{"id":18294,"date":"2022-05-18T18:58:05","date_gmt":"2022-05-18T18:58:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/?post_type=chapter&#038;p=18294"},"modified":"2022-05-18T18:58:07","modified_gmt":"2022-05-18T18:58:07","slug":"cr-11-factoring-trinomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/cr-11-factoring-trinomials\/","title":{"raw":"CR.11: Factoring Trinomials","rendered":"CR.11: Factoring Trinomials"},"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 1.<\/li>\r\n \t<li>Factor trinomials by grouping.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Factoring a Trinomial with Leading Coefficient 1<\/h2>\r\nAlthough we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].\r\n\r\nTrinomials of the form [latex]{x}^{2}+bx+c[\/latex] can be factored by finding two numbers with a product of [latex]c[\/latex] and a sum of [latex]b[\/latex]. The trinomial [latex]{x}^{2}+10x+16[\/latex], for example, can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex] because the product of these numbers is [latex]16[\/latex] and their sum is [latex]10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nA trinomial of the form [latex]{x}^{2}+bx+c[\/latex] can be written in factored form as [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex] where [latex]pq=c[\/latex] and [latex]p+q=b[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can every trinomial be factored as a product of binomials?<\/strong>\r\n\r\n<em>No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nFactor [latex]{x}^{2}+2x - 15[\/latex].\r\n\r\n[reveal-answer q=\"88306\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"88306\"]\r\n\r\nWe have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-15[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-15[\/latex]<\/td>\r\n<td>[latex]-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,15[\/latex]<\/td>\r\n<td>[latex]14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-5[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,5[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can check our work by multiplying. Use FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div><\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Does the order of the factors matter?<\/strong>\r\n\r\n<em>No. Multiplication is commutative, so the order of the factors does not matter.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]{x}^{2}-7x+6[\/latex].\r\n\r\n[reveal-answer q=\"823058\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"823058\"]\r\n\r\n[latex]\\left(x - 6\\right)\\left(x - 1\\right)[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]7897[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Factoring by Grouping<\/h2>\r\nTrinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Factoring by Grouping<\/h3>\r\nTo factor a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]{a}\\cdot{c}[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]{a}\\cdot{c}[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\r\n \t<li>Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\r\n \t<li>Pull out the GCF of [latex]qx+c[\/latex].<\/li>\r\n \t<li>Factor out the GCF of the expression.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Trinomial by Grouping<\/h3>\r\nFactor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.\r\n\r\n[reveal-answer q=\"806328\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"806328\"]\r\n\r\nWe have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-30[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-30[\/latex]<\/td>\r\n<td>[latex]-29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,30[\/latex]<\/td>\r\n<td>[latex]29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2,-15[\/latex]<\/td>\r\n<td>[latex]-13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,15[\/latex]<\/td>\r\n<td>[latex]13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-10[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,10[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]p=-3[\/latex] and [latex]q=10[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill &amp; \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill &amp; \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill &amp; \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].\r\n\r\n<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor the following.\r\n<ol>\r\n \t<li>[latex]2{x}^{2}+9x+9[\/latex]<\/li>\r\n \t<li>[latex]6{x}^{2}+x - 1[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"343485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"343485\"]\r\n<ol>\r\n \t<li>[latex]\\left(2x+3\\right)\\left(x+3\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(3x - 1\\right)\\left(2x+1\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n[ohm_question]7908[\/ohm_question]\r\n\r\n[ohm_question]1846[\/ohm_question]\r\n\r\n<\/div>\r\nIn the next video we see another example of how to factor a trinomial by grouping.\r\n\r\n[embed]https:\/\/youtu.be\/agDaQ_cZnNc[\/embed]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor a trinomial with leading coefficient 1.<\/li>\n<li>Factor trinomials by grouping.<\/li>\n<\/ul>\n<\/div>\n<h2>Factoring a Trinomial with Leading Coefficient 1<\/h2>\n<p>Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].<\/p>\n<p>Trinomials of the form [latex]{x}^{2}+bx+c[\/latex] can be factored by finding two numbers with a product of [latex]c[\/latex] and a sum of [latex]b[\/latex]. The trinomial [latex]{x}^{2}+10x+16[\/latex], for example, can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex] because the product of these numbers is [latex]16[\/latex] and their sum is [latex]10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>A trinomial of the form [latex]{x}^{2}+bx+c[\/latex] can be written in factored form as [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex] where [latex]pq=c[\/latex] and [latex]p+q=b[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can every trinomial be factored as a product of binomials?<\/strong><\/p>\n<p><em>No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\n<ol>\n<li>List factors of [latex]c[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>Factor [latex]{x}^{2}+2x - 15[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q88306\">Show Solution<\/span><\/p>\n<div id=\"q88306\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\n<thead>\n<tr>\n<th>Factors of [latex]-15[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-15[\/latex]<\/td>\n<td>[latex]-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,15[\/latex]<\/td>\n<td>[latex]14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,5[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Does the order of the factors matter?<\/strong><\/p>\n<p><em>No. Multiplication is commutative, so the order of the factors does not matter.<\/em><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q823058\">Show Solution<\/span><\/p>\n<div id=\"q823058\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(x - 6\\right)\\left(x - 1\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm7897\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7897&theme=oea&iframe_resize_id=ohm7897&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Factoring by Grouping<\/h2>\n<p>Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Factoring by Grouping<\/h3>\n<p>To factor a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping<\/h3>\n<ol>\n<li>List factors of [latex]{a}\\cdot{c}[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]{a}\\cdot{c}[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\n<li>Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\n<li>Pull out the GCF of [latex]qx+c[\/latex].<\/li>\n<li>Factor out the GCF of the expression.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Trinomial by Grouping<\/h3>\n<p>Factor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q806328\">Show Solution<\/span><\/p>\n<div id=\"q806328\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\n<thead>\n<tr>\n<th>Factors of [latex]-30[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-30[\/latex]<\/td>\n<td>[latex]-29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,30[\/latex]<\/td>\n<td>[latex]29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2,-15[\/latex]<\/td>\n<td>[latex]-13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,15[\/latex]<\/td>\n<td>[latex]13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-10[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,10[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]p=-3[\/latex] and [latex]q=10[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill & \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill & \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill & \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].<\/p>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor the following.<\/p>\n<ol>\n<li>[latex]2{x}^{2}+9x+9[\/latex]<\/li>\n<li>[latex]6{x}^{2}+x - 1[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q343485\">Show Solution<\/span><\/p>\n<div id=\"q343485\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\left(2x+3\\right)\\left(x+3\\right)[\/latex]<\/li>\n<li>[latex]\\left(3x - 1\\right)\\left(2x+1\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm7908\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7908&theme=oea&iframe_resize_id=ohm7908&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm1846\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1846&theme=oea&iframe_resize_id=ohm1846&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the next video we see another example of how to factor a trinomial by grouping.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/agDaQ_cZnNc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n","protected":false},"author":264444,"menu_order":11,"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-18294","chapter","type-chapter","status-publish","hentry"],"part":18142,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18294","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":7,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18294\/revisions"}],"predecessor-version":[{"id":18706,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/18294\/revisions\/18706"}],"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\/18294\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=18294"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=18294"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=18294"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=18294"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}