3}{2x^2} &amp;&amp; {\\color{blue}\\textsf{multiply and simplify}}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we present another example of dividing rational expressions.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=B1tigfgs268&amp;feature=youtu.be\r\n\r\nWe conclude with an example involving both multiplication and division of rational expressions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFor the following expression, state the restricted values of the variable, perform the operations, and simplify.\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{3x^2}{2x+4}\\div\\frac{5x}{x^2-4}\\cdot\\frac{x+1}{x-2}[\/latex]<\/p>\r\n[reveal-answer q=\"373866\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"373866\"]\r\n\r\nFor the restricted values of [latex]x[\/latex], we acknowledge that in addition to restricting the zeros of all denominators, we must prevent division by zero when dividing by [latex]\\dfrac{5x}{x^2-4}[\/latex] by restricting the zeros of the numerator of that expression. We solve four equations:\r\n<ul>\r\n \t<li>[latex]2x+4=0[\/latex], resulting in [latex]x=-2[\/latex],<\/li>\r\n \t<li>[latex]x^2-4=0[\/latex], equivalent to [latex](x-2)(x+2)=0[\/latex], resulting in [latex]x=2[\/latex] or [latex]x=-2[\/latex] by the Zero-Factor Property,<\/li>\r\n \t<li>[latex]x-2=0[\/latex], resulting in [latex]x=2[\/latex], and<\/li>\r\n \t<li>[latex]5x=0[\/latex], resulting in [latex]x=0[\/latex].<\/li>\r\n<\/ul>\r\nSummarizing, the restricted values of [latex]x[\/latex] are [latex]-2[\/latex], [latex]2[\/latex], and [latex]0[\/latex].\r\n\r\nRemembering that to divide means to multiply by the reciprocal, we have\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{3x^2}{2x+4}\\div\\frac{5x}{x^2-4}\\cdot\\frac{x+1}{x-2}&amp;=\\frac{3x^2}{2x+4}\\cdot\\frac{x^2-4}{5x}\\cdot\\frac{x+1}{x-2}\\\\[5pt]\r\n&amp;=\\frac{3\\cdot {\\color{red}\\cancel{\\color{black}{x}}} \\cdot x}{2\\cdot {\\color{red}\\cancel{\\color{black}{(x+2)}}}}\\cdot\\frac{{\\color{red}\\cancel{\\color{black}{(x-2)}}}\\cdot{\\color{red}\\cancel{\\color{black}{(x+2)}}}\\cdot 1}{5\\cdot {\\color{red}\\cancel{\\color{black}{x}}}}\\cdot\\frac{(x+1)}{{\\color{red}\\cancel{\\color{black}{(x-2)}}}\\cdot 1}\\\\[5pt]\r\n&amp;=\\frac{3x(x+1)}{10}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nRational expressions are multiplied and divided the same way as fractions. To multiply, first factor the numerators and denominators of the rational expressions acknowledging greatest common factors. Next, divide out common factors between the numerators and denominators. Then, multiply and simplify. To divide, first rewrite the division as multiplication by the reciprocal. After that, the steps are the same as those for multiplication.\r\n\r\nWhen expressing a product or quotient, it is important to remember to identify restricted values of the variable. It\u00a0will be very important when solving rational equations.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Multiply and simplify rational expressions.<\/li>\n<li>Divide and simplify rational expressions.<\/li>\n<\/ul>\n<\/div>\n<p>Just as we can multiply and divide fractions, we can multiply and divide <strong>rational expressions<\/strong>. In fact, we use the same processes for multiplying and dividing rational expressions as we use for multiplying and dividing fractions. The process is the same even though the expressions look different!<\/p>\n<h3>Multiply Rational Expressions<\/h3>\n<p>Remember that there are two ways to multiply fractions.<\/p>\n<p>One way is to multiply the numerators and the denominators separately, and then simplify the product, as shown here. Notice we factor out the <strong>greatest common factor<\/strong>\u00a0(between the numerator and denominator) for efficiency, though we could have worked with the prime factorization as well.<\/p>\n<p style=\"text-align: center;\">[latex]\\require{color} \\displaystyle \\frac{4}{5}\\cdot \\frac{9}{8}= \\frac{4\\cdot 9}{5\\cdot 8} = \\frac{36}{40}=\\frac{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 9}{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 10}=\\frac{9}{10}[\/latex]<\/p>\n<p>A second way is to factor and simplify between the fractions&#8217; numerators and denominators\u00a0<i>before<\/i> performing the multiplication.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{4}{5}\\cdot\\frac{9}{8}  =\\frac{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot 1}{5}\\cdot\\frac{9}{{\\color{red}\\cancel{\\color{black}{4}}}\\cdot2}  =\\frac{1\\cdot9}{5\\cdot2}=\\frac{9}{10}[\/latex]<\/p>\n<p>Although both methods result in the same product, it usually makes more sense to simplify fractions <em>before<\/em> multiplying.<\/p>\n<p>The same two approaches can be applied to rational expressions. Our first two\u00a0examples apply\u00a0both techniques to one expression. After that we will\u00a0let you decide which works best for you. You can factor out <i>any<\/i> common factors, but finding <em>the greatest one<\/em> will take fewer steps.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>For the following expression, multiply, and state the product in simplest form.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{5a^2}{14}\\cdot \\frac{7}{10a^3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q518862\">Show Solution<\/span><\/p>\n<div id=\"q518862\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{5a^2}{14}\\cdot \\frac{7}{10a^3} &= \\frac{5a^2\\cdot 7}{14\\cdot 10a^3} && {\\color{blue}\\textsf{multiply the numerators and denominators accordingly}}\\\\[5pt]  &= \\frac{35a^2}{140a^3} && {\\color{blue}\\textsf{simplify the numerator and denominator}}\\\\[5pt]  &= \\frac{35 \\cdot a^2}{35 \\cdot 4 \\cdot a^2\\cdot a} && {\\color{blue}\\textsf{factor the numerator and denominator; look for the GCF}}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{35}}}\\cdot {\\color{red}\\cancel{\\color{black}{a^2}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{35}}} \\cdot 4\\cdot {\\color{red}\\cancel{\\color{black}{a^2}}}\\cdot a} && {\\color{blue}\\textsf{divide out the GCF between the numerator and denominator}}\\\\[5pt]  &= \\frac{1}{4a} && {\\color{blue}\\textsf{simplify}}\\\\[5pt]  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Okay, that worked. This time, let us factor first, then simplify.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>For the following expression, multiply, and state the product in simplest form.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{5a^2}{14}\\cdot \\frac{7}{10a^3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q724339\">Show Solution<\/span><\/p>\n<div id=\"q724339\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{5a^2}{14}\\cdot \\frac{7}{10a^3} &= \\frac{5\\cdot a^2}{7\\cdot 2}\\cdot \\frac{7}{5\\cdot 2\\cdot a^2 \\cdot a} && {\\color{blue}\\textsf{factor the numerators and denominators; look for the GCFs}}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot {\\color{red}\\cancel{\\color{black}{a^2}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{7}}}\\cdot 2}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{7}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot 2\\cdot {\\color{red}\\cancel{\\color{black}{a^2}}} \\cdot a} && {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]  &= \\frac{1}{4a} && {\\color{blue}\\textsf{simplify}}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Both methods produced the same answer whereas the second method appeared faster and easier.<\/p>\n<p>Remember that when working with rational expressions, we should get into the habit of identifying the restricted values of the variable, any values that would result in division by\u00a0[latex]0[\/latex]. This will be very important later when we solve rational equations. In the example above, [latex]\\displaystyle \\frac{5{{a}^{2}}}{14}\\cdot \\frac{7}{10{{a}^{3}}}[\/latex], the only restricted value of [latex]a[\/latex] is [latex]0[\/latex]. When [latex]a=0[\/latex], the denominator of the\u00a0rational expression [latex]\\dfrac{7}{10a^3}[\/latex]\u00a0equals\u00a0[latex]0[\/latex], which will make the rational expression undefined.<\/p>\n<p>Rational expressions might contain multi-term polynomials. To multiply such rational expressions, the best approach is to first factor the polynomials and then look for common factors to divide out. Multiplying the numerators and denominators accordingly before factoring will often create complicated polynomials and then we will have to factor these polynomials anyway! For this reason, it is easier to multiply by factoring and then simplifying. Just take it step by step like in the examples below.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>For the following expression, state the restricted values of the variable, multiply, and\u00a0state the product in simplest form.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{a^2-a-2}{5a}\\cdot \\frac{10a}{a+1}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q794041\">Show Solution<\/span><\/p>\n<div id=\"q794041\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since [latex]a=0[\/latex] makes the denominator of the first rational expression equal to [latex]0[\/latex] and [latex]a=-1[\/latex] makes the denominator of the second rational expression equal to [latex]0[\/latex], the restricted values of [latex]a[\/latex] are [latex]0[\/latex] and [latex]-1[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{a^2-a-2}{5a}\\cdot \\frac{10a}{a+1} &= \\frac{(a-2)\\cdot (a+1)}{5\\cdot a}\\cdot \\frac{5\\cdot 2 \\cdot a}{(a+1)} && {\\color{blue}\\textsf{factor the numerators and denominators}}\\\\[5pt]  &= \\frac{(a-2)\\cdot {\\color{red}\\cancel{\\color{black}{(a+1)}}}}{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot {\\color{red}\\cancel{\\color{black}{a}}}\\cdot 1}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot 2 \\cdot {\\color{red}\\cancel{\\color{black}{a}}}}{{\\color{red}\\cancel{\\color{black}{(a+1)}}}\\cdot 1} && {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]  &= \\frac{2(a-2)}{1} && {\\color{blue}\\textsf{multiply}}\\\\[5pt]  &= 2(a-2) && {\\color{blue}\\textsf{simplify}}  \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">The final answer may also be written in expanded (distributed) form\u00a0[latex]2a-4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note that\u00a0the factors containing more than one term were put in parentheses. Make sure to remember to always do that.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>For the following expression, state the restricted values of the variable, multiply, and\u00a0state the product in simplest form.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{t^2+4t+4}{2t^2-t-10}\\cdot\\frac{t+5}{t^2+2t}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q980309\">Show Solution<\/span><\/p>\n<div id=\"q980309\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since [latex]2t^2-t-10=(2t-5)(t+2)[\/latex], using the Zero-Product Property, we acknowledge that [latex]t=\\dfrac{5}{2}[\/latex] and [latex]t=-2[\/latex] make the denominator of the first rational expression equal to [latex]0[\/latex]. Also, [latex]t^2+2t=t(t+2)[\/latex], so [latex]t=0[\/latex] and [latex]t=-2[\/latex] make\u00a0the denominator of the second rational expression equal to [latex]0[\/latex]. Hence, the restricted values of [latex]t[\/latex] are [latex]\\dfrac{5}{2}[\/latex], [latex]0[\/latex], and [latex]-2[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{t^2+4t+4}{2t^2-t-10}\\cdot\\frac{t+5}{t^2+2t} &= \\frac{(t+2)\\cdot (t+2)}{(2t-5)\\cdot (t+2)}\\cdot\\frac{(t+5)}{t\\cdot (t+2)} && {\\color{blue}\\textsf{factor the numerators and denominators}}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{(t+2)}}}\\cdot {\\color{red}\\cancel{\\color{black}{(t+2)}}}\\cdot1}{(2t-5)\\cdot {\\color{red}\\cancel{\\color{black}{(t+2)}}}}\\cdot\\frac{(t+5)}{t\\cdot {\\color{red}\\cancel{\\color{black}{(t+2)}}}} && {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]  &= \\frac{t+5}{t(2t-5)} && {\\color{blue}\\textsf{multiply and simplify}}  \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">The final answer may also be written in expanded (distributed) form\u00a0[latex]\\dfrac{t+5}{2t^2-5t}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\" style=\"text-align: center;\">\n<table style=\"font-size: 110%; width: 280px; background-color: transparent; border-style: none; border-spacing: 0pxpx;\" cellpadding=\"0px\">\n<tbody style=\"background-color: transparent;\">\n<tr style=\"height: 20px; background-color: transparent;\">\n<td style=\"width: 80px; background-color: transparent;\"><img decoding=\"async\" class=\"wp-image-2132\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" \/><\/td>\n<td style=\"width: 20px; background-color: transparent; vertical-align: middle; font-size: 130%;\"><strong>CAUTION!<\/strong><\/td>\n<td style=\"width: 80px; background-color: transparent;\"><img decoding=\"async\" class=\"wp-image-2132\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01182614\/traffic-sign-160659-300x265.png\" alt=\"Caution\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left;\"><strong><span style=\"color: #ff0000;\">DON&#8217;T EVEN THINK ABOUT<br \/>\n<\/span><\/strong><span style=\"color: #000000;\">simplifying the [latex]t[\/latex]s or [latex]5[\/latex]s between the numerator and denominator in the final answer in the example above!<\/span><\/p>\n<\/div>\n<p>In the following video, we present another example of multiplying rational expressions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Multiply Rational Expressions and Give the Domain\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Hj6gF1SNttk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The following example involves a difference of cubes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>For the following expression, state the restricted values of the variable, multiply, and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{8x^3-1}{x^2}\\cdot\\frac{3x}{4x^2-1}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q944439\">Show Solution<\/span><\/p>\n<div id=\"q944439\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since [latex]x^2=0[\/latex] only if [latex]x=0[\/latex], and [latex]4x^2-1=0[\/latex], equivalent to [latex](2x)^2-1^2=0[\/latex], equivalent to [latex](2x-1)(2x+1)=0[\/latex], results in [latex]x=\\dfrac{1}{2}[\/latex] or [latex]x=-\\dfrac{1}{2}[\/latex]\u00a0by the Zero-Factor Property, the restricted values of [latex]x[\/latex] are [latex]0[\/latex],\u00a0[latex]\\dfrac{1}{2}[\/latex], and [latex]-\\dfrac{1}{2}[\/latex].<\/p>\n<p>Now, referring to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/3-7-factoring-the-sum-or-difference-of-cubes-and-a-general-approach-to-factoring\/\" target=\"_blank\" rel=\"noopener\">Sec. 3.7<\/a> how to factor a difference of cubes, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{8x^3-1}{x^2}\\cdot\\frac{3x}{4x^2-1}&=\\frac{(2x)^3-1^3}{x\\cdot {\\color{red}\\cancel{\\color{black}{x}}}}\\cdot\\frac{3\\cdot {\\color{red}\\cancel{\\color{black}{x}}}}{(2x)^2-1^2}\\\\[5pt]  &=\\frac{{\\color{red}\\cancel{\\color{black}{(2x-1)}}}\\cdot\\left((2x)^2+2x\\cdot 1+1^2\\right)}{x}\\cdot\\frac{3}{{\\color{red}\\cancel{\\color{black}{(2x-1)}}}\\cdot(2x+1)}\\\\[5pt]  &=\\frac{4x^2+2x+1}{x}\\cdot\\frac{3}{2x+1}\\\\[5pt]  &=\\frac{3\\left(4x^2+2x+1\\right)}{x(2x+1)}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Divide Rational Expressions<\/h2>\n<p>You have seen that we multiply rational expressions as we multiply fractions. It should come as no surprise that we also divide rational expressions the same way we divide fractions. Recall that to divide means to multiply by the reciprocal. To divide rational expressions, keep the first rational expression, change the division sign to multiplication, and then take the reciprocal of the second rational expression.<\/p>\n<p>Let us begin by recalling division of fractions.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{2}{3}\\div\\frac{5}{9}=\\frac{2}{3}\\cdot\\frac{9}{5}=\\frac{2}{{\\color{red}\\cancel{\\color{black}{3}}}\\cdot 1}\\cdot\\frac{{\\color{red}\\cancel{\\color{black}{3}}}\\cdot 3}{5}=\\frac{6}{5}[\/latex]<\/p>\n<p>Use the same process to divide rational expressions. Once division is presented as multiplication by the reciprocal, use what you know about multiplication to simplify.<\/p>\n<p>We still need to think about the restricted values of the variable. There is a new consideration this time \u2013 because we divide by a rational expression, we also need to find the values that would make the\u00a0<i>numerator <\/i>of that expression equal zero. Have a look.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>For the following expression, state the restricted values of the variable, divide, and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{5x^2}{9}\\div\\frac{15x^3}{27}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q688236\">Show Solution<\/span><\/p>\n<div id=\"q688236\" class=\"hidden-answer\" style=\"display: none\">\n<p>For the restricted values of [latex]x[\/latex], first notice the two denominators, [latex]9[\/latex] and\u00a0[latex]27[\/latex]. These never equal\u00a0zero. Since we divide by\u00a0[latex]\\dfrac{15x^3}{27}[\/latex], the numerator of that rational expression cannot equal zero either. Hence, [latex]x\\ne 0[\/latex] and the\u00a0only restricted value of [latex]x[\/latex] is [latex]0[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{5x^2}{9}\\div\\frac{15x^3}{27} &= \\frac{5x^2}{9}\\cdot\\frac{27}{15x^3} && {\\color{blue}\\textsf{rewrite division as multiplication by the reciprocal}}\\\\[5pt]  &= \\frac{5\\cdot x^2}{9}\\cdot\\frac{9\\cdot 3}{5\\cdot 3 \\cdot x^2\\cdot x} && {\\color{blue}\\textsf{factor the numerators and denominators; look for the GCFs}}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot {\\color{red}\\cancel{\\color{black}{x^2}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{9}}}\\cdot 1}\\cdot\\frac{{\\color{red}\\cancel{\\color{black}{9}}}\\cdot {\\color{red}\\cancel{\\color{black}{3}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{5}}}\\cdot {\\color{red}\\cancel{\\color{black}{3}}} \\cdot {\\color{red}\\cancel{\\color{black}{x^2}}}\\cdot x} && {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]  &= \\frac{1}{x} && {\\color{blue}\\textsf{multiply and simplify}}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>For the following expression, state the restricted values of the variable, divide, and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{3x^2}{x+2}\\div\\frac{6x^4}{x^2+5x+6}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q53255\">Show Solution<\/span><\/p>\n<div id=\"q53255\" class=\"hidden-answer\" style=\"display: none\">\n<p>For the restricted values of [latex]x[\/latex], first notice that [latex]x=-2[\/latex] makes the denominator of the first rational expression, [latex]x+2[\/latex], equal to zero. Since [latex]x^2+5x+6=(x+3)(x+2)[\/latex], using the Zero-Product Property, we acknowledge that [latex]x=-3[\/latex] and [latex]x=-2[\/latex] make the denominator of the second rational expression equal to zero. Since we divide by\u00a0[latex]\\dfrac{6x^4}{x^2+5x+6}[\/latex], the numerator of that rational expression, [latex]6x^4[\/latex], cannot equal zero either. Hence, [latex]x\\ne 0[\/latex].\u00a0Summarizing, the restricted values of [latex]x[\/latex] are [latex]-3[\/latex], [latex]-2[\/latex], and [latex]0[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{3x^2}{x+2}\\div\\frac{6x^4}{x^2+5x+6} &= \\frac{3x^2}{x+2}\\cdot\\frac{x^2+5x+6}{6x^4} && {\\color{blue}\\textsf{rewrite division as multiplication by the reciprocal}}\\\\[5pt]  &= \\frac{3\\cdot x^2}{(x+2)}\\cdot\\frac{(x+3)\\cdot (x+2)}{3\\cdot 2\\cdot x^2\\cdot x^2} && {\\color{blue}\\textsf{factor the numerators and denominators; look for the GCFs}}\\\\[5pt]  &= \\frac{{\\color{red}\\cancel{\\color{black}{3}}}\\cdot {\\color{red}\\cancel{\\color{black}{x^2}}}\\cdot 1}{{\\color{red}\\cancel{\\color{black}{(x+2)}}}\\cdot 1}\\cdot\\frac{(x+3)\\cdot {\\color{red}\\cancel{\\color{black}{(x+2)}}}}{{\\color{red}\\cancel{\\color{black}{3}}}\\cdot 2\\cdot {\\color{red}\\cancel{\\color{black}{x^2}}}\\cdot x^2} && {\\color{blue}\\textsf{divide out common factors}}\\\\[5pt]  &= \\frac{x+3}{2x^2} && {\\color{blue}\\textsf{multiply and simplify}}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, we present another example of dividing rational expressions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Divide Rational Expressions and Give the Domain\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/B1tigfgs268?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We conclude with an example involving both multiplication and division of rational expressions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>For the following expression, state the restricted values of the variable, perform the operations, and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\frac{3x^2}{2x+4}\\div\\frac{5x}{x^2-4}\\cdot\\frac{x+1}{x-2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q373866\">Show Solution<\/span><\/p>\n<div id=\"q373866\" class=\"hidden-answer\" style=\"display: none\">\n<p>For the restricted values of [latex]x[\/latex], we acknowledge that in addition to restricting the zeros of all denominators, we must prevent division by zero when dividing by [latex]\\dfrac{5x}{x^2-4}[\/latex] by restricting the zeros of the numerator of that expression. We solve four equations:<\/p>\n<ul>\n<li>[latex]2x+4=0[\/latex], resulting in [latex]x=-2[\/latex],<\/li>\n<li>[latex]x^2-4=0[\/latex], equivalent to [latex](x-2)(x+2)=0[\/latex], resulting in [latex]x=2[\/latex] or [latex]x=-2[\/latex] by the Zero-Factor Property,<\/li>\n<li>[latex]x-2=0[\/latex], resulting in [latex]x=2[\/latex], and<\/li>\n<li>[latex]5x=0[\/latex], resulting in [latex]x=0[\/latex].<\/li>\n<\/ul>\n<p>Summarizing, the restricted values of [latex]x[\/latex] are [latex]-2[\/latex], [latex]2[\/latex], and [latex]0[\/latex].<\/p>\n<p>Remembering that to divide means to multiply by the reciprocal, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{3x^2}{2x+4}\\div\\frac{5x}{x^2-4}\\cdot\\frac{x+1}{x-2}&=\\frac{3x^2}{2x+4}\\cdot\\frac{x^2-4}{5x}\\cdot\\frac{x+1}{x-2}\\\\[5pt]  &=\\frac{3\\cdot {\\color{red}\\cancel{\\color{black}{x}}} \\cdot x}{2\\cdot {\\color{red}\\cancel{\\color{black}{(x+2)}}}}\\cdot\\frac{{\\color{red}\\cancel{\\color{black}{(x-2)}}}\\cdot{\\color{red}\\cancel{\\color{black}{(x+2)}}}\\cdot 1}{5\\cdot {\\color{red}\\cancel{\\color{black}{x}}}}\\cdot\\frac{(x+1)}{{\\color{red}\\cancel{\\color{black}{(x-2)}}}\\cdot 1}\\\\[5pt]  &=\\frac{3x(x+1)}{10}  \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Summary<\/h2>\n<p>Rational expressions are multiplied and divided the same way as fractions. To multiply, first factor the numerators and denominators of the rational expressions acknowledging greatest common factors. Next, divide out common factors between the numerators and denominators. Then, multiply and simplify. To divide, first rewrite the division as multiplication by the reciprocal. After that, the steps are the same as those for multiplication.<\/p>\n<p>When expressing a product or quotient, it is important to remember to identify restricted values of the variable. It\u00a0will be very important when solving rational equations.<\/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-175\">\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>Screenshot: Multiply and Divide. <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>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>Screenshot: Reciprocal Architecture. <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>Multiply Rational Expressions and Give the Domain. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Hj6gF1SNttk\">https:\/\/youtu.be\/Hj6gF1SNttk<\/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>Divide Rational Expressions and Give the Domain. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/B1tigfgs268\">https:\/\/youtu.be\/B1tigfgs268<\/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 15: Rational Expressions, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <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><\/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":395986,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot: Multiply and Divide\",\"author\":\"\",\"organization\":\"Lumen 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