{"id":2572,"date":"2025-08-13T18:06:39","date_gmt":"2025-08-13T18:06:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2572"},"modified":"2026-01-13T17:42:14","modified_gmt":"2026-01-13T17:42:14","slug":"polynomial-equations-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polynomial-equations-background-youll-need-1\/","title":{"raw":"Polynomial Equations: Background You'll Need 1","rendered":"Polynomial Equations: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Factor trinomials with a = 1<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Factoring a Trinomial with Leading Coefficient of 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\n<section class=\"textbox keyTakeaway\">\r\n<h3>Factoring a Trinomial (Leading Coefficient\u00a0 [latex]= 1[\/latex])<\/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<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Some polynomials cannot be factored. These polynomials are said to be prime.<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/strong>\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<\/section><section class=\"textbox example\">Factor [latex]{x}^{2}+2x - 15[\/latex].\r\n\r\n<hr \/>\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex].<\/li>\r\n \t<li>We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex].<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 40px;\">In the table, we list factors until we find a pair with the desired sum.<\/p>\r\n\r\n<table style=\"margin-left: 40px;\" 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><strong>[latex]-3,5[\/latex]<\/strong><\/td>\r\n<td><strong>[latex]2[\/latex]<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ul>\r\n \t<li>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].<\/li>\r\n<\/ul>\r\nThus, [latex]{x}^{2}+2x - 15 = \\left(x - 3\\right)\\left(x+5\\right)[\/latex]\r\n\r\n[reveal-answer q=\"523566\"]Does the order of the factors matter?[\/reveal-answer]\r\n[hidden-answer a=\"523566\"]No. Multiplication is commutative, so the order of the factors does not matter. [latex]{x}^{2}+2x - 15 = \\left(x - 3\\right)\\left(x+5\\right) = \\left(x+5\\right)\\left(x - 3\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318844[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Factor trinomials with a = 1<\/li>\n<\/ul>\n<\/section>\n<h2>Factoring a Trinomial with Leading Coefficient of 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<section class=\"textbox keyTakeaway\">\n<h3>Factoring a Trinomial (Leading Coefficient\u00a0 [latex]= 1[\/latex])<\/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<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Some polynomials cannot be factored. These polynomials are said to be prime.<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/strong><\/p>\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<\/section>\n<section class=\"textbox example\">Factor [latex]{x}^{2}+2x - 15[\/latex].<\/p>\n<hr \/>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex].<\/li>\n<li>We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex].<\/li>\n<\/ul>\n<p style=\"padding-left: 40px;\">In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table style=\"margin-left: 40px;\" 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><strong>[latex]-3,5[\/latex]<\/strong><\/td>\n<td><strong>[latex]2[\/latex]<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li>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].<\/li>\n<\/ul>\n<p>Thus, [latex]{x}^{2}+2x - 15 = \\left(x - 3\\right)\\left(x+5\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q523566\">Does the order of the factors matter?<\/button><\/p>\n<div id=\"q523566\" class=\"hidden-answer\" style=\"display: none\">No. Multiplication is commutative, so the order of the factors does not matter. [latex]{x}^{2}+2x - 15 = \\left(x - 3\\right)\\left(x+5\\right) = \\left(x+5\\right)\\left(x - 3\\right)[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318844\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318844&theme=lumen&iframe_resize_id=ohm318844&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":67,"menu_order":2,"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\/2572"}],"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":6,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2572\/revisions"}],"predecessor-version":[{"id":5317,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2572\/revisions\/5317"}],"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\/2572\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2572"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2572"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2572"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2572"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}