{"id":3131,"date":"2025-08-15T22:10:39","date_gmt":"2025-08-15T22:10:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3131"},"modified":"2026-01-13T18:58:53","modified_gmt":"2026-01-13T18:58:53","slug":"dividing-polynomials-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/dividing-polynomials-apply-it\/","title":{"raw":"Dividing Polynomials: Apply It","rendered":"Dividing Polynomials: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use long division to divide polynomials.<\/li>\r\n \t<li>Use synthetic division to divide polynomials.<\/li>\r\n \t<li><span data-sheets-root=\"1\">Evaluate a polynomial using the Remainder Theorem.<\/span><\/li>\r\n<\/ul>\r\n<\/section><section aria-label=\"Learning Goals\">\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h2>Remainder Theorem<\/h2>\r\nRemember how, in our synthetic division example, we found that dividing [latex]5x^2-3x-36[\/latex] by [latex]x-3[\/latex] resulted in a quotient of [latex]5x+12[\/latex] with a reminder of [latex]0[\/latex]? This connection between division and remainders brings us to the <strong>Remainder Theorem<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Remainder Theorem<\/h3>\r\nIf a polynomial [latex]f(x)[\/latex] is divided by [latex]x-k[\/latex], then the remainder is the value [latex]f(k)[\/latex].\r\n\r\n<\/section>Let's walk through the proof of the theorem.\r\n\r\n<section class=\"textbox connectIt\">Recall that the Division Algorithm states that, given a polynomial dividend [latex]f(x)[\/latex] and a non-zero polynomial divisor [latex]d(x)[\/latex], there exist unique polynomials [latex]g(x)[\/latex] and [latex]r(x)[\/latex] such that\r\n<p style=\"text-align: center;\">[latex]f(x) = d(x)g(x) + r(x)[\/latex]<\/p>\r\nand either [latex]r(x) = 0[\/latex] or the degree of [latex]r(x)[\/latex] is less than the degree of [latex]d(x)[\/latex]. In practice divisors, [latex]d(x)[\/latex] will have degrees less than or equal to the degree of [latex]f(x)[\/latex]. If the divisor, [latex]d(x)[\/latex], is [latex]x - k[\/latex], this takes the form\r\n<p style=\"text-align: center;\">[latex]f(x) = (x - k)g(x) + r[\/latex]<\/p>\r\nSince the divisor [latex]x - k[\/latex] is linear, the remainder will be a constant, [latex]r[\/latex]. And, if we evaluate this for [latex]x = k[\/latex], we have\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl} f(k) &amp; = (k - k)g(k) + r \\\\ &amp; = 0 \\cdot g(k) + r \\\\ &amp; = r \\end{array}[\/latex]<\/p>\r\nIn other words, [latex]f(k)[\/latex] is the remainder obtained by dividing [latex]f(x)[\/latex] by [latex]x - k[\/latex].\r\n\r\n<\/section><section class=\"textbox questionHelp\"><strong>How to: Given a polynomial function [latex]f[\/latex], evaluate [latex]f(x)[\/latex] at [latex]x=k[\/latex] using the Remainder Theorem.<\/strong>\r\n<ol id=\"fs-id1575825\" type=\"1\">\r\n \t<li>Use synthetic division to divide the polynomial by [latex]x\u2212k[\/latex].<\/li>\r\n \t<li>The remainder is the value [latex]f(k)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>This powerful theorem allows us to quickly find the remainder of polynomial division without completing the entire division process.\r\n\r\n<section class=\"textbox example\">For example, in our case, we divided [latex]5x^2-3x-36[\/latex] by [latex]x-3[\/latex]. According to the Remainder Theorem, the remainder is [latex]f(3)[\/latex].\r\n[latex]\\\\[\/latex]\r\nBy substituting [latex]3[\/latex] into the polynomial [latex]f(x) = 5x^2-3x-36[\/latex], we get:\r\n<p style=\"text-align: center;\">[latex]f(3) = 5(3)^2 - 3(3) - 36 = 45 - 9 - 36 = 0[\/latex]<\/p>\r\nThis confirms that the remainder is indeed [latex]0[\/latex], just as we found using synthetic division.\r\n\r\n<\/section><section class=\"textbox example\">Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7[\/latex]\u00a0at [latex]x=2[\/latex].[reveal-answer q=\"830571\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"830571\"]To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ 2\\overline{)\\begin{array}{lllllllll}6\\hfill &amp; -1\\hfill &amp; -15\\hfill &amp; 2\\hfill &amp; -7\\hfill \\\\ \\hfill &amp; \\text{ }12\\hfill &amp; \\text{ }\\text{ }\\text{ }22\\hfill &amp; 14\\hfill &amp; \\text{ }\\text{ }32\\hfill \\end{array}}\\\\ \\begin{array}{llllll}\\hfill &amp; \\text{}6\\hfill &amp; 11\\hfill &amp; \\text{ }\\text{ }\\text{ }7\\hfill &amp; \\text{ }\\text{ }16\\hfill &amp; \\text{ }\\text{ }25\\hfill \\end{array}\\end{array}[\/latex]<\/p>\r\nThe remainder is [latex]25[\/latex]. Therefore, [latex]f\\left(2\\right)=25[\/latex].\r\n\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nWe can check our answer by evaluating [latex]f\\left(2\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}f\\left(x\\right) &amp; =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\\\ f\\left(2\\right) &amp; =6{\\left(2\\right)}^{4}-{\\left(2\\right)}^{3}-15{\\left(2\\right)}^{2}+2\\left(2\\right)-7 \\\\ f\\left(2\\right) &amp; =25\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318862[\/ohm_question]<\/section><\/div>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use long division to divide polynomials.<\/li>\n<li>Use synthetic division to divide polynomials.<\/li>\n<li><span data-sheets-root=\"1\">Evaluate a polynomial using the Remainder Theorem.<\/span><\/li>\n<\/ul>\n<\/section>\n<section aria-label=\"Learning Goals\">\n<div class=\"bcc-box bcc-highlight\">\n<h2>Remainder Theorem<\/h2>\n<p>Remember how, in our synthetic division example, we found that dividing [latex]5x^2-3x-36[\/latex] by [latex]x-3[\/latex] resulted in a quotient of [latex]5x+12[\/latex] with a reminder of [latex]0[\/latex]? This connection between division and remainders brings us to the <strong>Remainder Theorem<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>Remainder Theorem<\/h3>\n<p>If a polynomial [latex]f(x)[\/latex] is divided by [latex]x-k[\/latex], then the remainder is the value [latex]f(k)[\/latex].<\/p>\n<\/section>\n<p>Let&#8217;s walk through the proof of the theorem.<\/p>\n<section class=\"textbox connectIt\">Recall that the Division Algorithm states that, given a polynomial dividend [latex]f(x)[\/latex] and a non-zero polynomial divisor [latex]d(x)[\/latex], there exist unique polynomials [latex]g(x)[\/latex] and [latex]r(x)[\/latex] such that<\/p>\n<p style=\"text-align: center;\">[latex]f(x) = d(x)g(x) + r(x)[\/latex]<\/p>\n<p>and either [latex]r(x) = 0[\/latex] or the degree of [latex]r(x)[\/latex] is less than the degree of [latex]d(x)[\/latex]. In practice divisors, [latex]d(x)[\/latex] will have degrees less than or equal to the degree of [latex]f(x)[\/latex]. If the divisor, [latex]d(x)[\/latex], is [latex]x - k[\/latex], this takes the form<\/p>\n<p style=\"text-align: center;\">[latex]f(x) = (x - k)g(x) + r[\/latex]<\/p>\n<p>Since the divisor [latex]x - k[\/latex] is linear, the remainder will be a constant, [latex]r[\/latex]. And, if we evaluate this for [latex]x = k[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl} f(k) & = (k - k)g(k) + r \\\\ & = 0 \\cdot g(k) + r \\\\ & = r \\end{array}[\/latex]<\/p>\n<p>In other words, [latex]f(k)[\/latex] is the remainder obtained by dividing [latex]f(x)[\/latex] by [latex]x - k[\/latex].<\/p>\n<\/section>\n<section class=\"textbox questionHelp\"><strong>How to: Given a polynomial function [latex]f[\/latex], evaluate [latex]f(x)[\/latex] at [latex]x=k[\/latex] using the Remainder Theorem.<\/strong><\/p>\n<ol id=\"fs-id1575825\" type=\"1\">\n<li>Use synthetic division to divide the polynomial by [latex]x\u2212k[\/latex].<\/li>\n<li>The remainder is the value [latex]f(k)[\/latex].<\/li>\n<\/ol>\n<\/section>\n<p>This powerful theorem allows us to quickly find the remainder of polynomial division without completing the entire division process.<\/p>\n<section class=\"textbox example\">For example, in our case, we divided [latex]5x^2-3x-36[\/latex] by [latex]x-3[\/latex]. According to the Remainder Theorem, the remainder is [latex]f(3)[\/latex].<br \/>\n[latex]\\\\[\/latex]<br \/>\nBy substituting [latex]3[\/latex] into the polynomial [latex]f(x) = 5x^2-3x-36[\/latex], we get:<\/p>\n<p style=\"text-align: center;\">[latex]f(3) = 5(3)^2 - 3(3) - 36 = 45 - 9 - 36 = 0[\/latex]<\/p>\n<p>This confirms that the remainder is indeed [latex]0[\/latex], just as we found using synthetic division.<\/p>\n<\/section>\n<section class=\"textbox example\">Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7[\/latex]\u00a0at [latex]x=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q830571\">Show Solution<\/button><\/p>\n<div id=\"q830571\" class=\"hidden-answer\" style=\"display: none\">To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ 2\\overline{)\\begin{array}{lllllllll}6\\hfill & -1\\hfill & -15\\hfill & 2\\hfill & -7\\hfill \\\\ \\hfill & \\text{ }12\\hfill & \\text{ }\\text{ }\\text{ }22\\hfill & 14\\hfill & \\text{ }\\text{ }32\\hfill \\end{array}}\\\\ \\begin{array}{llllll}\\hfill & \\text{}6\\hfill & 11\\hfill & \\text{ }\\text{ }\\text{ }7\\hfill & \\text{ }\\text{ }16\\hfill & \\text{ }\\text{ }25\\hfill \\end{array}\\end{array}[\/latex]<\/p>\n<p>The remainder is [latex]25[\/latex]. Therefore, [latex]f\\left(2\\right)=25[\/latex].<\/p>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>We can check our answer by evaluating [latex]f\\left(2\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}f\\left(x\\right) & =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\\\ f\\left(2\\right) & =6{\\left(2\\right)}^{4}-{\\left(2\\right)}^{3}-15{\\left(2\\right)}^{2}+2\\left(2\\right)-7 \\\\ f\\left(2\\right) & =25\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318862\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318862&theme=lumen&iframe_resize_id=ohm318862&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<\/section>\n","protected":false},"author":67,"menu_order":9,"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":"apply_it","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\/3131"}],"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\/3131\/revisions"}],"predecessor-version":[{"id":5327,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3131\/revisions\/5327"}],"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\/3131\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3131"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3131"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3131"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3131"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}