{"id":83,"date":"2025-02-13T22:43:33","date_gmt":"2025-02-13T22:43:33","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/zeros-of-polynomial-functions\/"},"modified":"2026-01-13T20:31:25","modified_gmt":"2026-01-13T20:31:25","slug":"zeros-of-polynomial-functions","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/zeros-of-polynomial-functions\/","title":{"raw":"Zeros of Polynomial Functions: Learn It 1","rendered":"Zeros of Polynomial Functions: Learn It 1"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\"><section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the Factor Theorem to solve a polynomial equation.<\/li>\r\n \t<li>Use the Rational Zero Theorem to find rational zeros.<\/li>\r\n \t<li>Find zeros of a polynomial function.<\/li>\r\n \t<li>Use the Linear Factorization Theorem to find polynomials with given zeros.<\/li>\r\n \t<li>Solve real-world applications of polynomial equations<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<div id=\"Example_03_06_01\" class=\"example\">\r\n<div id=\"fs-id1165135199549\" class=\"exercise\">\r\n<h2>Rational Zero Theorem<\/h2>\r\n<section class=\"textbox recall\" aria-label=\"Recall\">If a polynomial [latex]f(x)[\/latex] is divided by [latex]x-k[\/latex], then the remainder is the value [latex]f(k)[\/latex].<\/section>Another use for the remainder theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The <strong>rational zero theorem<\/strong> helps us narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial.\r\n\r\nConsider a quadratic function with two zeros, [latex]x = \\frac{2}{5}[\/latex] and [latex]x = \\frac{3}{4}[\/latex]. By the factor theorem, these zeros have factors associated with them. Let us set each factor equal to [latex]0[\/latex], and then construct the original quadratic function absent its stretching factor.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l l} x - \\frac{2}{5} = 0 \\text{ or } x - \\frac{3}{4} = 0 &amp; \\text{Set each factor equal to 0.} \\\\ 5x - 2 = 0 \\text{ or } 4x - 3 = 0 &amp; \\text{Multiply both sides of the equation to eliminate fractions.} \\\\ f(x) = (5x - 2)(4x - 3) &amp; \\text{Create the quadratic function, multiplying the factors.} \\\\ f(x) = 20x^2 - 23x + 6 &amp; \\text{Expand the polynomial.} \\\\ f(x) = (5 \\cdot 4)x^2 - 23x + (2 \\cdot 3) &amp; \\\\ \\end{array}[\/latex]<\/p>\r\nNotice that two of the factors of the constant term, [latex]6[\/latex], are the two numerators from the original rational roots: [latex]2[\/latex] and [latex]3[\/latex]. Similarly, two of the factors from the leading coefficient, [latex]20[\/latex], are the two denominators from the original rational roots: [latex]5[\/latex] and [latex]4[\/latex].\r\n\r\nWe can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of <strong>the rational zero theorem<\/strong>; it is a means to give us a pool of possible rational zeros.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>the rational zero theorem<\/h3>\r\nThe Rational Zero Theorem states that, if the polynomial [latex]f(x) = a_nx^n + a_{n-1}x^{n-1} + \\ldots + a_1x + a_0[\/latex] has integer coefficients, then every rational zero of [latex]f(x)[\/latex] has the form [latex]\\frac{p}{q}[\/latex] where [latex]p[\/latex] is a factor of the constant term [latex]a_0[\/latex] and [latex]q[\/latex] is a factor of the leading coefficient [latex]a_n[\/latex].\r\n\r\n&nbsp;\r\n\r\nWhen the leading coefficient is [latex]1[\/latex], the possible rational zeros are the factors of the constant term.\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox questionHelp\"><strong>How to: Given a polynomial function [latex]f(x)[\/latex], use the Rational Zero Theorem to find rational zeros.<\/strong>\r\n<ol>\r\n \t<li>Determine all factors of the constant term and all factors of the leading coefficient.<\/li>\r\n \t<li>Determine all possible values of [latex]\\frac{p}{q}[\/latex], where [latex]p[\/latex] is a factor of the constant term and [latex]q[\/latex] is a factor of the leading coefficient. Be sure to include both positive and negative candidates.<\/li>\r\n \t<li>Determine which possible zeros are actual zeros by evaluating each case of [latex]f\\left(\\frac{p}{q}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">List all possible rational zeros of [latex]f(x) = 2x^4 - 5x^3 + x^2 - 4[\/latex].[reveal-answer q=\"981343\"]Show Solution[\/reveal-answer] [hidden-answer a=\"981343\"]The only possible rational zeros of [latex]f(x)[\/latex] are the quotients of the factors of the last term, [latex]-4[\/latex], and the factors of the leading coefficient, [latex]2[\/latex].The constant term is [latex]-4[\/latex]; the factors of [latex]-4[\/latex] are [latex]p = \\pm1, \\pm2, \\pm4[\/latex].The leading coefficient is [latex]2[\/latex]; the factors of [latex]2[\/latex] are [latex]q = \\pm1, \\pm2[\/latex].If any of the four real zeros are rational zeros, then they will be one of the following factors of [latex]-4[\/latex] divided by one of the factors of [latex]2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\frac{p}{q} = \\pm \\frac{1}{1}, \\pm \\frac{1}{2}[\/latex] and [latex]\\frac{p}{q} = \\pm \\frac{2}{1}, \\pm \\frac{2}{2}[\/latex] and [latex]\\frac{p}{q} = \\pm \\frac{4}{1}, \\pm \\frac{4}{2}[\/latex]<\/p>\r\nNote that [latex]\\frac{2}{2} = 1[\/latex] and [latex]\\frac{4}{2} = 2[\/latex], which have already been listed. So we can shorten our list.\r\n<p style=\"text-align: center;\">[latex]\\frac{p}{q} = \\frac{\\text{Factors of the last}}{\\text{Factors of the first}} = \\pm1, \\pm2, \\pm4, \\pm\\frac{1}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318887[\/ohm_question]<\/section><section class=\"textbox tryIt\">[ohm_question hide_question_numbers=1]318889[\/ohm_question]<\/section><\/div>\r\n<\/div>\r\n<dl id=\"fs-id1165137938597\" class=\"definition\">\r\n \t<dd id=\"fs-id1165137938602\"><\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the Factor Theorem to solve a polynomial equation.<\/li>\n<li>Use the Rational Zero Theorem to find rational zeros.<\/li>\n<li>Find zeros of a polynomial function.<\/li>\n<li>Use the Linear Factorization Theorem to find polynomials with given zeros.<\/li>\n<li>Solve real-world applications of polynomial equations<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<div id=\"Example_03_06_01\" class=\"example\">\n<div id=\"fs-id1165135199549\" class=\"exercise\">\n<h2>Rational Zero Theorem<\/h2>\n<section class=\"textbox recall\" aria-label=\"Recall\">If a polynomial [latex]f(x)[\/latex] is divided by [latex]x-k[\/latex], then the remainder is the value [latex]f(k)[\/latex].<\/section>\n<p>Another use for the remainder theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The <strong>rational zero theorem<\/strong> helps us narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial.<\/p>\n<p>Consider a quadratic function with two zeros, [latex]x = \\frac{2}{5}[\/latex] and [latex]x = \\frac{3}{4}[\/latex]. By the factor theorem, these zeros have factors associated with them. Let us set each factor equal to [latex]0[\/latex], and then construct the original quadratic function absent its stretching factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l l} x - \\frac{2}{5} = 0 \\text{ or } x - \\frac{3}{4} = 0 & \\text{Set each factor equal to 0.} \\\\ 5x - 2 = 0 \\text{ or } 4x - 3 = 0 & \\text{Multiply both sides of the equation to eliminate fractions.} \\\\ f(x) = (5x - 2)(4x - 3) & \\text{Create the quadratic function, multiplying the factors.} \\\\ f(x) = 20x^2 - 23x + 6 & \\text{Expand the polynomial.} \\\\ f(x) = (5 \\cdot 4)x^2 - 23x + (2 \\cdot 3) & \\\\ \\end{array}[\/latex]<\/p>\n<p>Notice that two of the factors of the constant term, [latex]6[\/latex], are the two numerators from the original rational roots: [latex]2[\/latex] and [latex]3[\/latex]. Similarly, two of the factors from the leading coefficient, [latex]20[\/latex], are the two denominators from the original rational roots: [latex]5[\/latex] and [latex]4[\/latex].<\/p>\n<p>We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of <strong>the rational zero theorem<\/strong>; it is a means to give us a pool of possible rational zeros.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>the rational zero theorem<\/h3>\n<p>The Rational Zero Theorem states that, if the polynomial [latex]f(x) = a_nx^n + a_{n-1}x^{n-1} + \\ldots + a_1x + a_0[\/latex] has integer coefficients, then every rational zero of [latex]f(x)[\/latex] has the form [latex]\\frac{p}{q}[\/latex] where [latex]p[\/latex] is a factor of the constant term [latex]a_0[\/latex] and [latex]q[\/latex] is a factor of the leading coefficient [latex]a_n[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p>When the leading coefficient is [latex]1[\/latex], the possible rational zeros are the factors of the constant term.<\/p>\n<\/div>\n<\/section>\n<section class=\"textbox questionHelp\"><strong>How to: Given a polynomial function [latex]f(x)[\/latex], use the Rational Zero Theorem to find rational zeros.<\/strong><\/p>\n<ol>\n<li>Determine all factors of the constant term and all factors of the leading coefficient.<\/li>\n<li>Determine all possible values of [latex]\\frac{p}{q}[\/latex], where [latex]p[\/latex] is a factor of the constant term and [latex]q[\/latex] is a factor of the leading coefficient. Be sure to include both positive and negative candidates.<\/li>\n<li>Determine which possible zeros are actual zeros by evaluating each case of [latex]f\\left(\\frac{p}{q}\\right)[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">List all possible rational zeros of [latex]f(x) = 2x^4 - 5x^3 + x^2 - 4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q981343\">Show Solution<\/button> <\/p>\n<div id=\"q981343\" class=\"hidden-answer\" style=\"display: none\">The only possible rational zeros of [latex]f(x)[\/latex] are the quotients of the factors of the last term, [latex]-4[\/latex], and the factors of the leading coefficient, [latex]2[\/latex].The constant term is [latex]-4[\/latex]; the factors of [latex]-4[\/latex] are [latex]p = \\pm1, \\pm2, \\pm4[\/latex].The leading coefficient is [latex]2[\/latex]; the factors of [latex]2[\/latex] are [latex]q = \\pm1, \\pm2[\/latex].If any of the four real zeros are rational zeros, then they will be one of the following factors of [latex]-4[\/latex] divided by one of the factors of [latex]2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{p}{q} = \\pm \\frac{1}{1}, \\pm \\frac{1}{2}[\/latex] and [latex]\\frac{p}{q} = \\pm \\frac{2}{1}, \\pm \\frac{2}{2}[\/latex] and [latex]\\frac{p}{q} = \\pm \\frac{4}{1}, \\pm \\frac{4}{2}[\/latex]<\/p>\n<p>Note that [latex]\\frac{2}{2} = 1[\/latex] and [latex]\\frac{4}{2} = 2[\/latex], which have already been listed. So we can shorten our list.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{p}{q} = \\frac{\\text{Factors of the last}}{\\text{Factors of the first}} = \\pm1, \\pm2, \\pm4, \\pm\\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318887\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318887&theme=lumen&iframe_resize_id=ohm318887&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm318889\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=318889&theme=lumen&iframe_resize_id=ohm318889&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<\/div>\n<dl id=\"fs-id1165137938597\" class=\"definition\">\n<dd id=\"fs-id1165137938602\"><\/dd>\n<\/dl>\n","protected":false},"author":6,"menu_order":17,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":506,"module-header":"learn_it","content_attributions":[{"type":"cc-attribution","description":"Precalculus","author":"OpenStax College","organization":"OpenStax","url":"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface","project":"","license":"cc-by","license_terms":""}],"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\/83"}],"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\/6"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/83\/revisions"}],"predecessor-version":[{"id":5333,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/83\/revisions\/5333"}],"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\/83\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=83"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=83"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=83"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=83"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}