{"id":3773,"date":"2021-05-12T20:24:07","date_gmt":"2021-05-12T20:24:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-section-2-3-the-limit-laws\/"},"modified":"2021-06-28T18:43:40","modified_gmt":"2021-06-28T18:43:40","slug":"review-for-section-2-3-the-limit-laws","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-section-2-3-the-limit-laws\/","title":{"raw":"Skills Review for The Limit Laws","rendered":"Skills Review for The Limit Laws"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor the greatest common factor (monomial) of a polynomial<\/li>\r\n \t<li>Factor a trinomial with a leading coefficient of 1<\/li>\r\n \t<li>Rewrite a trinomial as a four term polynomial and factor by grouping terms<\/li>\r\n \t<li>Factor difference of squares<\/li>\r\n \t<li>Factor sum or difference of cubes<\/li>\r\n \t<li>Simplify a rational expression<\/li>\r\n \t<li>Remove radicals from a multiple term denominator<\/li>\r\n \t<li>Simplify complex rational expressions<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe Limit Laws section will introduce yet another way to calculate a limit, using limit laws. Basically, calculating limits of functions algebraically will be the topic of focus in the section. Several techniques that are used to simplify functions are sometimes needed to successfully calculate a function's limit. Here we will review factoring, simplifying rational expressions, rationalizing radical expressions, and simplifying complex rational expressions.\r\n<h2>Factor Polynomials<\/h2>\r\nRecall that the <strong>greatest common factor<\/strong> (GCF) of two numbers is the largest number that divides evenly into both numbers. For example, [latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest number that divides evenly into both [latex]16[\/latex] and [latex]20[\/latex]. The GCF of polynomials works the same way: [latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest polynomial that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].\r\n\r\nFinding and factoring out a GCF from a polynomial is the first skill involved in factoring polynomials.\r\n<h3>Factor a GCF out of a Polynomial<\/h3>\r\nWhen factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term. To make it less challenging to find this GCF of the polynomial terms, first look for the GCF of the coefficients, and then look for the GCF of the variables.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Greatest Common Factor<\/h3>\r\nThe <strong>greatest common factor<\/strong> (GCF) of a polynomial is the largest polynomial that divides evenly into each term of the polynomial.\r\n\r\n<\/div>\r\nTo factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property \"backwards\" to rewrite the polynomial in a factored form. Recall that the <strong>distributive property of multiplication over addition<\/strong> states that a product of a number and a sum is the same as the sum of the products.\r\n<div class=\"textbox shaded\">\r\n<h4>Distributive Property Forward and Backward<\/h4>\r\n<p style=\"text-align: left;\">Forward:\u00a0<em>We distribute [latex]a[\/latex] over [latex]b+c[\/latex]<\/em>.<\/p>\r\n<p style=\"text-align: center;\">[latex]a\\left(b+c\\right)=ab+ac[\/latex].<\/p>\r\nBackward:\u00a0<em>We factor [latex]a[\/latex] out of [latex]ab+ac[\/latex].<\/em>\r\n<p style=\"text-align: center;\">[latex]ab+ac=a\\left(b+c\\right)[\/latex].<\/p>\r\nWe have seen that we can distribute a factor over a sum or difference. Now we see that we can \"undo\" the distributive property with factoring.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Factoring The Greatest Common Factor<\/h3>\r\nFactor [latex]25b^{3}+10b^{2}[\/latex].\r\n\r\n[reveal-answer q=\"716902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"716902\"]Find the GCF.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}=5b^{2}\\end{array}[\/latex]<\/p>\r\nRewrite each term with the GCF as one factor.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3} = 5b^{2}\\cdot5b\\\\10b^{2}=5b^{2}\\cdot2\\end{array}[\/latex]<\/p>\r\nRewrite the polynomial using the factored terms in place of the original terms.\r\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b\\right)+5b^{2}\\left(2\\right)[\/latex]<\/p>\r\nFactor out the [latex]5b^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b+2\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]5b^{2}\\left(5b+2\\right)[\/latex]\r\n<h4>Analysis<\/h4>\r\nThe factored form of the polynomial [latex]25b^{3}+10b^{2}[\/latex] is [latex]5b^{2}\\left(5b+2\\right)[\/latex]. You can check this by doing the multiplication. [latex]5b^{2}\\left(5b+2\\right)=25b^{3}+10b^{2}[\/latex].\r\n\r\nNote that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.\r\n\r\nFor example:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3}+10b^{2}=5\\left(5b^{3}+2b^{2}\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }5.\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5b^{2}\\left(5b+2\\right) \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }b^{2}.\\end{array}[\/latex]<\/p>\r\nNotice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polynomial expression, factor out the greatest common factor<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Identify the GCF of the coefficients.<\/li>\r\n \t<li>Identify the GCF of the variables.<\/li>\r\n \t<li>Combine to find the GCF of the expression.<\/li>\r\n \t<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\r\n \t<li>Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring the Greatest Common Factor<\/h3>\r\nFactor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].\r\n\r\n[reveal-answer q=\"113189\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"113189\"]\r\n\r\nFirst find the GCF of the expression. The GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions of the form [latex]{x}^{n}[\/latex] will always be the lowest exponent.) The GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. Combine these to find the GCF of the polynomial, [latex]3xy[\/latex].\r\n\r\nNext, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that [latex]3xy\\left(2{x}^{2}{y}^{2}\\right)=6{x}^{3}{y}^{3}, 3xy\\left(15xy\\right)=45{x}^{2}{y}^{2}[\/latex], and [latex]3xy\\left(7\\right)=21xy[\/latex].\r\n\r\nFinally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.\r\n<div style=\"text-align: center;\">[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/div>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\nAfter factoring, we can check our work by multiplying. Use the distributive property to confirm that [latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].\r\n\r\n<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch this video to see more examples of how to factor the GCF from a trinomial.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454717&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=3f1RFTIw2Ng&amp;video_target=tpm-plugin-8plr8d8s-3f1RFTIw2Ng\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/Ex2IdentifyGCFAndFactorATrinomial_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ex 2: Identify GCF and Factor a Trinomial\" here (opens in new window)<\/a>.\r\n<h3>Factor a Trinomial with Leading Coefficient 1<\/h3>\r\nAlthough we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].\r\n\r\nTrinomials of the form [latex]{x}^{2}+bx+c[\/latex] can be factored by finding two numbers with a product of [latex]c[\/latex] and a sum of [latex]b[\/latex]. The trinomial [latex]{x}^{2}+10x+16[\/latex], for example, can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex] because the product of these numbers is [latex]16[\/latex] and their sum is [latex]10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nA trinomial of the form [latex]{x}^{2}+bx+c[\/latex] can be written in factored form as [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex] where [latex]pq=c[\/latex] and [latex]p+q=b[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can every trinomial be factored as a product of binomials?<\/strong>\r\n\r\n<em>No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nFactor [latex]{x}^{2}+2x - 15[\/latex].\r\n\r\n[reveal-answer q=\"88306\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"88306\"]\r\n\r\nWe have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-15[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-15[\/latex]<\/td>\r\n<td>[latex]-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,15[\/latex]<\/td>\r\n<td>[latex]14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-5[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,5[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can check our work by multiplying. Use FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Does the order of the factors matter?<\/strong>\r\n\r\n<em>No. Multiplication is commutative, so the order of the factors does not matter.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7897[\/ohm_question]\r\n\r\n<\/div>\r\n<h3>Factor by Grouping<\/h3>\r\nTrinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Factor by Grouping<\/h3>\r\nTo factor a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]ac[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]ac[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\r\n \t<li>Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\r\n \t<li>Pull out the GCF of [latex]qx+c[\/latex].<\/li>\r\n \t<li>Factor out the GCF of the expression.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Trinomial by Grouping<\/h3>\r\nFactor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.\r\n\r\n[reveal-answer q=\"806328\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"806328\"]\r\n\r\nWe have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-30[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-30[\/latex]<\/td>\r\n<td>[latex]-29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,30[\/latex]<\/td>\r\n<td>[latex]29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2,-15[\/latex]<\/td>\r\n<td>[latex]-13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,15[\/latex]<\/td>\r\n<td>[latex]13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-10[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,10[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]p=-3[\/latex] and [latex]q=10[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill &amp; \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill &amp; \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill &amp; \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].\r\n\r\n<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7908[\/ohm_question]\r\n\r\n<\/div>\r\nIn the next video, we see another example of how to factor a trinomial by grouping.\r\n\r\n[embed]https:\/\/youtu.be\/agDaQ_cZnNc[\/embed]\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/FactorATrinomialInTheFormUsingTheGroupingTechniqe_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique\" here (opens in new window)<\/a>.\r\n<h3>Factor a Difference of Squares<\/h3>\r\nA difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n\r\nWe can use this equation to factor any differences of squares.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Differences of Squares<\/h3>\r\nA difference of squares can be rewritten as two factors containing the same terms but opposite signs.\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a difference of squares, factor it into binomials<\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Difference of Squares<\/h3>\r\nFactor [latex]9{x}^{2}-25[\/latex].\r\n\r\n[reveal-answer q=\"830417\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"830417\"]\r\n\r\nNotice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex]. The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7929[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Is there a formula to factor the sum of squares?<\/strong>\r\n\r\n<em>No. A sum of squares cannot be factored.<\/em>\r\n\r\n<\/div>\r\nWatch this video to see another example of how to factor a difference of squares.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454718&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Li9IBp5HrFA&amp;video_target=tpm-plugin-llp8qg9s-Li9IBp5HrFA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExFactorADifferenceOfSquares_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ex: Factor a Difference of Squares\" here (opens in new window)<\/a>.\r\n<h3>Factor the Sum and Difference of Cubes<\/h3>\r\nNow we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.\r\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n\r\nSimilarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.\r\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n\r\nWe can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: <strong>S<\/strong>ame <strong>O<\/strong>pposite <strong>A<\/strong>lways <strong>P<\/strong>ositive. For example, consider the following example.\r\n<div style=\"text-align: center;\">[latex]{x}^{3}-{2}^{3}=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]<\/div>\r\nThe sign of the first 2 is the <em>same<\/em> as the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. The sign of the [latex]2x[\/latex] term is <em>opposite<\/em> the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. And the sign of the last term, 4, is <em>always positive<\/em>.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Sum and Difference of Cubes<\/h3>\r\nWe can factor the sum of two cubes as\r\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\r\nWe can factor the difference of two cubes as\r\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a sum of cubes or difference of cubes, factor it<\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are cubes, [latex]{a}^{3}+{b}^{3}[\/latex] or [latex]{a}^{3}-{b}^{3}[\/latex].<\/li>\r\n \t<li>For a sum of cubes, write the factored form as [latex]\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]. For a difference of cubes, write the factored form as [latex]\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Sum of Cubes<\/h3>\r\nFactor [latex]{x}^{3}+512[\/latex].\r\n\r\n[reveal-answer q=\"22673\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"22673\"]\r\n\r\nNotice that [latex]{x}^{3}[\/latex] and [latex]512[\/latex] are cubes because [latex]{8}^{3}=512[\/latex]. Rewrite the sum of cubes as [latex]\\left(x+8\\right)\\left({x}^{2}-8x+64\\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nAfter writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor the sum of cubes [latex]216{a}^{3}+{b}^{3}[\/latex].\r\n\r\n[reveal-answer q=\"496204\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"496204\"]\r\n\r\n[latex]\\left(6a+b\\right)\\left(36{a}^{2}-6ab+{b}^{2}\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Difference of Cubes<\/h3>\r\nFactor [latex]8{x}^{3}-125[\/latex].\r\n\r\n[reveal-answer q=\"741836\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"741836\"]\r\n\r\nNotice that [latex]8{x}^{3}[\/latex] and [latex]125[\/latex] are cubes because [latex]8{x}^{3}={\\left(2x\\right)}^{3}[\/latex] and [latex]125={5}^{3}[\/latex]. Write the difference of cubes as [latex]\\left(2x - 5\\right)\\left(4{x}^{2}+10x+25\\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nJust as with the sum of cubes, we will not be able to further factor the trinomial portion.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor the difference of cubes: [latex]1,000{x}^{3}-1[\/latex].\r\n\r\n[reveal-answer q=\"510077\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"510077\"]\r\n\r\n[latex]\\left(10x - 1\\right)\\left(100{x}^{2}+10x+1\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7922[\/ohm_question]\r\n\r\n<\/div>\r\nIn the following two video examples, we show more binomials that can be factored as a sum or difference of cubes.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454719&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=tFSEpOB262M&amp;video_target=tpm-plugin-je39dfag-tFSEpOB262M\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/Ex1FactorASumOrDifferenceOfCubes_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ex 1: Factor a Sum or Difference of Cubes\" here (opens in new window)<\/a>.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454720&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=J_0ctMrl5_0&amp;video_target=tpm-plugin-cnyddfw5-J_0ctMrl5_0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/Ex3FactorASumOrDifferenceOfCubes_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ex 3: Factor a Sum or Difference of Cubes\" here (opens in new window)<\/a>.\r\n<h2>Simplify Rational Expressions<\/h2>\r\nThe quotient of two polynomial expressions is called a <strong>rational expression<\/strong>. We can apply the properties of fractions to rational expressions such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let\u2019s start with the rational expression shown.\r\n<p style=\"text-align: center;\">[latex]\\frac{{x}^{2}+8x+16}{{x}^{2}+11x+28}[\/latex]<\/p>\r\n[latex]\\\\[\/latex]We can factor the numerator and denominator to rewrite the expression as [latex]\\frac{{\\left(x+4\\right)}^{2}}{\\left(x+4\\right)\\left(x+7\\right)}[\/latex]\r\n<div><\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Then we can simplify the expression by canceling the common factor [latex]\\left(x+4\\right)[\/latex] to get [latex]\\frac{x+4}{x+7}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a rational expression, simplify it<\/h3>\r\n<ol>\r\n \t<li>Factor the numerator and denominator.<\/li>\r\n \t<li>Cancel any common factors.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Simplifying Rational Expressions<\/h3>\r\nSimplify [latex]\\frac{{x}^{2}-9}{{x}^{2}+4x+3}[\/latex].\r\n\r\n[reveal-answer q=\"568949\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"568949\"]\r\n\r\n[latex]\\begin{array}{lllllllll}\\frac{\\left(x+3\\right)\\left(x - 3\\right)}{\\left(x+3\\right)\\left(x+1\\right)}\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Factor the numerator and the denominator}.\\hfill \\\\ \\frac{x - 3}{x+1}\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Cancel common factor }\\left(x+3\\right).\\hfill \\end{array}[\/latex]\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can cancel the common factor because any expression divided by itself is equal to 1.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can the [latex]{x}^{2}[\/latex] term be cancelled in the above example?<\/strong>\r\n\r\n<em>No. A factor is an expression that is multiplied by another expression. The [latex]{x}^{2}[\/latex] term is not a factor of the numerator or the denominator.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify [latex]\\frac{x - 6}{{x}^{2}-36}[\/latex].\r\n\r\n[reveal-answer q=\"17752\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"17752\"]\r\n\r\n[latex]\\frac{1}{x+6}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]110917-110916[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Rationalize Radical Expressions<\/h2>\r\nWhen an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called <em>rationalizing the denominator<\/em>.\r\n\r\nWe know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by a form of 1 that will eliminate the radical.\r\n\r\nFor a denominator containing a binomial where at least one of the terms is a square root, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign in the middle of the binomial. If the denominator is [latex]a+b\\sqrt{c}[\/latex], then the conjugate is [latex]a-b\\sqrt{c}[\/latex].\r\n<div class=\"textbox\">\r\n<h3>How To: Given an expression with a Binomial containing a square root in the denominator, rationalize the denominator<\/h3>\r\n<ol>\r\n \t<li>Find the conjugate of the denominator.<\/li>\r\n \t<li>Multiply the numerator and denominator by the conjugate.<\/li>\r\n \t<li>Use the distributive property.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Rationalizing a Denominator with a binomial Containing a square root<\/h3>\r\nRationalize [latex]\\dfrac{4}{1+\\sqrt{5}}[\/latex].\r\n\r\n[reveal-answer q=\"726340\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"726340\"]\r\n\r\nBegin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of [latex]1+\\sqrt{5}[\/latex] is [latex]1-\\sqrt{5}[\/latex]. Then multiply the fraction by [latex]\\dfrac{1-\\sqrt{5}}{1-\\sqrt{5}}[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\frac{4}{1+\\sqrt{5}}\\cdot \\frac{1-\\sqrt{5}}{1-\\sqrt{5}} &amp;= \\frac{4 - 4\\sqrt{5}}{-4} &amp;&amp; \\text{Use the distributive property}. \\\\ &amp;=\\sqrt{5}-1 &amp;&amp; \\text{Simplify}. \\end{align}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite [latex]\\dfrac{7}{2+\\sqrt{3}}[\/latex] in simplest form.\r\n\r\n[reveal-answer q=\"132932\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"132932\"]\r\n\r\n[latex]14 - 7\\sqrt{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]3441[\/ohm_question]\r\n\r\n<\/div>\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454721&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=vINRIRgeKqU&amp;video_target=tpm-plugin-cw1h5y35-vINRIRgeKqU\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExRationalizeTheDenominatorOfARadicalExpression_Conjugate_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ex: Rationalize the Denominator of a Radical Expression - Conjugate\" here (opens in new window)<\/a>.\r\n<h2>Simplify Complex Rational Expressions<\/h2>\r\n<\/div>\r\nA complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator as single rational expressions and dividing. The complex rational expression [latex]\\dfrac{a}{\\dfrac{1}{b}+c}[\/latex] can be simplified by rewriting the numerator as the fraction [latex]\\dfrac{a}{1}[\/latex] and combining the expressions in the denominator as [latex]\\dfrac{1+bc}{b}[\/latex]. We can then rewrite the expression as a multiplication problem using the reciprocal of the denominator. We get [latex]\\dfrac{a}{1}\\cdot \\dfrac{b}{1+bc}[\/latex] which is equal to [latex]\\dfrac{ab}{1+bc}[\/latex].\r\n<div class=\"textbox\">\r\n<h3>How To: Given a complex rational expression, simplify it<\/h3>\r\n<ol>\r\n \t<li>Combine the expressions in the numerator into a single rational expression by adding or subtracting.<\/li>\r\n \t<li>Combine the expressions in the denominator into a single rational expression by adding or subtracting.<\/li>\r\n \t<li>Rewrite as the numerator divided by the denominator.<\/li>\r\n \t<li>Rewrite as multiplication.<\/li>\r\n \t<li>Multiply.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Simplifying Complex Rational Expressions<\/h3>\r\nSimplify: [latex]\\dfrac{y+\\dfrac{1}{x}}{\\dfrac{x}{y}}[\/latex] .\r\n\r\n[reveal-answer q=\"967019\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"967019\"]\r\n\r\nBegin by combining the expressions in the numerator into one expression.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}y\\cdot \\dfrac{x}{x}+\\dfrac{1}{x}\\hfill &amp; \\text{Multiply by }\\dfrac{x}{x}\\text{to get LCD as denominator}.\\hfill \\\\ \\dfrac{xy}{x}+\\dfrac{1}{x}\\hfill &amp; \\\\ \\dfrac{xy+1}{x}\\hfill &amp; \\text{Add numerators}.\\hfill \\end{array}[\/latex]<\/div>\r\nNow the numerator is a single rational expression and the denominator is a single rational expression.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{\\dfrac{xy+1}{x}}{\\dfrac{x}{y}}[\/latex]<\/div>\r\nWe can rewrite this as division and then multiplication.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}\\dfrac{xy+1}{x}\\div \\dfrac{x}{y}\\hfill &amp; \\\\ \\dfrac{xy+1}{x}\\cdot \\dfrac{y}{x}\\hfill &amp; \\text{Rewrite as multiplication}\\text{.}\\hfill \\\\ \\dfrac{y\\left(xy+1\\right)}{{x}^{2}}\\hfill &amp; \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]3078-3080-59554[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor the greatest common factor (monomial) of a polynomial<\/li>\n<li>Factor a trinomial with a leading coefficient of 1<\/li>\n<li>Rewrite a trinomial as a four term polynomial and factor by grouping terms<\/li>\n<li>Factor difference of squares<\/li>\n<li>Factor sum or difference of cubes<\/li>\n<li>Simplify a rational expression<\/li>\n<li>Remove radicals from a multiple term denominator<\/li>\n<li>Simplify complex rational expressions<\/li>\n<\/ul>\n<\/div>\n<p>The Limit Laws section will introduce yet another way to calculate a limit, using limit laws. Basically, calculating limits of functions algebraically will be the topic of focus in the section. Several techniques that are used to simplify functions are sometimes needed to successfully calculate a function&#8217;s limit. Here we will review factoring, simplifying rational expressions, rationalizing radical expressions, and simplifying complex rational expressions.<\/p>\n<h2>Factor Polynomials<\/h2>\n<p>Recall that the <strong>greatest common factor<\/strong> (GCF) of two numbers is the largest number that divides evenly into both numbers. For example, [latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest number that divides evenly into both [latex]16[\/latex] and [latex]20[\/latex]. The GCF of polynomials works the same way: [latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest polynomial that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].<\/p>\n<p>Finding and factoring out a GCF from a polynomial is the first skill involved in factoring polynomials.<\/p>\n<h3>Factor a GCF out of a Polynomial<\/h3>\n<p>When factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term. To make it less challenging to find this GCF of the polynomial terms, first look for the GCF of the coefficients, and then look for the GCF of the variables.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Greatest Common Factor<\/h3>\n<p>The <strong>greatest common factor<\/strong> (GCF) of a polynomial is the largest polynomial that divides evenly into each term of the polynomial.<\/p>\n<\/div>\n<p>To factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property &#8220;backwards&#8221; to rewrite the polynomial in a factored form. Recall that the <strong>distributive property of multiplication over addition<\/strong> states that a product of a number and a sum is the same as the sum of the products.<\/p>\n<div class=\"textbox shaded\">\n<h4>Distributive Property Forward and Backward<\/h4>\n<p style=\"text-align: left;\">Forward:\u00a0<em>We distribute [latex]a[\/latex] over [latex]b+c[\/latex]<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]a\\left(b+c\\right)=ab+ac[\/latex].<\/p>\n<p>Backward:\u00a0<em>We factor [latex]a[\/latex] out of [latex]ab+ac[\/latex].<\/em><\/p>\n<p style=\"text-align: center;\">[latex]ab+ac=a\\left(b+c\\right)[\/latex].<\/p>\n<p>We have seen that we can distribute a factor over a sum or difference. Now we see that we can &#8220;undo&#8221; the distributive property with factoring.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Factoring The Greatest Common Factor<\/h3>\n<p>Factor [latex]25b^{3}+10b^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q716902\">Show Solution<\/span><\/p>\n<div id=\"q716902\" class=\"hidden-answer\" style=\"display: none\">Find the GCF.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}=5b^{2}\\end{array}[\/latex]<\/p>\n<p>Rewrite each term with the GCF as one factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3} = 5b^{2}\\cdot5b\\\\10b^{2}=5b^{2}\\cdot2\\end{array}[\/latex]<\/p>\n<p>Rewrite the polynomial using the factored terms in place of the original terms.<\/p>\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b\\right)+5b^{2}\\left(2\\right)[\/latex]<\/p>\n<p>Factor out the [latex]5b^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b+2\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]5b^{2}\\left(5b+2\\right)[\/latex]<\/p>\n<h4>Analysis<\/h4>\n<p>The factored form of the polynomial [latex]25b^{3}+10b^{2}[\/latex] is [latex]5b^{2}\\left(5b+2\\right)[\/latex]. You can check this by doing the multiplication. [latex]5b^{2}\\left(5b+2\\right)=25b^{3}+10b^{2}[\/latex].<\/p>\n<p>Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.<\/p>\n<p>For example:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3}+10b^{2}=5\\left(5b^{3}+2b^{2}\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }5.\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5b^{2}\\left(5b+2\\right) \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }b^{2}.\\end{array}[\/latex]<\/p>\n<p>Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a polynomial expression, factor out the greatest common factor<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Identify the GCF of the coefficients.<\/li>\n<li>Identify the GCF of the variables.<\/li>\n<li>Combine to find the GCF of the expression.<\/li>\n<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\n<li>Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring the Greatest Common Factor<\/h3>\n<p>Factor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q113189\">Show Solution<\/span><\/p>\n<div id=\"q113189\" class=\"hidden-answer\" style=\"display: none\">\n<p>First find the GCF of the expression. The GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions of the form [latex]{x}^{n}[\/latex] will always be the lowest exponent.) The GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. Combine these to find the GCF of the polynomial, [latex]3xy[\/latex].<\/p>\n<p>Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that [latex]3xy\\left(2{x}^{2}{y}^{2}\\right)=6{x}^{3}{y}^{3}, 3xy\\left(15xy\\right)=45{x}^{2}{y}^{2}[\/latex], and [latex]3xy\\left(7\\right)=21xy[\/latex].<\/p>\n<p>Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.<\/p>\n<div style=\"text-align: center;\">[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/div>\n<div>\n<h4>Analysis of the Solution<\/h4>\n<p>After factoring, we can check our work by multiplying. Use the distributive property to confirm that [latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].<\/p>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch this video to see more examples of how to factor the GCF from a trinomial.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454717&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=3f1RFTIw2Ng&amp;video_target=tpm-plugin-8plr8d8s-3f1RFTIw2Ng\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/Ex2IdentifyGCFAndFactorATrinomial_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ex 2: Identify GCF and Factor a Trinomial&#8221; here (opens in new window)<\/a>.<\/p>\n<h3>Factor a Trinomial with Leading Coefficient 1<\/h3>\n<p>Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].<\/p>\n<p>Trinomials of the form [latex]{x}^{2}+bx+c[\/latex] can be factored by finding two numbers with a product of [latex]c[\/latex] and a sum of [latex]b[\/latex]. The trinomial [latex]{x}^{2}+10x+16[\/latex], for example, can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex] because the product of these numbers is [latex]16[\/latex] and their sum is [latex]10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>A trinomial of the form [latex]{x}^{2}+bx+c[\/latex] can be written in factored form as [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex] where [latex]pq=c[\/latex] and [latex]p+q=b[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can every trinomial be factored as a product of binomials?<\/strong><\/p>\n<p><em>No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\n<ol>\n<li>List factors of [latex]c[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>Factor [latex]{x}^{2}+2x - 15[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q88306\">Show Solution<\/span><\/p>\n<div id=\"q88306\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\n<thead>\n<tr>\n<th>Factors of [latex]-15[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-15[\/latex]<\/td>\n<td>[latex]-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,15[\/latex]<\/td>\n<td>[latex]14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,5[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Does the order of the factors matter?<\/strong><\/p>\n<p><em>No. Multiplication is commutative, so the order of the factors does not matter.<\/em><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7897\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7897&theme=oea&iframe_resize_id=ohm7897&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3>Factor by Grouping<\/h3>\n<p>Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Factor by Grouping<\/h3>\n<p>To factor a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping<\/h3>\n<ol>\n<li>List factors of [latex]ac[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]ac[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\n<li>Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\n<li>Pull out the GCF of [latex]qx+c[\/latex].<\/li>\n<li>Factor out the GCF of the expression.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Trinomial by Grouping<\/h3>\n<p>Factor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q806328\">Show Solution<\/span><\/p>\n<div id=\"q806328\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\n<thead>\n<tr>\n<th>Factors of [latex]-30[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-30[\/latex]<\/td>\n<td>[latex]-29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,30[\/latex]<\/td>\n<td>[latex]29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2,-15[\/latex]<\/td>\n<td>[latex]-13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,15[\/latex]<\/td>\n<td>[latex]13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-10[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,10[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]p=-3[\/latex] and [latex]q=10[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill & \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill & \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill & \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].<\/p>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7908\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7908&theme=oea&iframe_resize_id=ohm7908&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the next video, we see another example of how to factor a trinomial by grouping.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/agDaQ_cZnNc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/FactorATrinomialInTheFormUsingTheGroupingTechniqe_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique&#8221; here (opens in new window)<\/a>.<\/p>\n<h3>Factor a Difference of Squares<\/h3>\n<p>A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>We can use this equation to factor any differences of squares.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Differences of Squares<\/h3>\n<p>A difference of squares can be rewritten as two factors containing the same terms but opposite signs.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a difference of squares, factor it into binomials<\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Difference of Squares<\/h3>\n<p>Factor [latex]9{x}^{2}-25[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q830417\">Show Solution<\/span><\/p>\n<div id=\"q830417\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex]. The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7929\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7929&theme=oea&iframe_resize_id=ohm7929&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Is there a formula to factor the sum of squares?<\/strong><\/p>\n<p><em>No. A sum of squares cannot be factored.<\/em><\/p>\n<\/div>\n<p>Watch this video to see another example of how to factor a difference of squares.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454718&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Li9IBp5HrFA&amp;video_target=tpm-plugin-llp8qg9s-Li9IBp5HrFA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExFactorADifferenceOfSquares_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ex: Factor a Difference of Squares&#8221; here (opens in new window)<\/a>.<\/p>\n<h3>Factor the Sum and Difference of Cubes<\/h3>\n<p>Now we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>Similarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: <strong>S<\/strong>ame <strong>O<\/strong>pposite <strong>A<\/strong>lways <strong>P<\/strong>ositive. For example, consider the following example.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{3}-{2}^{3}=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]<\/div>\n<p>The sign of the first 2 is the <em>same<\/em> as the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. The sign of the [latex]2x[\/latex] term is <em>opposite<\/em> the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. And the sign of the last term, 4, is <em>always positive<\/em>.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Sum and Difference of Cubes<\/h3>\n<p>We can factor the sum of two cubes as<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\n<p>We can factor the difference of two cubes as<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a sum of cubes or difference of cubes, factor it<\/h3>\n<ol>\n<li>Confirm that the first and last term are cubes, [latex]{a}^{3}+{b}^{3}[\/latex] or [latex]{a}^{3}-{b}^{3}[\/latex].<\/li>\n<li>For a sum of cubes, write the factored form as [latex]\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]. For a difference of cubes, write the factored form as [latex]\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Sum of Cubes<\/h3>\n<p>Factor [latex]{x}^{3}+512[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q22673\">Show Solution<\/span><\/p>\n<div id=\"q22673\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]{x}^{3}[\/latex] and [latex]512[\/latex] are cubes because [latex]{8}^{3}=512[\/latex]. Rewrite the sum of cubes as [latex]\\left(x+8\\right)\\left({x}^{2}-8x+64\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor the sum of cubes [latex]216{a}^{3}+{b}^{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q496204\">Show Solution<\/span><\/p>\n<div id=\"q496204\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(6a+b\\right)\\left(36{a}^{2}-6ab+{b}^{2}\\right)[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Difference of Cubes<\/h3>\n<p>Factor [latex]8{x}^{3}-125[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q741836\">Show Solution<\/span><\/p>\n<div id=\"q741836\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]8{x}^{3}[\/latex] and [latex]125[\/latex] are cubes because [latex]8{x}^{3}={\\left(2x\\right)}^{3}[\/latex] and [latex]125={5}^{3}[\/latex]. Write the difference of cubes as [latex]\\left(2x - 5\\right)\\left(4{x}^{2}+10x+25\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Just as with the sum of cubes, we will not be able to further factor the trinomial portion.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor the difference of cubes: [latex]1,000{x}^{3}-1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q510077\">Show Solution<\/span><\/p>\n<div id=\"q510077\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(10x - 1\\right)\\left(100{x}^{2}+10x+1\\right)[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7922\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7922&theme=oea&iframe_resize_id=ohm7922&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following two video examples, we show more binomials that can be factored as a sum or difference of cubes.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454719&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=tFSEpOB262M&amp;video_target=tpm-plugin-je39dfag-tFSEpOB262M\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/Ex1FactorASumOrDifferenceOfCubes_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ex 1: Factor a Sum or Difference of Cubes&#8221; here (opens in new window)<\/a>.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454720&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=J_0ctMrl5_0&amp;video_target=tpm-plugin-cnyddfw5-J_0ctMrl5_0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/Ex3FactorASumOrDifferenceOfCubes_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ex 3: Factor a Sum or Difference of Cubes&#8221; here (opens in new window)<\/a>.<\/p>\n<h2>Simplify Rational Expressions<\/h2>\n<p>The quotient of two polynomial expressions is called a <strong>rational expression<\/strong>. We can apply the properties of fractions to rational expressions such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let\u2019s start with the rational expression shown.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{{x}^{2}+8x+16}{{x}^{2}+11x+28}[\/latex]<\/p>\n<p>[latex]\\\\[\/latex]We can factor the numerator and denominator to rewrite the expression as [latex]\\frac{{\\left(x+4\\right)}^{2}}{\\left(x+4\\right)\\left(x+7\\right)}[\/latex]<\/p>\n<div><\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div>Then we can simplify the expression by canceling the common factor [latex]\\left(x+4\\right)[\/latex] to get [latex]\\frac{x+4}{x+7}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a rational expression, simplify it<\/h3>\n<ol>\n<li>Factor the numerator and denominator.<\/li>\n<li>Cancel any common factors.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Simplifying Rational Expressions<\/h3>\n<p>Simplify [latex]\\frac{{x}^{2}-9}{{x}^{2}+4x+3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q568949\">Show Solution<\/span><\/p>\n<div id=\"q568949\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{lllllllll}\\frac{\\left(x+3\\right)\\left(x - 3\\right)}{\\left(x+3\\right)\\left(x+1\\right)}\\hfill & \\hfill & \\hfill & \\hfill & \\text{Factor the numerator and the denominator}.\\hfill \\\\ \\frac{x - 3}{x+1}\\hfill & \\hfill & \\hfill & \\hfill & \\text{Cancel common factor }\\left(x+3\\right).\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can cancel the common factor because any expression divided by itself is equal to 1.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can the [latex]{x}^{2}[\/latex] term be cancelled in the above example?<\/strong><\/p>\n<p><em>No. A factor is an expression that is multiplied by another expression. The [latex]{x}^{2}[\/latex] term is not a factor of the numerator or the denominator.<\/em><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify [latex]\\frac{x - 6}{{x}^{2}-36}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q17752\">Show Solution<\/span><\/p>\n<div id=\"q17752\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{1}{x+6}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm110917\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110917-110916&theme=oea&iframe_resize_id=ohm110917&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Rationalize Radical Expressions<\/h2>\n<p>When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called <em>rationalizing the denominator<\/em>.<\/p>\n<p>We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by a form of 1 that will eliminate the radical.<\/p>\n<p>For a denominator containing a binomial where at least one of the terms is a square root, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign in the middle of the binomial. If the denominator is [latex]a+b\\sqrt{c}[\/latex], then the conjugate is [latex]a-b\\sqrt{c}[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an expression with a Binomial containing a square root in the denominator, rationalize the denominator<\/h3>\n<ol>\n<li>Find the conjugate of the denominator.<\/li>\n<li>Multiply the numerator and denominator by the conjugate.<\/li>\n<li>Use the distributive property.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Rationalizing a Denominator with a binomial Containing a square root<\/h3>\n<p>Rationalize [latex]\\dfrac{4}{1+\\sqrt{5}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q726340\">Show Solution<\/span><\/p>\n<div id=\"q726340\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of [latex]1+\\sqrt{5}[\/latex] is [latex]1-\\sqrt{5}[\/latex]. Then multiply the fraction by [latex]\\dfrac{1-\\sqrt{5}}{1-\\sqrt{5}}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\frac{4}{1+\\sqrt{5}}\\cdot \\frac{1-\\sqrt{5}}{1-\\sqrt{5}} &= \\frac{4 - 4\\sqrt{5}}{-4} && \\text{Use the distributive property}. \\\\ &=\\sqrt{5}-1 && \\text{Simplify}. \\end{align}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write [latex]\\dfrac{7}{2+\\sqrt{3}}[\/latex] in simplest form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q132932\">Show Solution<\/span><\/p>\n<div id=\"q132932\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]14 - 7\\sqrt{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm3441\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3441&theme=oea&iframe_resize_id=ohm3441&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454721&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=vINRIRgeKqU&amp;video_target=tpm-plugin-cw1h5y35-vINRIRgeKqU\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExRationalizeTheDenominatorOfARadicalExpression_Conjugate_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ex: Rationalize the Denominator of a Radical Expression &#8211; Conjugate&#8221; here (opens in new window)<\/a>.<\/p>\n<h2>Simplify Complex Rational Expressions<\/h2>\n<\/div>\n<p>A complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator as single rational expressions and dividing. The complex rational expression [latex]\\dfrac{a}{\\dfrac{1}{b}+c}[\/latex] can be simplified by rewriting the numerator as the fraction [latex]\\dfrac{a}{1}[\/latex] and combining the expressions in the denominator as [latex]\\dfrac{1+bc}{b}[\/latex]. We can then rewrite the expression as a multiplication problem using the reciprocal of the denominator. We get [latex]\\dfrac{a}{1}\\cdot \\dfrac{b}{1+bc}[\/latex] which is equal to [latex]\\dfrac{ab}{1+bc}[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a complex rational expression, simplify it<\/h3>\n<ol>\n<li>Combine the expressions in the numerator into a single rational expression by adding or subtracting.<\/li>\n<li>Combine the expressions in the denominator into a single rational expression by adding or subtracting.<\/li>\n<li>Rewrite as the numerator divided by the denominator.<\/li>\n<li>Rewrite as multiplication.<\/li>\n<li>Multiply.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Simplifying Complex Rational Expressions<\/h3>\n<p>Simplify: [latex]\\dfrac{y+\\dfrac{1}{x}}{\\dfrac{x}{y}}[\/latex] .<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q967019\">Show Solution<\/span><\/p>\n<div id=\"q967019\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by combining the expressions in the numerator into one expression.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}y\\cdot \\dfrac{x}{x}+\\dfrac{1}{x}\\hfill & \\text{Multiply by }\\dfrac{x}{x}\\text{to get LCD as denominator}.\\hfill \\\\ \\dfrac{xy}{x}+\\dfrac{1}{x}\\hfill & \\\\ \\dfrac{xy+1}{x}\\hfill & \\text{Add numerators}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Now the numerator is a single rational expression and the denominator is a single rational expression.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{\\dfrac{xy+1}{x}}{\\dfrac{x}{y}}[\/latex]<\/div>\n<p>We can rewrite this as division and then multiplication.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}\\dfrac{xy+1}{x}\\div \\dfrac{x}{y}\\hfill & \\\\ \\dfrac{xy+1}{x}\\cdot \\dfrac{y}{x}\\hfill & \\text{Rewrite as multiplication}\\text{.}\\hfill \\\\ \\dfrac{y\\left(xy+1\\right)}{{x}^{2}}\\hfill & \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm3078\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3078-3080-59554&theme=oea&iframe_resize_id=ohm3078&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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-3773\">\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>Modification and Revision. <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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/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>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/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 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