{"id":18301,"date":"2022-04-14T22:04:35","date_gmt":"2022-04-14T22:04:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/?post_type=chapter&#038;p=18301"},"modified":"2022-04-28T23:16:07","modified_gmt":"2022-04-28T23:16:07","slug":"cr-12-factoring-higher-power-polynomials-and-special-polynomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/cr-12-factoring-higher-power-polynomials-and-special-polynomials\/","title":{"raw":"CR.12: Factoring Higher Power Polynomials and Special Polynomials","rendered":"CR.12: Factoring Higher Power Polynomials and Special Polynomials"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor a perfect square trinomial.<\/li>\r\n \t<li>Factor a difference of squares.<\/li>\r\n \t<li>Factor a sum and difference of cubes.<\/li>\r\n \t<li>Factor higher power polynomials using substitution<\/li>\r\n \t<li>Factor polynomials completely using all methods<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Factoring a Perfect Square Trinomial<\/h2>\r\nA perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}&amp; =&amp; {\\left(a+b\\right)}^{2}\\hfill \\\\ &amp; \\text{and}&amp; \\\\ \\hfill {a}^{2}-2ab+{b}^{2}&amp; =&amp; {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div style=\"text-align: left;\">We can use this equation to factor any perfect square trinomial.<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Perfect Square Trinomials<\/h3>\r\nA perfect square trinomial can be written as the square of a binomial:\r\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a perfect square trinomial, factor it into the square of a binomial<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\r\n \t<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Perfect Square Trinomial<\/h3>\r\nFactor [latex]25{x}^{2}+20x+4[\/latex].\r\n\r\n[reveal-answer q=\"114092\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"114092\"]\r\n\r\nNotice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex]. Then check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(5x+2\\right)}^{2}[\/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\nFactor [latex]49{x}^{2}-14x+1[\/latex].\r\n\r\n[reveal-answer q=\"530221\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530221\"]\r\n\r\n[latex]{\\left(7x - 1\\right)}^{2}[\/latex][\/hidden-answer]\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7919&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Factoring a Difference of Squares<\/h2>\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\nFactor [latex]81{y}^{2}-100[\/latex].\r\n\r\n[reveal-answer q=\"979538\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979538\"]\r\n\r\n[latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7929&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\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\nhttps:\/\/youtu.be\/Li9IBp5HrFA\r\n<h2>Factoring the Sum and Difference of Cubes<\/h2>\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<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7922&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\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\nhttps:\/\/youtu.be\/tFSEpOB262M\r\n\r\nhttps:\/\/youtu.be\/J_0ctMrl5_0\r\n<h2>Factor Using Substitution<\/h2>\r\nWe are going to move back to factoring polynomials; our exponents will be positive integers. Sometimes we encounter a polynomial that looks similar to something we know how to factor but is not quite the same. Substitution is a useful tool that can be used to \u201cmask\u201d a term or expression to make algebraic operations easier.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]x^4+3x^2+2[\/latex].\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q98597\">Show Solution<\/span>\r\n<div id=\"q98597\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nThis looks a lot like a trinomial that we know how to factor: [latex]x^2+3x+2=(x+2)(x+1)[\/latex]. The only thing different is the exponents.\r\n\r\nIf we substitute [latex]u=x^2[\/latex] and recognize that [latex]u^2=(x^2)^2=x^4[\/latex], we may be able to factor this beast!\r\n\r\nEverywhere there is a [latex]x^2[\/latex] we will replace it with a [latex]u[\/latex] then factor.\r\n\r\n[latex]u^2+3u+2=(u+1)(u+2)[\/latex]\r\n\r\nWe are not quite done yet. We want to factor the original polynomial which had [latex]x[\/latex] as its variable, so we need to replace [latex]x^2=u[\/latex] now that we are done factoring.\r\n\r\n[latex](u+1)(u+2)=(x^2+1)(x^2+2)[\/latex]\r\n\r\nWe conclude that [latex]x^4+3x^2+2=(x^2+1)(x^2+2)[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]1366[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]x^4-81[\/latex].\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q98600\">Show Solution<\/span>\r\n<div id=\"q98600\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nThis looks a lot like a difference of squares that we know how to factor: [latex]x^2-81=(x+9)(x-9)[\/latex]. The only thing different is the exponents.\r\n\r\nIf we substitute [latex]u=x^2[\/latex] and recognize that [latex]u^2=(x^2)^2=x^4[\/latex], we will be able to factor.\r\n\r\nEverywhere there is a [latex]x^2[\/latex] we will replace it with a [latex]u[\/latex] then factor.\r\n\r\n[latex]u^2-81=(u+9)(u-9)[\/latex]\r\n\r\nWe are not quite done yet. We want to factor the original polynomial which had [latex]x[\/latex] as its variable, so we need to replace [latex]x^2=u[\/latex].\r\n\r\n[latex](u+9)(u-9)=(x^2+9)(x^2-9)[\/latex]\r\n\r\nHowever, this is not fully factored. We notice that [latex]x^2-9[\/latex] can be factored into [latex](x+3)(x-3)[\/latex]. You always want to factor as much as possible.\r\n\r\nTherefore, we conclude that [latex]x^4-81=(x^2+9)(x+3)(x-3)[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]248356[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Factor Completely<\/h2>\r\nSometimes you may encounter a polynomial that takes an extra step to factor. In our next example, we will first find the GCF of a trinomial, and after factoring it out, we will be able to factor again so that we end up with a product of a monomial and two binomials.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6m^2k-3mk-3k[\/latex] completely.\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q698742\">Show Solution<\/span>\r\n<div id=\"q698742\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nWhenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is [latex]3k[\/latex].\r\n\r\nFactor [latex]3k[\/latex] from the trinomial:\r\n\r\n[latex]6m^2k-3mk-3k=3k\\left(2m^2-m-1\\right)[\/latex]\r\n\r\nWe are left with a trinomial that can be factored using your choice of factoring methods. We will create a table to find the factors of [latex]2\\cdot{-1}=-2[\/latex] that sum to [latex]-1[\/latex]\r\n<table class=\" aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 and Sum of Factors. The entries in the first column are: 1, 6; -1, -6; 2,3; and -2,-3. The entries in the second column are: 7, -7, 5, and -5.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]2\\cdot-1=-2[\/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]2,-1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,1[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nOur factors are [latex]-2,1[\/latex] which will allow us to factor by grouping:\r\n\r\nRewrite the middle term with the factors we found:\r\n<p style=\"text-align: center;\">[latex](2m^2-m-1)=2m^2-2m+m-1[\/latex]<\/p>\r\nRegroup and find the GCF of each group:\r\n<p style=\"text-align: center;\">[latex](2m^2-2m)+(m-1)=2m(m-1)+1(m-1)[\/latex]<\/p>\r\nNow factor [latex](m-1)[\/latex] from each term:\r\n<p style=\"text-align: center;\">[latex]2m^2-m-1=(m-1)(2m+1)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Do not forget the original GCF that we factored out! Our final factored form is:<\/p>\r\n<p style=\"text-align: center;\">[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]2x^4-144x^2+2592[\/latex] completely.\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q698760\">Show Solution<\/span>\r\n<div id=\"q698760\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nWhenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is [latex]2[\/latex].\r\n\r\nFactor [latex]2[\/latex] from the binomial:\r\n\r\n[latex]2(x^4-72x^2+1296)[\/latex]\r\n\r\nWe are left with a trinomial that can be factored using perfect squares. Notice that [latex]x^4=(x^2)^2[\/latex] and [latex](36)^2=1296[\/latex]. Also notice that the middle term, [latex]-72x^2[\/latex], of the trinomial is equal to [latex]-2[\/latex] times the product of [latex]x^2[\/latex] and [latex]36[\/latex]. Therefore, [latex]-2 \\cdot x^2 \\cdot 36=-72x^2[\/latex]. We can conclude that the trinomial [latex]x^4-72x^2+1296[\/latex] is a perfect square trinomial.\r\n\r\nSince the middle term is negative, use the formula [latex]A^2-2AB+B^2=(A-B)^2[\/latex] to factor the perfect square trinomial.\r\n[latex]x^4-72x^2+1296=(x^2)^2-2 \\cdot x^2 \\cdot 36+(36)^2=(x^2-36)^2[\/latex].\r\n\r\nNotice that [latex]x^2-36[\/latex] is a difference of two squares. We can factor this into [latex](x+6)(x-6)[\/latex]. Since this is squared, then [latex](x^2-36)^2=[(x+6)(x-6)]^2=(x+6)^2(x-6)^2[\/latex].\r\n\r\nPutting this all together, [latex]2x^4-144x^2+2592=2(x+6)^2(x-6)^2[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]248371[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]54x^{17}+2x^{14}[\/latex] completely.\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q698750\">Show Solution<\/span>\r\n<div id=\"q698750\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nWhenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is [latex]2x^{14}[\/latex].\r\n\r\nFactor [latex]2x^{14}[\/latex] from the binomial:\r\n\r\n[latex]2x^{14}(27x^3+1)[\/latex]\r\n\r\nWe are left with a binomial that can be factored using the sum of cubes. Notice that [latex]27x^3[\/latex] and 1 are cubes because [latex]27x^3=(3x)^3[\/latex] and [latex]1=1^3[\/latex]. Therefore, [latex]27x^3+1=(3x+1)(9x^2-3x+1)[\/latex]\r\n\r\nOur final answer must include the original GCF we already factored out: [latex]2x^{14}(3x+1)(9x^2-3x+1)[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]248369[\/ohm_question]\r\n\r\n<\/div>\r\nIn our last example, we show why it is important to factor out a GCF, if there is one, before you begin using the techniques shown in this section.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/hMAImz2BuPc?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor a perfect square trinomial.<\/li>\n<li>Factor a difference of squares.<\/li>\n<li>Factor a sum and difference of cubes.<\/li>\n<li>Factor higher power polynomials using substitution<\/li>\n<li>Factor polynomials completely using all methods<\/li>\n<\/ul>\n<\/div>\n<h2>Factoring a Perfect Square Trinomial<\/h2>\n<p>A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}& =& {\\left(a+b\\right)}^{2}\\hfill \\\\ & \\text{and}& \\\\ \\hfill {a}^{2}-2ab+{b}^{2}& =& {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div style=\"text-align: left;\">We can use this equation to factor any perfect square trinomial.<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Perfect Square Trinomials<\/h3>\n<p>A perfect square trinomial can be written as the square of a binomial:<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a perfect square trinomial, factor it into the square of a binomial<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\n<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Perfect Square Trinomial<\/h3>\n<p>Factor [latex]25{x}^{2}+20x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q114092\">Show Solution<\/span><\/p>\n<div id=\"q114092\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex]. Then check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(5x+2\\right)}^{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor [latex]49{x}^{2}-14x+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530221\">Show Solution<\/span><\/p>\n<div id=\"q530221\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\left(7x - 1\\right)}^{2}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7919&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Factoring a Difference of Squares<\/h2>\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>Factor [latex]81{y}^{2}-100[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q979538\">Show Solution<\/span><\/p>\n<div id=\"q979538\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7929&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/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\" id=\"oembed-1\" title=\"Ex:  Factor a Difference of Squares\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Li9IBp5HrFA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Factoring the Sum and Difference of Cubes<\/h2>\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<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7922&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/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\" id=\"oembed-2\" title=\"Ex 1:  Factor a Sum or Difference of Cubes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tFSEpOB262M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 3:  Factor a Sum or Difference of Cubes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/J_0ctMrl5_0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Factor Using Substitution<\/h2>\n<p>We are going to move back to factoring polynomials; our exponents will be positive integers. Sometimes we encounter a polynomial that looks similar to something we know how to factor but is not quite the same. Substitution is a useful tool that can be used to \u201cmask\u201d a term or expression to make algebraic operations easier.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]x^4+3x^2+2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q98597\">Show Solution<\/span><\/p>\n<div id=\"q98597\" class=\"hidden-answer\" style=\"display: none;\">\n<p>This looks a lot like a trinomial that we know how to factor: [latex]x^2+3x+2=(x+2)(x+1)[\/latex]. The only thing different is the exponents.<\/p>\n<p>If we substitute [latex]u=x^2[\/latex] and recognize that [latex]u^2=(x^2)^2=x^4[\/latex], we may be able to factor this beast!<\/p>\n<p>Everywhere there is a [latex]x^2[\/latex] we will replace it with a [latex]u[\/latex] then factor.<\/p>\n<p>[latex]u^2+3u+2=(u+1)(u+2)[\/latex]<\/p>\n<p>We are not quite done yet. We want to factor the original polynomial which had [latex]x[\/latex] as its variable, so we need to replace [latex]x^2=u[\/latex] now that we are done factoring.<\/p>\n<p>[latex](u+1)(u+2)=(x^2+1)(x^2+2)[\/latex]<\/p>\n<p>We conclude that [latex]x^4+3x^2+2=(x^2+1)(x^2+2)[\/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=\"ohm1366\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1366&theme=oea&iframe_resize_id=ohm1366&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]x^4-81[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q98600\">Show Solution<\/span><\/p>\n<div id=\"q98600\" class=\"hidden-answer\" style=\"display: none;\">\n<p>This looks a lot like a difference of squares that we know how to factor: [latex]x^2-81=(x+9)(x-9)[\/latex]. The only thing different is the exponents.<\/p>\n<p>If we substitute [latex]u=x^2[\/latex] and recognize that [latex]u^2=(x^2)^2=x^4[\/latex], we will be able to factor.<\/p>\n<p>Everywhere there is a [latex]x^2[\/latex] we will replace it with a [latex]u[\/latex] then factor.<\/p>\n<p>[latex]u^2-81=(u+9)(u-9)[\/latex]<\/p>\n<p>We are not quite done yet. We want to factor the original polynomial which had [latex]x[\/latex] as its variable, so we need to replace [latex]x^2=u[\/latex].<\/p>\n<p>[latex](u+9)(u-9)=(x^2+9)(x^2-9)[\/latex]<\/p>\n<p>However, this is not fully factored. We notice that [latex]x^2-9[\/latex] can be factored into [latex](x+3)(x-3)[\/latex]. You always want to factor as much as possible.<\/p>\n<p>Therefore, we conclude that [latex]x^4-81=(x^2+9)(x+3)(x-3)[\/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=\"ohm248356\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=248356&theme=oea&iframe_resize_id=ohm248356&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Factor Completely<\/h2>\n<p>Sometimes you may encounter a polynomial that takes an extra step to factor. In our next example, we will first find the GCF of a trinomial, and after factoring it out, we will be able to factor again so that we end up with a product of a monomial and two binomials.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]6m^2k-3mk-3k[\/latex] completely.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q698742\">Show Solution<\/span><\/p>\n<div id=\"q698742\" class=\"hidden-answer\" style=\"display: none;\">\n<p>Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is [latex]3k[\/latex].<\/p>\n<p>Factor [latex]3k[\/latex] from the trinomial:<\/p>\n<p>[latex]6m^2k-3mk-3k=3k\\left(2m^2-m-1\\right)[\/latex]<\/p>\n<p>We are left with a trinomial that can be factored using your choice of factoring methods. We will create a table to find the factors of [latex]2\\cdot{-1}=-2[\/latex] that sum to [latex]-1[\/latex]<\/p>\n<table class=\"aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 and Sum of Factors. The entries in the first column are: 1, 6; -1, -6; 2,3; and -2,-3. The entries in the second column are: 7, -7, 5, and -5.\">\n<thead>\n<tr>\n<th>Factors of [latex]2\\cdot-1=-2[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]2,-1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,1[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Our factors are [latex]-2,1[\/latex] which will allow us to factor by grouping:<\/p>\n<p>Rewrite the middle term with the factors we found:<\/p>\n<p style=\"text-align: center;\">[latex](2m^2-m-1)=2m^2-2m+m-1[\/latex]<\/p>\n<p>Regroup and find the GCF of each group:<\/p>\n<p style=\"text-align: center;\">[latex](2m^2-2m)+(m-1)=2m(m-1)+1(m-1)[\/latex]<\/p>\n<p>Now factor [latex](m-1)[\/latex] from each term:<\/p>\n<p style=\"text-align: center;\">[latex]2m^2-m-1=(m-1)(2m+1)[\/latex]<\/p>\n<p style=\"text-align: left;\">Do not forget the original GCF that we factored out! Our final factored form is:<\/p>\n<p style=\"text-align: center;\">[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]2x^4-144x^2+2592[\/latex] completely.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q698760\">Show Solution<\/span><\/p>\n<div id=\"q698760\" class=\"hidden-answer\" style=\"display: none;\">\n<p>Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is [latex]2[\/latex].<\/p>\n<p>Factor [latex]2[\/latex] from the binomial:<\/p>\n<p>[latex]2(x^4-72x^2+1296)[\/latex]<\/p>\n<p>We are left with a trinomial that can be factored using perfect squares. Notice that [latex]x^4=(x^2)^2[\/latex] and [latex](36)^2=1296[\/latex]. Also notice that the middle term, [latex]-72x^2[\/latex], of the trinomial is equal to [latex]-2[\/latex] times the product of [latex]x^2[\/latex] and [latex]36[\/latex]. Therefore, [latex]-2 \\cdot x^2 \\cdot 36=-72x^2[\/latex]. We can conclude that the trinomial [latex]x^4-72x^2+1296[\/latex] is a perfect square trinomial.<\/p>\n<p>Since the middle term is negative, use the formula [latex]A^2-2AB+B^2=(A-B)^2[\/latex] to factor the perfect square trinomial.<br \/>\n[latex]x^4-72x^2+1296=(x^2)^2-2 \\cdot x^2 \\cdot 36+(36)^2=(x^2-36)^2[\/latex].<\/p>\n<p>Notice that [latex]x^2-36[\/latex] is a difference of two squares. We can factor this into [latex](x+6)(x-6)[\/latex]. Since this is squared, then [latex](x^2-36)^2=[(x+6)(x-6)]^2=(x+6)^2(x-6)^2[\/latex].<\/p>\n<p>Putting this all together, [latex]2x^4-144x^2+2592=2(x+6)^2(x-6)^2[\/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=\"ohm248371\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=248371&theme=oea&iframe_resize_id=ohm248371&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]54x^{17}+2x^{14}[\/latex] completely.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q698750\">Show Solution<\/span><\/p>\n<div id=\"q698750\" class=\"hidden-answer\" style=\"display: none;\">\n<p>Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is [latex]2x^{14}[\/latex].<\/p>\n<p>Factor [latex]2x^{14}[\/latex] from the binomial:<\/p>\n<p>[latex]2x^{14}(27x^3+1)[\/latex]<\/p>\n<p>We are left with a binomial that can be factored using the sum of cubes. Notice that [latex]27x^3[\/latex] and 1 are cubes because [latex]27x^3=(3x)^3[\/latex] and [latex]1=1^3[\/latex]. Therefore, [latex]27x^3+1=(3x+1)(9x^2-3x+1)[\/latex]<\/p>\n<p>Our final answer must include the original GCF we already factored out: [latex]2x^{14}(3x+1)(9x^2-3x+1)[\/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=\"ohm248369\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=248369&theme=oea&iframe_resize_id=ohm248369&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In our last example, we show why it is important to factor out a GCF, if there is one, before you begin using the techniques shown in this section.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/hMAImz2BuPc?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n","protected":false},"author":264444,"menu_order":12,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-18301","chapter","type-chapter","status-publish","hentry"],"part":18142,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18301","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":20,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18301\/revisions"}],"predecessor-version":[{"id":18707,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18301\/revisions\/18707"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/18142"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18301\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=18301"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=18301"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=18301"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=18301"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}