{"id":2576,"date":"2025-08-13T18:06:51","date_gmt":"2025-08-13T18:06:51","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2576"},"modified":"2026-01-13T17:44:07","modified_gmt":"2026-01-13T17:44:07","slug":"polynomial-equations-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polynomial-equations-background-youll-need-2\/","title":{"raw":"Polynomial Equations: Background You'll Need 2","rendered":"Polynomial Equations: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Factor trinomials where a is not 1<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Factoring by Grouping (Factoring a Trinomial with Leading Coefficient of Not 1)<\/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 [latex]x[\/latex] term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Factoring Trinomial (Leading Coefficient [latex]\\ne 1[\/latex])<\/h3>\r\n<strong>Factoring by grouping<\/strong> is a method used to decompose a trinomial of the form [latex]ax^2+bx+c[\/latex] into a product of two binomials.\r\n\r\n&nbsp;\r\n\r\nThis method involves finding two numbers that combine to give the product of the leading coefficient and the constant term ([latex]a \\times c[\/latex]) and the sum of the middle coefficient ([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<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping<\/strong>\r\n<ol>\r\n \t<li>List factors of [latex]a \\times c[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]a \\times 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<\/section><section class=\"textbox example\">Factor [latex]5{x}^{2}+7x - 6[\/latex].\r\n\r\n<hr \/>\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>List factors of [latex]a \\times c[\/latex].\r\n<ul>\r\n \t<li>Calculate [latex]a \\times c[\/latex]: [latex]a = 5[\/latex] and [latex]c = -6[\/latex], so [latex]a \\times c = -30[\/latex].<\/li>\r\n \t<li>Factors of [latex]-30[\/latex]:\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\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]a \\times c[\/latex] with a sum of [latex]b[\/latex].\r\n<ul>\r\n \t<li>Based on the table above, the correct pair is [latex]p = -3[\/latex] and [latex]q = 10[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]5{x}^{2}+7x - 6 = 5{x}^{2} -3x+10x-6[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>Group the first 2 terms and the last 2 terms. Then, pull out the GCF of each group.<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]5{x}^{2}+7x - 6 = (5{x}^{2} -3x)+(10x-6) = x(5x-3)+2(5x-3)[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>Factor out the GCF of the expression.<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]5{x}^{2}+7x - 6 = (x+2)(5x-3)[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Factor 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][\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318845[\/ohm_question]<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318846[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Factor trinomials where a is not 1<\/li>\n<\/ul>\n<\/section>\n<h2>Factoring by Grouping (Factoring a Trinomial with Leading Coefficient of Not 1)<\/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 [latex]x[\/latex] term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Factoring Trinomial (Leading Coefficient [latex]\\ne 1[\/latex])<\/h3>\n<p><strong>Factoring by grouping<\/strong> is a method used to decompose a trinomial of the form [latex]ax^2+bx+c[\/latex] into a product of two binomials.<\/p>\n<p>&nbsp;<\/p>\n<p>This method involves finding two numbers that combine to give the product of the leading coefficient and the constant term ([latex]a \\times c[\/latex]) and the sum of the middle coefficient ([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<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping<\/strong><\/p>\n<ol>\n<li>List factors of [latex]a \\times c[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]a \\times 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<\/section>\n<section class=\"textbox example\">Factor [latex]5{x}^{2}+7x - 6[\/latex].<\/p>\n<hr \/>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>List factors of [latex]a \\times c[\/latex].\n<ul>\n<li>Calculate [latex]a \\times c[\/latex]: [latex]a = 5[\/latex] and [latex]c = -6[\/latex], so [latex]a \\times c = -30[\/latex].<\/li>\n<li>Factors of [latex]-30[\/latex]:<br \/>\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<\/li>\n<\/ul>\n<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]a \\times c[\/latex] with a sum of [latex]b[\/latex].\n<ul>\n<li>Based on the table above, the correct pair is [latex]p = -3[\/latex] and [latex]q = 10[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]5{x}^{2}+7x - 6 = 5{x}^{2} -3x+10x-6[\/latex]<\/p>\n<ul>\n<li>Group the first 2 terms and the last 2 terms. Then, pull out the GCF of each group.<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]5{x}^{2}+7x - 6 = (5{x}^{2} -3x)+(10x-6) = x(5x-3)+2(5x-3)[\/latex]<\/p>\n<ul>\n<li>Factor out the GCF of the expression.<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]5{x}^{2}+7x - 6 = (x+2)(5x-3)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q343485\">Show Solution<\/button><\/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]<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318845\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318845&theme=lumen&iframe_resize_id=ohm318845&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318846\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318846&theme=lumen&iframe_resize_id=ohm318846&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":67,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":506,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2576"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2576\/revisions"}],"predecessor-version":[{"id":5319,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2576\/revisions\/5319"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/506"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2576\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2576"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2576"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2576"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2576"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}