{"id":2718,"date":"2016-07-18T20:17:05","date_gmt":"2016-07-18T20:17:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2718"},"modified":"2019-08-06T18:56:06","modified_gmt":"2019-08-06T18:56:06","slug":"read-factor-a-trinomial-with-leading-coefficient-1","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-factor-a-trinomial-with-leading-coefficient-1\/","title":{"raw":"Factor a Trinomial with Leading Coefficient = 1","rendered":"Factor a Trinomial with Leading Coefficient = 1"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor a trinomial with leading coefficient [latex]= 1[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nTrinomials are polynomials with three terms. We are going to show you a method for factoring a trinomial whose leading coefficient is\u00a0[latex]1[\/latex]. \u00a0Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that trinomials\u00a0can be factored. The trinomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of\u00a0[latex]1[\/latex], but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].\r\n\r\nRecall how to use the distributive property to multiply two binomials:\r\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x+3\\right) = x^2+3x+2x+6=x^2+5x+6[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We can reverse the distributive property and return [latex]x^2+5x+6\\text{ to }\\left(x+2\\right)\\left(x+3\\right) [\/latex]\u00a0by finding two numbers with a product of [latex]6[\/latex] and a sum of [latex]5[\/latex].<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nIn general, for a trinomial of the form [latex]{x}^{2}+bx+c[\/latex], you can factor a trinomial with leading coefficient\u00a0[latex]1[\/latex] by finding two numbers, [latex]p[\/latex] and [latex]q[\/latex]\u00a0whose product is c and whose sum is b.\r\n\r\n<\/div>\r\nLet us put this idea to practice with the following example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]{x}^{2}+2x - 15[\/latex].\r\n[reveal-answer q=\"44696\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44696\"]\r\n\r\nWe have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table 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 of [latex]-15[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-15[\/latex]<\/td>\r\n<td>[latex]-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,15[\/latex]<\/td>\r\n<td>[latex]14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-5[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,5[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we present two more examples of factoring a trinomial with a leading coefficient of 1.\r\n\r\nhttps:\/\/youtu.be\/-SVBVVYVNTM\r\n\r\nTo summarize our process, consider the following steps:\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\r\n<\/ol>\r\nWe will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.\r\n\r\n<\/div>\r\nIn our next example, we show that when c is negative, either p or q will be negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]x^{2}+x\u201312[\/latex].\r\n\r\n[reveal-answer q=\"205737\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"205737\"]\r\n\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 whose product is [latex]\u221212[\/latex]<\/th>\r\n<th>Sum of the factors<\/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>[latex]4\\cdot\u22123=\u221212[\/latex]<\/td>\r\n<td>[latex]4+\u22123=1[\/latex]<\/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]p=4[\/latex] and [latex]q=\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[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nWhich property of multiplication can be used to describe why\u00a0[latex]\\left(x+4\\right)\\left(x-3\\right) =\\left(x-3\\right)\\left(x+4\\right)[\/latex]. Use the textbox below to write down your ideas before you look at the answer.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n[reveal-answer q=\"177955\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"177955\"]\r\n\r\nThe <strong>commutative property of multiplication<\/strong> states that factors may be multiplied in any order without affecting the product.\r\n<div class=\"bcc-box bcc-success\">\r\n<div style=\"text-align: center;\">[latex]a\\cdot b=b\\cdot a[\/latex]<\/div>\r\n<\/div>\r\n[\/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 exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]{x}^{2}-7x+6[\/latex].\r\n[reveal-answer q=\"662468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"662468\"]\r\n\r\nList the factors of\u00a0[latex]6[\/latex]. Note that the b term is negative, 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 of [latex]6[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,6[\/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>[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<\/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[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the last example, the b\u00a0term was negative and the c term was positive. This will always mean that if it can be factored, p and q\u00a0will both be negative.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nCan every trinomial be factored as a product of binomials?\r\n\r\nMathematicians often use a counterexample to prove\u00a0or disprove a question. A counterexample means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient\u00a0[latex]1[\/latex] that\u00a0<em>cannot\u00a0<\/em>be factored as a product of binomials?\r\n\r\nUse the textbox below to write your ideas.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n[reveal-answer q=\"776075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"776075\"]\r\n\r\nCan every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime.\r\n\r\nA counterexample would be: [latex]x^2+3x+7[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor a trinomial with leading coefficient [latex]= 1[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>Trinomials are polynomials with three terms. We are going to show you a method for factoring a trinomial whose leading coefficient is\u00a0[latex]1[\/latex]. \u00a0Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that trinomials\u00a0can be factored. The trinomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of\u00a0[latex]1[\/latex], but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].<\/p>\n<p>Recall how to use the distributive property to multiply two binomials:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x+3\\right) = x^2+3x+2x+6=x^2+5x+6[\/latex]<\/p>\n<p style=\"text-align: left;\">We can reverse the distributive property and return [latex]x^2+5x+6\\text{ to }\\left(x+2\\right)\\left(x+3\\right)[\/latex]\u00a0by finding two numbers with a product of [latex]6[\/latex] and a sum of [latex]5[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>In general, for a trinomial of the form [latex]{x}^{2}+bx+c[\/latex], you can factor a trinomial with leading coefficient\u00a0[latex]1[\/latex] by finding two numbers, [latex]p[\/latex] and [latex]q[\/latex]\u00a0whose product is c and whose sum is b.<\/p>\n<\/div>\n<p>Let us put this idea to practice with the following example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]{x}^{2}+2x - 15[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44696\">Show Solution<\/span><\/p>\n<div id=\"q44696\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table 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 of [latex]-15[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-15[\/latex]<\/td>\n<td>[latex]-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,15[\/latex]<\/td>\n<td>[latex]14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,5[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we present two more examples of factoring a trinomial with a leading coefficient of 1.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-SVBVVYVNTM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>To summarize our process, consider the following steps:<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\n<ol>\n<li>List factors of [latex]c[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\n<\/ol>\n<p>We will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.<\/p>\n<\/div>\n<p>In our next example, we show that when c is negative, either p or q will be negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]x^{2}+x\u201312[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q205737\">Show Solution<\/span><\/p>\n<div id=\"q205737\" class=\"hidden-answer\" style=\"display: none\">\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 whose product is [latex]\u221212[\/latex]<\/th>\n<th>Sum of the factors<\/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>[latex]4\\cdot\u22123=\u221212[\/latex]<\/td>\n<td>[latex]4+\u22123=1[\/latex]<\/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]p=4[\/latex] and [latex]q=\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 key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Which property of multiplication can be used to describe why\u00a0[latex]\\left(x+4\\right)\\left(x-3\\right) =\\left(x-3\\right)\\left(x+4\\right)[\/latex]. Use the textbox below to write down your ideas before you look at the answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"2\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q177955\">Show Solution<\/span><\/p>\n<div id=\"q177955\" class=\"hidden-answer\" style=\"display: none\">\n<p>The <strong>commutative property of multiplication<\/strong> states that factors may be multiplied in any order without affecting the product.<\/p>\n<div class=\"bcc-box bcc-success\">\n<div style=\"text-align: center;\">[latex]a\\cdot b=b\\cdot a[\/latex]<\/div>\n<\/div>\n<\/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 exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]{x}^{2}-7x+6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q662468\">Show Solution<\/span><\/p>\n<div id=\"q662468\" class=\"hidden-answer\" style=\"display: none\">\n<p>List the factors of\u00a0[latex]6[\/latex]. Note that the b term is negative, 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 of [latex]6[\/latex]<\/th>\n<th>Sum of Factors<\/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>[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<\/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<\/div>\n<p>In the last example, the b\u00a0term was negative and the c term was positive. This will always mean that if it can be factored, p and q\u00a0will both be negative.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Can every trinomial be factored as a product of binomials?<\/p>\n<p>Mathematicians often use a counterexample to prove\u00a0or disprove a question. A counterexample means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient\u00a0[latex]1[\/latex] that\u00a0<em>cannot\u00a0<\/em>be factored as a product of binomials?<\/p>\n<p>Use the textbox below to write your ideas.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"2\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q776075\">Show Solution<\/span><\/p>\n<div id=\"q776075\" class=\"hidden-answer\" style=\"display: none\">\n<p>Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/p>\n<p>A counterexample would be: [latex]x^2+3x+7[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2718\">\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>Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/-SVBVVYVNTM\">https:\/\/youtu.be\/-SVBVVYVNTM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>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":21,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/-SVBVVYVNTM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 12: Factoring, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"07ef4bb0-949b-44b6-acd3-2ce58a0a9c83","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2718","chapter","type-chapter","status-publish","hentry"],"part":146,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2718","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":22,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2718\/revisions"}],"predecessor-version":[{"id":5599,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2718\/revisions\/5599"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/parts\/146"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2718\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/media?parent=2718"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2718"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/contributor?post=2718"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/license?post=2718"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}