{"id":1640,"date":"2022-04-18T17:34:33","date_gmt":"2022-04-18T17:34:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=1640"},"modified":"2026-01-06T18:26:42","modified_gmt":"2026-01-06T18:26:42","slug":"3-5-2-2-factor-trinomials-the-synthetic-method","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/3-5-2-2-factor-trinomials-the-synthetic-method\/","title":{"raw":"3.5.3: Factoring Trinomials - The ac-Method","rendered":"3.5.3: Factoring Trinomials &#8211; The ac-Method"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"wrapper\">\r\n<div id=\"wrap\">\r\n<div id=\"content\" role=\"main\">\r\n<div id=\"post-148\" class=\"standard post-148 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcome<\/h1>\r\n<ul>\r\n \t<li>Factor a trinomial using the ac-method<\/li>\r\n<\/ul>\r\n<\/div>\r\nA method that is often used when the coefficient of the leading term (i.e., the coefficient of [latex]x^2[\/latex]) is not 1 is called the <em><strong>ac-method<\/strong><\/em>. This method splits the [latex]x[\/latex]-term into the sum of two terms so that we can use <em><strong>grouping<\/strong><\/em> to factor the function. Determining how to split up the [latex]x[\/latex]-term is done by finding two numbers whose product is [latex]ac[\/latex] and whose sum is [latex]b[\/latex], where [latex]a, b, c[\/latex] are real number coefficients of the function [latex]f(x)=ax^2+bx+c[\/latex].\r\n\r\nSuppose we wish to factor [latex]f(x)=2x^2+5x+3[\/latex]. The first step is to figure out how to write the\u00a0[latex]x[\/latex]-term as the sum of two new terms. We are looking for two numbers, [latex]s, t[\/latex], with a product of [latex]ac=6[\/latex] and a sum of [latex]b=5[\/latex]. We start by listing factors of 6, then look at their sum. A table can be useful to keep everything organized.\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 style=\"text-align: right; width: 158.991px;\">Factors: [latex]st=6[\/latex]<\/th>\r\n<th style=\"text-align: right; width: 147.969px;\">Sum: [latex]s+t=5[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: right; width: 158.991px;\">[latex]1,6[\/latex]<\/td>\r\n<td style=\"text-align: right; width: 147.969px;\">[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: right; width: 158.991px;\">[latex]-1,-6[\/latex]<\/td>\r\n<td style=\"text-align: right; width: 147.969px;\">[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: right; width: 158.991px;\"><span style=\"color: #0000ff;\">[latex]2,3[\/latex]<\/span><\/td>\r\n<td style=\"text-align: right; width: 147.969px;\"><span style=\"color: #0000ff;\">[latex]5[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"text-align: right; width: 158.991px;\">[latex]-2,-3[\/latex]<\/td>\r\n<td style=\"text-align: right; width: 147.969px;\">[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe pair [latex]s=2[\/latex] and [latex]t=3[\/latex] gives the correct\u00a0[latex]x[\/latex]-term of [latex]5x[\/latex], so we will write [latex]5x=2x+3x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=2{x}^{2}+\\color{blue}{5x}+3\\\\&amp;=2x^2+\\color{blue}{2x+3x}+3\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we can group the polynomial into two binomials,<\/p>\r\n<p style=\"text-align: center;\">[latex]f(x)=(2x^2+2x)+(3x+3)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">then factor by grouping:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=(2x^2+2x)+(3x+3)\\\\&amp;=2x(x+1)+3(x+1)\\\\&amp;=(x+1)(2x+3)\\end{aligned}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\" style=\"text-align: center;\">\r\n<h2 style=\"text-align: left;\">Factoring using the ac-method<\/h2>\r\n<p style=\"text-align: left;\">To factor a trinomial function of the form [latex]f(x)=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 we factor out the GCF of the entire expression.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nFactor [latex]f(x)=5{x}^{2}+7x - 6[\/latex].\r\n<h4>Solution<\/h4>\r\nWe have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex].\r\n\r\n[latex]ac=-30[\/latex] so we need to find two numbers [latex]s, t[\/latex], 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 style=\"height: 136px; width: 231px;\" 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.\" width=\"339\">\r\n<thead>\r\n<tr style=\"height: 35px;\">\r\n<th style=\"text-align: right; width: 208.438px; height: 35px;\">Factors: [latex]st=-30[\/latex]<\/th>\r\n<th style=\"text-align: right; width: 184.787px; height: 35px;\">Sum: [latex]s+t=7[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 17px;\">\r\n<th style=\"text-align: right; width: 208.438px; height: 17px;\">[latex]1,-30[\/latex]<\/th>\r\n<th style=\"text-align: right; width: 184.787px; height: 17px;\">[latex]-29[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 17px;\">\r\n<th style=\"text-align: right; width: 208.438px; height: 17px;\">[latex]-1,30[\/latex]<\/th>\r\n<th style=\"text-align: right; width: 184.787px; height: 17px;\">[latex]29[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 17px;\">\r\n<th style=\"text-align: right; width: 208.438px; height: 17px;\">[latex]2,-15[\/latex]<\/th>\r\n<th style=\"text-align: right; width: 184.787px; height: 17px;\">[latex]-13[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 16px;\">\r\n<th style=\"text-align: right; width: 208.438px; height: 16px;\">[latex]-2,15[\/latex]<\/th>\r\n<th style=\"text-align: right; width: 184.787px; height: 16px;\">[latex]13[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 17px;\">\r\n<th style=\"text-align: right; width: 208.438px; height: 17px;\">[latex]3,-10[\/latex]<\/th>\r\n<th style=\"text-align: right; width: 184.787px; height: 17px;\">[latex]-7[\/latex]<\/th>\r\n<\/tr>\r\n<tr style=\"height: 17px;\">\r\n<th style=\"text-align: right; width: 208.438px; height: 17px;\"><span style=\"color: #0000ff;\">[latex]-3,10[\/latex]<\/span><\/th>\r\n<th style=\"text-align: right; width: 184.787px; height: 17px;\"><span style=\"color: #0000ff;\">[latex]7[\/latex]<\/span><\/th>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]s=-3[\/latex] and [latex]t=10[\/latex].\r\n\r\nNow write [latex]7x=-3x+10x[\/latex] in the function and factor by grouping:\r\n\r\n[latex]\\begin{aligned}f(x)&amp;=5{x}^{2}+7x - 6\\\\&amp;=5{x}^{2}-3x+10x - 6 \\hfill &amp; \\text{Write the original trinomial as }a{x}^{2}+sx+tx+c\\hfill \\\\ &amp;=x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill &amp; \\text{Factor out the GCF of each pair}\\hfill \\\\ &amp;=\\left(5x - 3\\right)\\left(x+2\\right)\\hfill &amp; \\text{Factor out the GCF}\\text{ }\\text{ of the expression}\\hfill \\end{aligned}[\/latex]\r\n\r\n&nbsp;\r\n\r\nWe can check our work by multiplying. Use the distributive property 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>\r\n<\/div>\r\n<div>\r\n\r\nIn this example, when we wrote\u00a0[latex]7x=-3x+10x[\/latex], we can just as easily write [latex]7x=10x-3x[\/latex]. We will end up with the same factorization:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=5{x}^{2}+7x - 6\\\\&amp;=5{x}^{2}+10x-3x - 6 \\\\ &amp;=5x\\left(x + 2\\right)-3\\left(x + 2\\right)\\\\ &amp;=\\left(x+2\\right)\\left(5x - 3\\right)\\end{aligned}[\/latex]<\/p>\r\nThe <em><strong>commutative property of multiplication<\/strong><\/em> says [latex]ab=ba[\/latex], so we have the same result as before.\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">We can summarize our process in the following way:<\/span>\r\n<div class=\"textbox\">\r\n<h3>factoring a trinomial function by the ac-method<\/h3>\r\nTo factor a trinomial function of the form [latex]f(x)=a{x}^{2}+bx+c[\/latex],\r\n<ol>\r\n \t<li>List factors of [latex]ac[\/latex].<\/li>\r\n \t<li>Find [latex]s[\/latex] and [latex]t[\/latex], factors of [latex]ac[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Write the original expression as [latex]a{x}^{2}+sx+tx+c[\/latex].<\/li>\r\n \t<li>Factor by grouping.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nFactoring trinomials whose leading coefficient is not\u00a0[latex]1[\/latex] becomes quick and kind of fun once you get the idea.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nFactor\u00a0[latex]g(x)=2{x}^{2}+9x+9[\/latex].\r\n<h4>Solution<\/h4>\r\nFind two numbers, [latex]s, t[\/latex], such that [latex]st=18[\/latex] and [latex]s + t = 9[\/latex].\r\n\r\n[latex]9[\/latex] and [latex]18[\/latex] are both positive, so we will only consider positive factors.\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]st=18[\/latex]<\/th>\r\n<th>[latex]s+t=9[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1, 18[\/latex]<\/td>\r\n<td>[latex]19[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><span style=\"color: #0000ff;\">[latex]3,6[\/latex]<\/span><\/td>\r\n<td><span style=\"color: #0000ff;\">[latex]9[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can stop because we have found our factors.\r\n\r\nWrite the [latex]9x[\/latex] as [latex]3x+6x[\/latex], then factor by grouping:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}g(x)&amp;=2x^2+\\color{blue}{9x}+9\\\\&amp;=2x^2+\\color{blue}{3x+6x}+9\\\\&amp;=(2x^2+3x)+(6x+9)\\\\&amp;=x(2x+3)+3(2x+3)\\\\&amp;=(2x+3)(x+3)\\end{aligned}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n[latex]g(x)=(x+3)(2x+3)[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nFactor [latex]f(x)=6{x}^{2}+x - 1[\/latex].\r\n\r\n[reveal-answer q=\"hjm108\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm108\"][latex]f(x)=(2x+1)(3x-1)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"font-size: 1rem; text-align: initial; background-color: #ffffff;\">If the leading term is negative, we can pull out \u20131 as a GCF before we factor the trinomial.<\/span>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nFactor [latex]f(x)=-4x^2+8x+21[\/latex].\r\n<h4>Solution<\/h4>\r\nStart by pulling \u20131 as a GCF. This will change all of the signs of the coefficients:\u00a0\u00a0[latex]f(x)=-1(4x^2-8x-21)[\/latex]\r\n\r\nNow, [latex]ac=4(-21)=-84[\/latex] so we need two numbers, [latex]s, t[\/latex], whose product is \u201384 and whose sum is \u20138.\r\n<table style=\"border-collapse: collapse; width: 48.8842%; height: 89px;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 50%;\">Product<\/th>\r\n<th style=\"width: 50%;\">Sum<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">[latex]st=-84[\/latex]<\/td>\r\n<td style=\"width: 50%;\">[latex]s+t=-8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">[latex]-12, 7[\/latex]<\/td>\r\n<td style=\"width: 50%;\">[latex]-12+7=-5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">[latex]-28, 3[\/latex]<\/td>\r\n<td style=\"width: 50%;\">[latex]-28+3=-25[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\"><span style=\"color: #0000ff;\">[latex]-14, 6[\/latex]<\/span><\/td>\r\n<td style=\"width: 50%;\"><span style=\"color: #0000ff;\">[latex]-14+6=-8[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWrite [latex]8x[\/latex] as [latex]-14x+6x[\/latex] then factor by grouping:\r\n\r\n[latex]\\begin{aligned}f(x)&amp;=-4x^2+8x+21\\\\&amp;=-1(4x^2-8x-21)\\\\&amp;=-1\\left (4x^2-14x+6x-21\\right )\\\\&amp;=-1\\left [2x(2x-7)+3(2x-7)\\right ]\\\\&amp;=-1(2x-7)(2x+3)\\end{aligned}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nFactor [latex]f(x)=-6x^2-7x+20[\/latex].\r\n\r\n[reveal-answer q=\"hjm583\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm583\"][latex]f(x)=-1(3x-4)(2x+5)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nSometimes we encounter polynomials that, despite our best efforts, cannot be factored into the product of two binomials. Such polynomials are said to be\u00a0<em><strong>prime<\/strong><\/em>.\r\n<div class=\"textbox examples\" style=\"text-align: left;\">\r\n<h3>Example 4<\/h3>\r\nFactor [latex]f(x)=7x^{2}-16x\u20135[\/latex].\r\n<h4>Solution<\/h4>\r\n[latex]ac=7(-5)=-35[\/latex], [latex]b=-16[\/latex]. We need to find two numbers [latex]s, t[\/latex], where [latex]st=-35[\/latex] and [latex]s+t=-16[\/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: [latex]st=-35[\/latex]<\/th>\r\n<th>Sum: [latex]s+t=-16[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]-1, 35[\/latex]<\/td>\r\n<td>[latex]34[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1, -35[\/latex]<\/td>\r\n<td>[latex]-34[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-5, 7[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-7,5[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNone of the factors add up to [latex]-16[\/latex] so the trinomial cannot be factored.\r\n\r\n[latex]f(x)=7x^{2}-16x\u20135[\/latex] is prime.\r\n\r\n<\/div>\r\n<!-- .entry-content -->\r\n\r\n<!-- #post-## -->\r\n\r\n<!-- CITATIONS AND ATTRIBUTIONS -->\r\n\r\n<section class=\"citations-section\" role=\"contentinfo\"><\/section>Before we jump head first into factoring a trinomial, we should always look to see if there is a GCF that can be \"pulled out\".\r\n<div class=\"textbox examples\">\r\n<h3>Example 5<\/h3>\r\nFactor [latex]f(x)=6x^3+21x^2-27x[\/latex].\r\n<h4>Solution<\/h4>\r\nThe terms in the function have a GCF of 3x, so we will pull out the GCF first:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=6x^3+21x^2-27x\\\\&amp;=3x(2x^2+7x-9)\\end{aligned}[\/latex]<\/p>\r\nTo factor the trinomial, [latex]ac=2(-9)=-18[\/latex] and [latex]b=7[\/latex] so, we need two numbers that multiply to -18 and add to 7:\u00a0 9 and \u20132.\r\n\r\nWrite [latex]7x=9x-2x[\/latex] then factor by grouping:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=3x(2x^2+7x-9)\\\\&amp;=3x(2x^2+9x-2x-9)\\\\&amp;=3x\\left [x(2x+9)-1(2x+9)\\right ]\\\\&amp;=3x(2x+9)(x-1)\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nFactor [latex]f(x)=-6x^3+14x^2+40x[\/latex]\r\n\r\n[reveal-answer q=\"hjm877\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm877\"][latex]f(x)=-2x(x-4)(3x+5)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nRemember to always look for a GCF before jumping in to factoring any trinomial.\r\n<div class=\"textbox examples\">\r\n<h3>Example 6<\/h3>\r\nFactor [latex]f(x)=72x^3+168x^2-144x[\/latex].\r\n<h4>Solution<\/h4>\r\nFirst notice that [latex]x[\/latex] is common to all terms in the trinomial. Also 72, 168, and 144 have at least 2 in common. Let's factor them to find the GCF:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}72 &amp;= 2^3\\cdot 3^2\\\\168&amp;=2^3\\cdot 3\\cdot 7\\\\144&amp;=2^4\\cdot 3^2\\\\\\hline GCF&amp;=2^3\\cdot 3\\\\&amp;=24\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Factor out the GCF of [latex]24x[\/latex] from the trinomial:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=72x^3+168x^2-144x\\\\&amp;=24x(3x^2+7x-6)\\end{aligned}[\/latex]<\/p>\r\nNow use the [latex]ac[\/latex]-method:\r\n\r\n[latex]ac=-18[\/latex] and [latex]b=+7[\/latex]:\u00a0 [latex]9\\cdot -2=-18[\/latex] and [latex]9-2=7[\/latex]\r\n\r\nWrite [latex]7x=9x-2x[\/latex] then factor by grouping:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&amp;=24x(3x^2+7x-6)\\\\&amp;=24x(3x^2+9x-2x-6)\\\\&amp;=24x[3x(x+3)-2(x+3)]\\\\&amp;=24x(x+3)(3x-2)\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nFactor [latex]g(x)=140x^3-161x^2+42x[\/latex]\r\n\r\n[reveal-answer q=\"hjm647\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm647\"][latex]g(x)=7x(5x-2)(4x-3)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"wrapper\">\n<div id=\"wrap\">\n<div id=\"content\" role=\"main\">\n<div id=\"post-148\" class=\"standard post-148 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h1>Learning Outcome<\/h1>\n<ul>\n<li>Factor a trinomial using the ac-method<\/li>\n<\/ul>\n<\/div>\n<p>A method that is often used when the coefficient of the leading term (i.e., the coefficient of [latex]x^2[\/latex]) is not 1 is called the <em><strong>ac-method<\/strong><\/em>. This method splits the [latex]x[\/latex]-term into the sum of two terms so that we can use <em><strong>grouping<\/strong><\/em> to factor the function. Determining how to split up the [latex]x[\/latex]-term is done by finding two numbers whose product is [latex]ac[\/latex] and whose sum is [latex]b[\/latex], where [latex]a, b, c[\/latex] are real number coefficients of the function [latex]f(x)=ax^2+bx+c[\/latex].<\/p>\n<p>Suppose we wish to factor [latex]f(x)=2x^2+5x+3[\/latex]. The first step is to figure out how to write the\u00a0[latex]x[\/latex]-term as the sum of two new terms. We are looking for two numbers, [latex]s, t[\/latex], with a product of [latex]ac=6[\/latex] and a sum of [latex]b=5[\/latex]. We start by listing factors of 6, then look at their sum. A table can be useful to keep everything organized.<\/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 style=\"text-align: right; width: 158.991px;\">Factors: [latex]st=6[\/latex]<\/th>\n<th style=\"text-align: right; width: 147.969px;\">Sum: [latex]s+t=5[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"text-align: right; width: 158.991px;\">[latex]1,6[\/latex]<\/td>\n<td style=\"text-align: right; width: 147.969px;\">[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: right; width: 158.991px;\">[latex]-1,-6[\/latex]<\/td>\n<td style=\"text-align: right; width: 147.969px;\">[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: right; width: 158.991px;\"><span style=\"color: #0000ff;\">[latex]2,3[\/latex]<\/span><\/td>\n<td style=\"text-align: right; width: 147.969px;\"><span style=\"color: #0000ff;\">[latex]5[\/latex]<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: right; width: 158.991px;\">[latex]-2,-3[\/latex]<\/td>\n<td style=\"text-align: right; width: 147.969px;\">[latex]-5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The pair [latex]s=2[\/latex] and [latex]t=3[\/latex] gives the correct\u00a0[latex]x[\/latex]-term of [latex]5x[\/latex], so we will write [latex]5x=2x+3x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=2{x}^{2}+\\color{blue}{5x}+3\\\\&=2x^2+\\color{blue}{2x+3x}+3\\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we can group the polynomial into two binomials,<\/p>\n<p style=\"text-align: center;\">[latex]f(x)=(2x^2+2x)+(3x+3)[\/latex]<\/p>\n<p style=\"text-align: left;\">then factor by grouping:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=(2x^2+2x)+(3x+3)\\\\&=2x(x+1)+3(x+1)\\\\&=(x+1)(2x+3)\\end{aligned}[\/latex]<\/p>\n<div class=\"textbox\" style=\"text-align: center;\">\n<h2 style=\"text-align: left;\">Factoring using the ac-method<\/h2>\n<p style=\"text-align: left;\">To factor a trinomial function of the form [latex]f(x)=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 we factor out the GCF of the entire expression.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Factor [latex]f(x)=5{x}^{2}+7x - 6[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>We have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex].<\/p>\n<p>[latex]ac=-30[\/latex] so we need to find two numbers [latex]s, t[\/latex], 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 style=\"height: 136px; width: 231px; width: 339px;\" 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 style=\"height: 35px;\">\n<th style=\"text-align: right; width: 208.438px; height: 35px;\">Factors: [latex]st=-30[\/latex]<\/th>\n<th style=\"text-align: right; width: 184.787px; height: 35px;\">Sum: [latex]s+t=7[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 17px;\">\n<th style=\"text-align: right; width: 208.438px; height: 17px;\">[latex]1,-30[\/latex]<\/th>\n<th style=\"text-align: right; width: 184.787px; height: 17px;\">[latex]-29[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 17px;\">\n<th style=\"text-align: right; width: 208.438px; height: 17px;\">[latex]-1,30[\/latex]<\/th>\n<th style=\"text-align: right; width: 184.787px; height: 17px;\">[latex]29[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 17px;\">\n<th style=\"text-align: right; width: 208.438px; height: 17px;\">[latex]2,-15[\/latex]<\/th>\n<th style=\"text-align: right; width: 184.787px; height: 17px;\">[latex]-13[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 16px;\">\n<th style=\"text-align: right; width: 208.438px; height: 16px;\">[latex]-2,15[\/latex]<\/th>\n<th style=\"text-align: right; width: 184.787px; height: 16px;\">[latex]13[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 17px;\">\n<th style=\"text-align: right; width: 208.438px; height: 17px;\">[latex]3,-10[\/latex]<\/th>\n<th style=\"text-align: right; width: 184.787px; height: 17px;\">[latex]-7[\/latex]<\/th>\n<\/tr>\n<tr style=\"height: 17px;\">\n<th style=\"text-align: right; width: 208.438px; height: 17px;\"><span style=\"color: #0000ff;\">[latex]-3,10[\/latex]<\/span><\/th>\n<th style=\"text-align: right; width: 184.787px; height: 17px;\"><span style=\"color: #0000ff;\">[latex]7[\/latex]<\/span><\/th>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]s=-3[\/latex] and [latex]t=10[\/latex].<\/p>\n<p>Now write [latex]7x=-3x+10x[\/latex] in the function and factor by grouping:<\/p>\n<p>[latex]\\begin{aligned}f(x)&=5{x}^{2}+7x - 6\\\\&=5{x}^{2}-3x+10x - 6 \\hfill & \\text{Write the original trinomial as }a{x}^{2}+sx+tx+c\\hfill \\\\ &=x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill & \\text{Factor out the GCF of each pair}\\hfill \\\\ &=\\left(5x - 3\\right)\\left(x+2\\right)\\hfill & \\text{Factor out the GCF}\\text{ }\\text{ of the expression}\\hfill \\end{aligned}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>We can check our work by multiplying. Use the distributive property to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<p>In this example, when we wrote\u00a0[latex]7x=-3x+10x[\/latex], we can just as easily write [latex]7x=10x-3x[\/latex]. We will end up with the same factorization:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=5{x}^{2}+7x - 6\\\\&=5{x}^{2}+10x-3x - 6 \\\\ &=5x\\left(x + 2\\right)-3\\left(x + 2\\right)\\\\ &=\\left(x+2\\right)\\left(5x - 3\\right)\\end{aligned}[\/latex]<\/p>\n<p>The <em><strong>commutative property of multiplication<\/strong><\/em> says [latex]ab=ba[\/latex], so we have the same result as before.<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">We can summarize our process in the following way:<\/span><\/p>\n<div class=\"textbox\">\n<h3>factoring a trinomial function by the ac-method<\/h3>\n<p>To factor a trinomial function of the form [latex]f(x)=a{x}^{2}+bx+c[\/latex],<\/p>\n<ol>\n<li>List factors of [latex]ac[\/latex].<\/li>\n<li>Find [latex]s[\/latex] and [latex]t[\/latex], factors of [latex]ac[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Write the original expression as [latex]a{x}^{2}+sx+tx+c[\/latex].<\/li>\n<li>Factor by grouping.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>Factoring trinomials whose leading coefficient is not\u00a0[latex]1[\/latex] becomes quick and kind of fun once you get the idea.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Factor\u00a0[latex]g(x)=2{x}^{2}+9x+9[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Find two numbers, [latex]s, t[\/latex], such that [latex]st=18[\/latex] and [latex]s + t = 9[\/latex].<\/p>\n<p>[latex]9[\/latex] and [latex]18[\/latex] are both positive, so we will only consider positive factors.<\/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]st=18[\/latex]<\/th>\n<th>[latex]s+t=9[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1, 18[\/latex]<\/td>\n<td>[latex]19[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #0000ff;\">[latex]3,6[\/latex]<\/span><\/td>\n<td><span style=\"color: #0000ff;\">[latex]9[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can stop because we have found our factors.<\/p>\n<p>Write the [latex]9x[\/latex] as [latex]3x+6x[\/latex], then factor by grouping:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}g(x)&=2x^2+\\color{blue}{9x}+9\\\\&=2x^2+\\color{blue}{3x+6x}+9\\\\&=(2x^2+3x)+(6x+9)\\\\&=x(2x+3)+3(2x+3)\\\\&=(2x+3)(x+3)\\end{aligned}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>[latex]g(x)=(x+3)(2x+3)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Factor [latex]f(x)=6{x}^{2}+x - 1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm108\">Show Answer<\/span><\/p>\n<div id=\"qhjm108\" class=\"hidden-answer\" style=\"display: none\">[latex]f(x)=(2x+1)(3x-1)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p><span style=\"font-size: 1rem; text-align: initial; background-color: #ffffff;\">If the leading term is negative, we can pull out \u20131 as a GCF before we factor the trinomial.<\/span><\/p>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Factor [latex]f(x)=-4x^2+8x+21[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Start by pulling \u20131 as a GCF. This will change all of the signs of the coefficients:\u00a0\u00a0[latex]f(x)=-1(4x^2-8x-21)[\/latex]<\/p>\n<p>Now, [latex]ac=4(-21)=-84[\/latex] so we need two numbers, [latex]s, t[\/latex], whose product is \u201384 and whose sum is \u20138.<\/p>\n<table style=\"border-collapse: collapse; width: 48.8842%; height: 89px;\">\n<tbody>\n<tr>\n<th style=\"width: 50%;\">Product<\/th>\n<th style=\"width: 50%;\">Sum<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">[latex]st=-84[\/latex]<\/td>\n<td style=\"width: 50%;\">[latex]s+t=-8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">[latex]-12, 7[\/latex]<\/td>\n<td style=\"width: 50%;\">[latex]-12+7=-5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">[latex]-28, 3[\/latex]<\/td>\n<td style=\"width: 50%;\">[latex]-28+3=-25[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\"><span style=\"color: #0000ff;\">[latex]-14, 6[\/latex]<\/span><\/td>\n<td style=\"width: 50%;\"><span style=\"color: #0000ff;\">[latex]-14+6=-8[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Write [latex]8x[\/latex] as [latex]-14x+6x[\/latex] then factor by grouping:<\/p>\n<p>[latex]\\begin{aligned}f(x)&=-4x^2+8x+21\\\\&=-1(4x^2-8x-21)\\\\&=-1\\left (4x^2-14x+6x-21\\right )\\\\&=-1\\left [2x(2x-7)+3(2x-7)\\right ]\\\\&=-1(2x-7)(2x+3)\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Factor [latex]f(x)=-6x^2-7x+20[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm583\">Show Answer<\/span><\/p>\n<div id=\"qhjm583\" class=\"hidden-answer\" style=\"display: none\">[latex]f(x)=-1(3x-4)(2x+5)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Sometimes we encounter polynomials that, despite our best efforts, cannot be factored into the product of two binomials. Such polynomials are said to be\u00a0<em><strong>prime<\/strong><\/em>.<\/p>\n<div class=\"textbox examples\" style=\"text-align: left;\">\n<h3>Example 4<\/h3>\n<p>Factor [latex]f(x)=7x^{2}-16x\u20135[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>[latex]ac=7(-5)=-35[\/latex], [latex]b=-16[\/latex]. We need to find two numbers [latex]s, t[\/latex], where [latex]st=-35[\/latex] and [latex]s+t=-16[\/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: [latex]st=-35[\/latex]<\/th>\n<th>Sum: [latex]s+t=-16[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]-1, 35[\/latex]<\/td>\n<td>[latex]34[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1, -35[\/latex]<\/td>\n<td>[latex]-34[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-5, 7[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-7,5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>None of the factors add up to [latex]-16[\/latex] so the trinomial cannot be factored.<\/p>\n<p>[latex]f(x)=7x^{2}-16x\u20135[\/latex] is prime.<\/p>\n<\/div>\n<p><!-- .entry-content --><\/p>\n<p><!-- #post-## --><\/p>\n<p><!-- CITATIONS AND ATTRIBUTIONS --><\/p>\n<section class=\"citations-section\" role=\"contentinfo\"><\/section>\n<p>Before we jump head first into factoring a trinomial, we should always look to see if there is a GCF that can be &#8220;pulled out&#8221;.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Factor [latex]f(x)=6x^3+21x^2-27x[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>The terms in the function have a GCF of 3x, so we will pull out the GCF first:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=6x^3+21x^2-27x\\\\&=3x(2x^2+7x-9)\\end{aligned}[\/latex]<\/p>\n<p>To factor the trinomial, [latex]ac=2(-9)=-18[\/latex] and [latex]b=7[\/latex] so, we need two numbers that multiply to -18 and add to 7:\u00a0 9 and \u20132.<\/p>\n<p>Write [latex]7x=9x-2x[\/latex] then factor by grouping:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=3x(2x^2+7x-9)\\\\&=3x(2x^2+9x-2x-9)\\\\&=3x\\left [x(2x+9)-1(2x+9)\\right ]\\\\&=3x(2x+9)(x-1)\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Factor [latex]f(x)=-6x^3+14x^2+40x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm877\">Show Answer<\/span><\/p>\n<div id=\"qhjm877\" class=\"hidden-answer\" style=\"display: none\">[latex]f(x)=-2x(x-4)(3x+5)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Remember to always look for a GCF before jumping in to factoring any trinomial.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 6<\/h3>\n<p>Factor [latex]f(x)=72x^3+168x^2-144x[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>First notice that [latex]x[\/latex] is common to all terms in the trinomial. Also 72, 168, and 144 have at least 2 in common. Let&#8217;s factor them to find the GCF:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}72 &= 2^3\\cdot 3^2\\\\168&=2^3\\cdot 3\\cdot 7\\\\144&=2^4\\cdot 3^2\\\\\\hline GCF&=2^3\\cdot 3\\\\&=24\\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: left;\">Factor out the GCF of [latex]24x[\/latex] from the trinomial:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=72x^3+168x^2-144x\\\\&=24x(3x^2+7x-6)\\end{aligned}[\/latex]<\/p>\n<p>Now use the [latex]ac[\/latex]-method:<\/p>\n<p>[latex]ac=-18[\/latex] and [latex]b=+7[\/latex]:\u00a0 [latex]9\\cdot -2=-18[\/latex] and [latex]9-2=7[\/latex]<\/p>\n<p>Write [latex]7x=9x-2x[\/latex] then factor by grouping:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}f(x)&=24x(3x^2+7x-6)\\\\&=24x(3x^2+9x-2x-6)\\\\&=24x[3x(x+3)-2(x+3)]\\\\&=24x(x+3)(3x-2)\\end{aligned}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Factor [latex]g(x)=140x^3-161x^2+42x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm647\">Show Answer<\/span><\/p>\n<div id=\"qhjm647\" class=\"hidden-answer\" style=\"display: none\">[latex]g(x)=7x(5x-2)(4x-3)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1640\">\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>Revision and Adaptation. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Try It: hjm583; hjm108; hjm877; hjm647 Examples 5 and 6. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":422608,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Try It: hjm583; 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