{"id":1589,"date":"2022-04-15T00:43:08","date_gmt":"2022-04-15T00:43:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=1589"},"modified":"2026-01-06T18:18:30","modified_gmt":"2026-01-06T18:18:30","slug":"3-5-2-factoring-trinomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/3-5-2-factoring-trinomials\/","title":{"raw":"3.5.2: Factoring Trinomials by Trial and Error","rendered":"3.5.2: Factoring Trinomials by Trial and Error"},"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-147\" class=\"standard post-147 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcomes<\/h1>\r\n<ul>\r\n \t<li>Factor a trinomial function using the trial and error method<\/li>\r\n \t<li>Explain \"prime polynomials\"<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Trial and Error Method<\/h2>\r\nSo far we have seen that we can factor polynomial functions by \"pulling out\" the greatest common factor from all terms, or by \"pulling out\" the greatest common factor from pairs of terms using factoring by grouping.\u00a0 In this section we are going to concentrate on factoring trinomials; polynomials with three terms.\r\n\r\nRecall how we used the distributive property to multiply two binomials:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(x+2\\right)\\left(x+3\\right) &amp;= x^2+3x+2x+6\\\\&amp;=x^2+5x+6\\end{aligned}[\/latex]<\/p>\r\nThe product of two binomial generally results in a trinomial. To factor a trinomial of the form [latex]f(x)=x^2+bx+c[\/latex], we want to\u00a0reverse the distributive property and return [latex]x^2+5x+6[\/latex] to [latex]\\left(x+2\\right)\\left(x+3\\right) [\/latex].\r\n\r\nSuppose the polynomial function [latex]f(x)=x^2+bx+c[\/latex] factors into [latex](x+s)(x+t)[\/latex].\r\n\r\nMultiplied out using the distributive property,\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(x+s)(x+t)&amp;=x^2+sx+tx+st\\\\&amp;=x^2+(s+t)x+st\\end{aligned}[\/latex].<\/p>\r\nThis means that [latex]s+t=b[\/latex] and [latex]st=c[\/latex]\r\n\r\nSo, to factor a polynomial function of the form [latex]f(x)=x^2+bx+c[\/latex] we will have to find two numbers [latex]s[\/latex] and [latex]t[\/latex] such that [latex]st=c[\/latex] and [latex]s+t=b[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nTo factor a polynomial function of the form [latex]f(x)=x^2+bx+c[\/latex] we have to find two numbers [latex]s[\/latex] and [latex]t[\/latex] such that [latex]st=c[\/latex] and [latex]s+t=b[\/latex]. Then\u00a0[latex]f(x)=x^2+bx+c=(x+s)(x+t)[\/latex]\r\n\r\nPolynomial functions that do not factor are referred to as\u00a0<em><strong>prime<\/strong><\/em><strong>.<\/strong>\r\n\r\n<\/div>\r\nLet us put this idea into practice.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nFactor [latex]f(x)={x}^{2}+2x - 15[\/latex].\r\n<h4>Solution<\/h4>\r\nWe have a trinomial with leading coefficient [latex]a=1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers, [latex]s, t[\/latex] with a product of [latex]st =-15[\/latex] and a sum of [latex]s+t=2[\/latex].\r\n\r\nIn the table, we list factors of the product \u201315 until we find a pair with the desired sum of 2.\r\n<table style=\"width: 20%;\" 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>Product: [latex]st=-15[\/latex]<\/th>\r\n<th>Sum: [latex](s+t)=2[\/latex]<\/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><span style=\"color: #0000ff;\">[latex]-3,5[\/latex]<\/span><\/td>\r\n<td><span style=\"color: #0000ff;\">[latex]2[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that we have identified [latex]s[\/latex] and [latex]t[\/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\r\nSo, [latex]f(x)=(x-3)(x+5)[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nTo summarize our process, consider the following steps:\r\n<div class=\"textbox shaded\">\r\n<h3>factoring a trinomial function<\/h3>\r\nFor a trinomial in the form [latex]f(x)=x^2+bx+c[\/latex] (i.e.,[latex]a=1[\/latex]),\r\n<ol>\r\n \t<li>List factors of [latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]s[\/latex] and [latex]t[\/latex], whose product is [latex]c[\/latex] and sum [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored expression [latex]\\left(x+s\\right)\\left(x+t\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nFactor [latex]g(x)=x^2+7x+6[\/latex].\r\n\r\n[reveal-answer q=\"hjm397\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm397\"][latex]g(x)=(x+6)(x+1)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nThe next example shows that when [latex]c[\/latex] is negative, either s or t (but not both) will be negative.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nFactor\u00a0[latex]x^{2}+x\u201312[\/latex].\r\n<h4>Solution<\/h4>\r\nConsider all the combinations of numbers whose product is\u00a0[latex]-12[\/latex] and list their sum.\r\n<table style=\"width: 30%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors: [latex]st=\u221212[\/latex]<\/th>\r\n<th>Sum: [latex]s+t=1[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot\u221212=\u221212[\/latex]<\/td>\r\n<td>[latex]1+\u221212=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot\u22126=\u221212[\/latex]<\/td>\r\n<td>[latex]2+\u22126=\u22124[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\\cdot\u22124=\u221212[\/latex]<\/td>\r\n<td>[latex]3+\u22124=\u22121[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><span style=\"color: #0000ff;\">[latex]4\\cdot\u22123=\u221212[\/latex]<\/span><\/td>\r\n<td><span style=\"color: #0000ff;\">[latex]4+\u22123=1[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6\\cdot\u22122=\u221212[\/latex]<\/td>\r\n<td>[latex]6+\u22122=4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]12\\cdot\u22121=\u221212[\/latex]<\/td>\r\n<td>[latex]12+\u22121=11[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nChoose the values whose sum is\u00a0[latex]+1[\/latex]:\u00a0\u00a0[latex]s=4[\/latex] and [latex]t=\u22123[\/latex], and place them into a product of binomials.\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nFactor [latex]h(x)=x^2+6x-16[\/latex].\r\n\r\n[reveal-answer q=\"hjm317\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm317\"][latex]h(x)=(x+8)(x-2)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last example, we will show how to factor a trinomial whose b term is negative.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nFactor [latex]{x}^{2}-7x+6[\/latex].\r\n<h4>Solution<\/h4>\r\nList the factors of\u00a0[latex]6[\/latex]. Note that [latex]b=-7[\/latex], so we will need to consider negative numbers in our list.\r\n<table style=\"width: 20%;\" 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: [latex]st=6[\/latex]<\/th>\r\n<th>Sum: [latex]s+t=-7[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,6[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2, 3[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><span style=\"color: #0000ff;\">[latex]-1, -6[\/latex]<\/span><\/td>\r\n<td><span style=\"color: #0000ff;\">[latex]-7[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2, -3[\/latex]<\/td>\r\n<td>[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nChoose the pair that sum to [latex]-7[\/latex], which is\u00a0[latex]-1, -6[\/latex]\r\n\r\nWrite the pair as constant terms in a product of binomials.\r\n\r\n[latex]\\left(x-1\\right)\\left(x-6\\right)[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\nIn the last example, the [latex]b[\/latex] was negative and [latex]c[\/latex] was positive. This will always mean that if it can be factored, s and t will both be negative.\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nFactor [latex]f(x)=x^2-4x+4[\/latex].\r\n\r\n[reveal-answer q=\"hjm877\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm877\"][latex]f(x)=(x-2)(x-2)=(x-2)^2[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nFactor [latex]g(x)=x^2+3x+4[\/latex].\r\n\r\n[reveal-answer q=\"hjm720\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm720\"][latex]g(x)=x^2+3x+4[\/latex] is prime.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Trinomials with [latex]a\\ne1[\/latex]<\/h3>\r\nWhen the leading coefficient of a trinomial is not equal to 1, factoring by trial and error becomes more difficult as there are more pairs of numbers that could work.\r\n\r\nFor example, suppose we want to factor [latex]f(x)=12x^2-x-20[\/latex]. First we need two numbers that multiply to 12: 1(12); 2(6); 3(4); 4(3); 6(2); 12(1). Then we need two numbers that multiply to \u201320:\u00a0 1(\u201320); \u201320(1); \u20131(20); 20(\u20131); 2(\u201310); \u201310(2); \u20132(10); 10(\u20132); 4(\u20135); \u20135(4); 5(\u20134); \u20134(5). Then from all these possible combinations we need the [latex]x[\/latex]-terms to add up to \u20131. There are 84 different options to try!\u00a0<span style=\"font-size: 1em;\">Let's find a more efficient method.<\/span>\r\n\r\n[reveal-answer q=\"hjm083\"]Try to factor [latex]f(x)=12x^2-x^2-20[\/latex] using trial and error.[\/reveal-answer]\r\n[hidden-answer a=\"hjm083\"][latex]f(x)=(3x-4)(4x+5)[\/latex][\/hidden-answer]\r\n\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-147\" class=\"standard post-147 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h1>Learning Outcomes<\/h1>\n<ul>\n<li>Factor a trinomial function using the trial and error method<\/li>\n<li>Explain &#8220;prime polynomials&#8221;<\/li>\n<\/ul>\n<\/div>\n<h2>The Trial and Error Method<\/h2>\n<p>So far we have seen that we can factor polynomial functions by &#8220;pulling out&#8221; the greatest common factor from all terms, or by &#8220;pulling out&#8221; the greatest common factor from pairs of terms using factoring by grouping.\u00a0 In this section we are going to concentrate on factoring trinomials; polynomials with three terms.<\/p>\n<p>Recall how we used the distributive property to multiply two binomials:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(x+2\\right)\\left(x+3\\right) &= x^2+3x+2x+6\\\\&=x^2+5x+6\\end{aligned}[\/latex]<\/p>\n<p>The product of two binomial generally results in a trinomial. To factor a trinomial of the form [latex]f(x)=x^2+bx+c[\/latex], we want to\u00a0reverse the distributive property and return [latex]x^2+5x+6[\/latex] to [latex]\\left(x+2\\right)\\left(x+3\\right)[\/latex].<\/p>\n<p>Suppose the polynomial function [latex]f(x)=x^2+bx+c[\/latex] factors into [latex](x+s)(x+t)[\/latex].<\/p>\n<p>Multiplied out using the distributive property,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}(x+s)(x+t)&=x^2+sx+tx+st\\\\&=x^2+(s+t)x+st\\end{aligned}[\/latex].<\/p>\n<p>This means that [latex]s+t=b[\/latex] and [latex]st=c[\/latex]<\/p>\n<p>So, to factor a polynomial function of the form [latex]f(x)=x^2+bx+c[\/latex] we will have to find two numbers [latex]s[\/latex] and [latex]t[\/latex] such that [latex]st=c[\/latex] and [latex]s+t=b[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>To factor a polynomial function of the form [latex]f(x)=x^2+bx+c[\/latex] we have to find two numbers [latex]s[\/latex] and [latex]t[\/latex] such that [latex]st=c[\/latex] and [latex]s+t=b[\/latex]. Then\u00a0[latex]f(x)=x^2+bx+c=(x+s)(x+t)[\/latex]<\/p>\n<p>Polynomial functions that do not factor are referred to as\u00a0<em><strong>prime<\/strong><\/em><strong>.<\/strong><\/p>\n<\/div>\n<p>Let us put this idea into practice.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Factor [latex]f(x)={x}^{2}+2x - 15[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>We have a trinomial with leading coefficient [latex]a=1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers, [latex]s, t[\/latex] with a product of [latex]st =-15[\/latex] and a sum of [latex]s+t=2[\/latex].<\/p>\n<p>In the table, we list factors of the product \u201315 until we find a pair with the desired sum of 2.<\/p>\n<table style=\"width: 20%;\" 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>Product: [latex]st=-15[\/latex]<\/th>\n<th>Sum: [latex](s+t)=2[\/latex]<\/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><span style=\"color: #0000ff;\">[latex]-3,5[\/latex]<\/span><\/td>\n<td><span style=\"color: #0000ff;\">[latex]2[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have identified [latex]s[\/latex] and [latex]t[\/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<p>So, [latex]f(x)=(x-3)(x+5)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>To summarize our process, consider the following steps:<\/p>\n<div class=\"textbox shaded\">\n<h3>factoring a trinomial function<\/h3>\n<p>For a trinomial in the form [latex]f(x)=x^2+bx+c[\/latex] (i.e.,[latex]a=1[\/latex]),<\/p>\n<ol>\n<li>List factors of [latex]c[\/latex].<\/li>\n<li>Find [latex]s[\/latex] and [latex]t[\/latex], whose product is [latex]c[\/latex] and sum [latex]b[\/latex].<\/li>\n<li>Write the factored expression [latex]\\left(x+s\\right)\\left(x+t\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Factor [latex]g(x)=x^2+7x+6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm397\">Show Answer<\/span><\/p>\n<div id=\"qhjm397\" class=\"hidden-answer\" style=\"display: none\">[latex]g(x)=(x+6)(x+1)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>The next example shows that when [latex]c[\/latex] is negative, either s or t (but not both) will be negative.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Factor\u00a0[latex]x^{2}+x\u201312[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Consider all the combinations of numbers whose product is\u00a0[latex]-12[\/latex] and list their sum.<\/p>\n<table style=\"width: 30%;\">\n<thead>\n<tr>\n<th>Factors: [latex]st=\u221212[\/latex]<\/th>\n<th>Sum: [latex]s+t=1[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot\u221212=\u221212[\/latex]<\/td>\n<td>[latex]1+\u221212=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot\u22126=\u221212[\/latex]<\/td>\n<td>[latex]2+\u22126=\u22124[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\\cdot\u22124=\u221212[\/latex]<\/td>\n<td>[latex]3+\u22124=\u22121[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #0000ff;\">[latex]4\\cdot\u22123=\u221212[\/latex]<\/span><\/td>\n<td><span style=\"color: #0000ff;\">[latex]4+\u22123=1[\/latex]<\/span><\/td>\n<\/tr>\n<tr>\n<td>[latex]6\\cdot\u22122=\u221212[\/latex]<\/td>\n<td>[latex]6+\u22122=4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]12\\cdot\u22121=\u221212[\/latex]<\/td>\n<td>[latex]12+\u22121=11[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Choose the values whose sum is\u00a0[latex]+1[\/latex]:\u00a0\u00a0[latex]s=4[\/latex] and [latex]t=\u22123[\/latex], and place them into a product of binomials.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Factor [latex]h(x)=x^2+6x-16[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm317\">Show Answer<\/span><\/p>\n<div id=\"qhjm317\" class=\"hidden-answer\" style=\"display: none\">[latex]h(x)=(x+8)(x-2)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>In our last example, we will show how to factor a trinomial whose b term is negative.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Factor [latex]{x}^{2}-7x+6[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>List the factors of\u00a0[latex]6[\/latex]. Note that [latex]b=-7[\/latex], so we will need to consider negative numbers in our list.<\/p>\n<table style=\"width: 20%;\" 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: [latex]st=6[\/latex]<\/th>\n<th>Sum: [latex]s+t=-7[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,6[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2, 3[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><span style=\"color: #0000ff;\">[latex]-1, -6[\/latex]<\/span><\/td>\n<td><span style=\"color: #0000ff;\">[latex]-7[\/latex]<\/span><\/td>\n<\/tr>\n<tr>\n<td>[latex]-2, -3[\/latex]<\/td>\n<td>[latex]-5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Choose the pair that sum to [latex]-7[\/latex], which is\u00a0[latex]-1, -6[\/latex]<\/p>\n<p>Write the pair as constant terms in a product of binomials.<\/p>\n<p>[latex]\\left(x-1\\right)\\left(x-6\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>In the last example, the [latex]b[\/latex] was negative and [latex]c[\/latex] was positive. This will always mean that if it can be factored, s and t will both be negative.<\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Factor [latex]f(x)=x^2-4x+4[\/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)=(x-2)(x-2)=(x-2)^2[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Factor [latex]g(x)=x^2+3x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm720\">Show Answer<\/span><\/p>\n<div id=\"qhjm720\" class=\"hidden-answer\" style=\"display: none\">[latex]g(x)=x^2+3x+4[\/latex] is prime.<\/div>\n<\/div>\n<\/div>\n<h3>Trinomials with [latex]a\\ne1[\/latex]<\/h3>\n<p>When the leading coefficient of a trinomial is not equal to 1, factoring by trial and error becomes more difficult as there are more pairs of numbers that could work.<\/p>\n<p>For example, suppose we want to factor [latex]f(x)=12x^2-x-20[\/latex]. First we need two numbers that multiply to 12: 1(12); 2(6); 3(4); 4(3); 6(2); 12(1). Then we need two numbers that multiply to \u201320:\u00a0 1(\u201320); \u201320(1); \u20131(20); 20(\u20131); 2(\u201310); \u201310(2); \u20132(10); 10(\u20132); 4(\u20135); \u20135(4); 5(\u20134); \u20134(5). Then from all these possible combinations we need the [latex]x[\/latex]-terms to add up to \u20131. There are 84 different options to try!\u00a0<span style=\"font-size: 1em;\">Let&#8217;s find a more efficient method.<\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm083\">Try to factor [latex]f(x)=12x^2-x^2-20[\/latex] using trial and error.<\/span><\/p>\n<div id=\"qhjm083\" class=\"hidden-answer\" style=\"display: none\">[latex]f(x)=(3x-4)(4x+5)[\/latex]<\/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-1589\">\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>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: hjm317; hjm877; hjm397; hjm720 hjm083.  Example 3. Trinomials with au22601.. <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":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Try It: hjm317; hjm877; hjm397; hjm720 hjm083.  Example 3. 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