{"id":13872,"date":"2018-08-24T22:26:12","date_gmt":"2018-08-24T22:26:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalcone\/?post_type=chapter&#038;p=13872"},"modified":"2025-02-05T05:19:17","modified_gmt":"2025-02-05T05:19:17","slug":"zeros-of-polynomial-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/zeros-of-polynomial-functions\/","title":{"raw":"Zeros of Polynomial Functions","rendered":"Zeros of Polynomial Functions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate a polynomial using the Remainder Theorem.<\/li>\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>Use Descartes\u2019 Rule of Signs.<\/li>\r\n \t<li>Solve real-world applications of polynomial equations<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137758829\">A new bakery offers decorated sheet cakes for children\u2019s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?<\/p>\r\n<p id=\"eip-151\">This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations.<\/p>\r\n\r\n<h2>Evaluate a polynomial using the Remainder Theorem<\/h2>\r\n<p id=\"fs-id1165135471230\">In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the <strong>Remainder Theorem<\/strong>. If the polynomial is divided by <em>x<\/em> \u2013\u00a0<em>k<\/em>, the remainder may be found quickly by evaluating the polynomial function at <em>k<\/em>, that is, <em>f<\/em>(<em>k<\/em>)\u00a0Let\u2019s walk through the proof of the theorem.<\/p>\r\n<p id=\"fs-id1165134085965\">Recall that the <strong>Division Algorithm<\/strong> states that, given a polynomial dividend <em>f<\/em>(<em>x<\/em>)\u00a0and a non-zero polynomial divisor <em>d<\/em>(<em>x<\/em>)\u00a0where the degree of\u00a0<em>d<\/em>(<em>x<\/em>) is less than or equal to the degree of <em>f<\/em>(<em>x<\/em>), there exist unique polynomials <em>q<\/em>(<em>x<\/em>)\u00a0and <em>r<\/em>(<em>x<\/em>)\u00a0such that<\/p>\r\n\r\n<div id=\"eip-753\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=d\\left(x\\right)q\\left(x\\right)+r\\left(x\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165134094600\">If the divisor, <em>d<\/em>(<em>x<\/em>), is <em>x<\/em> \u2013\u00a0<em>k<\/em>, this takes the form<\/p>\r\n\r\n<div id=\"eip-567\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex]<\/div>\r\n<p id=\"fs-id1165137447771\">Since the divisor <em>x<\/em> \u2013\u00a0<em>k<\/em>\u00a0is linear, the remainder will be a constant, <em>r<\/em>. And, if we evaluate this for <em>x<\/em> =\u00a0<em>k<\/em>, we have<\/p>\r\n\r\n<div id=\"eip-791\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}f\\left(k\\right)&amp;=\\left(k-k\\right)q\\left(k\\right)+r \\\\ &amp;=0\\cdot q\\left(k\\right)+r \\\\ &amp;=r \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135572088\">In other words, <em>f<\/em>(<em>k<\/em>)\u00a0is the remainder obtained by dividing <em>f<\/em>(<em>x<\/em>)\u00a0by <em>x<\/em> \u2013\u00a0<em>k<\/em>.<\/p>\r\n\r\n<div id=\"fs-id1165134042705\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Remainder Theorem<\/h3>\r\n<p id=\"fs-id1165137725230\">If a polynomial [latex]f\\left(x\\right)[\/latex] is divided by <em>x<\/em> \u2013\u00a0<em>k<\/em>, then the remainder is the value [latex]f\\left(k\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137823272\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165134130150\">How To: Given a polynomial function [latex]f[\/latex], evaluate [latex]f\\left(x\\right)[\/latex] at [latex]x=k[\/latex] using the Remainder Theorem.<\/h3>\r\n<ol id=\"fs-id1165137870981\">\r\n \t<li>Use synthetic division to divide the polynomial by [latex]x-k[\/latex].<\/li>\r\n \t<li>The remainder is the value [latex]f\\left(k\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_06_01\" class=\"example\">\r\n<div id=\"fs-id1165135199549\" class=\"exercise\">\r\n<div id=\"fs-id1165137727226\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Using the Remainder Theorem to Evaluate a Polynomial<\/h3>\r\n<p id=\"fs-id1165137855032\">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>\r\n[reveal-answer q=\"540399\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"540399\"]\r\n<p id=\"fs-id1165137771250\">To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[\/latex].<\/p>\r\n<p style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02010113\/Synthetic-Division-Example-1.jpg\"><img class=\"alignnone wp-image-15126\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02010113\/Synthetic-Division-Example-1.jpg\" alt=\"Synthetic division with divisor 2 and quotient {6,-1, -15, 2, -7}. Solution is {6,11,7, 16,25}\" width=\"255\" height=\"113\" \/><\/a><\/p>\r\n<p id=\"fs-id1165137932598\">The remainder is 25. Therefore, [latex]f\\left(2\\right)=25[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135397290\">We can check our answer by evaluating [latex]f\\left(2\\right)[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}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 \\\\ &amp;=25 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137892503\">Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=2{x}^{5}-3{x}^{4}-9{x}^{3}+8{x}^{2}+2[\/latex]\r\nat [latex]x=-3[\/latex].<\/p>\r\n[reveal-answer q=\"947133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"947133\"]\r\n\r\n[latex]f\\left(-3\\right)=-412[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Use the Factor Theorem to solve a polynomial equation<\/h2>\r\n<p id=\"fs-id1165137459796\">The <strong>Factor Theorem <\/strong>is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm tells us<\/p>\r\n\r\n<div id=\"eip-10\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex].<\/div>\r\n<p id=\"fs-id1165137463592\">If <em>k<\/em>\u00a0is a zero, then the remainder <em>r<\/em>\u00a0is [latex]f\\left(k\\right)=0[\/latex]\u00a0and [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+0[\/latex]\u00a0or [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165135176357\">Notice, written in this form, <em>x<\/em>\u00a0\u2013\u00a0<em>k<\/em> is a factor of [latex]f\\left(x\\right)[\/latex]. We can conclude if <em>k\u00a0<\/em>is a zero of [latex]f\\left(x\\right)[\/latex], then [latex]x-k[\/latex] is a factor of [latex]f\\left(x\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165135684373\">Similarly, if [latex]x-k[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex],\u00a0then the remainder of the Division Algorithm [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex]\u00a0is 0. This tells us that <em>k<\/em>\u00a0is a zero.<\/p>\r\n<p id=\"fs-id1165132943504\">This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree <em>n<\/em>\u00a0in the complex number system will have <em>n<\/em>\u00a0zeros. We can use the Factor Theorem to completely factor a polynomial into the product of <em>n<\/em>\u00a0factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.<\/p>\r\n\r\n<div id=\"fs-id1165135173601\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Factor Theorem<\/h3>\r\n<p id=\"fs-id1165135575975\">According to the <strong>Factor Theorem<\/strong>, <em>k<\/em>\u00a0is a zero of [latex]f\\left(x\\right)[\/latex]\u00a0if and only if [latex]\\left(x-k\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137809910\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165133356017\">How To: Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial.<strong>\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id1165135383467\">\r\n \t<li>Use synthetic division to divide the polynomial by [latex]\\left(x-k\\right)[\/latex].<\/li>\r\n \t<li>Confirm that the remainder is 0.<\/li>\r\n \t<li>Write the polynomial as the product of [latex]\\left(x-k\\right)[\/latex] and the quadratic quotient.<\/li>\r\n \t<li>If possible, factor the quadratic.<\/li>\r\n \t<li>Write the polynomial as the product of factors.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_06_02\" class=\"example\">\r\n<div id=\"fs-id1165134273636\" class=\"exercise\">\r\n<div id=\"fs-id1165134273638\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Using the Factor Theorem to Solve a Polynomial Equation<\/h3>\r\n<p id=\"fs-id1165134198723\">Show that [latex]\\left(x+2\\right)[\/latex]\u00a0is a factor of [latex]{x}^{3}-6{x}^{2}-x+30[\/latex]. Find the remaining factors. Use the factors to determine the zeros of the <strong>polynomial<\/strong>.<\/p>\r\n[reveal-answer q=\"884250\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"884250\"]\r\n<p id=\"fs-id1165134109661\">We can use synthetic division to show that [latex]\\left(x+2\\right)[\/latex] is a factor of the polynomial.<\/p>\r\n\r\n<div id=\"eip-id1165134503086\" class=\"equation unnumbered\" style=\"text-align: center;\">\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.03.13-PM.png\"><img class=\"aligncenter wp-image-13110 \" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.03.13-PM.png\" alt=\"Synthetic division with divisor -2 and quotient {1, 6, -1, 30}. Solution is {1, -8, 15, 0}\" width=\"229\" height=\"143\" \/><\/a>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165134261634\">The remainder is zero, so [latex]\\left(x+2\\right)[\/latex] is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left({x}^{2}-8x+15\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135440193\">We can factor the quadratic factor to write the polynomial as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x - 3\\right)\\left(x - 5\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137605065\">By the Factor Theorem, the zeros of [latex]{x}^{3}-6{x}^{2}-x+30[\/latex] are \u20132, 3, and 5.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134569514\">Use the Factor Theorem to find the zeros of [latex]f\\left(x\\right)={x}^{3}+4{x}^{2}-4x - 16[\/latex]\u00a0given that [latex]\\left(x - 2\\right)[\/latex]\u00a0is a factor of the polynomial.<\/p>\r\n[reveal-answer q=\"536719\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"536719\"]\r\n\r\nThe zeros are 2, \u20132, and \u20134.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]2367[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use the Rational Zero Theorem to find rational zeros<\/h2>\r\n<p id=\"fs-id1165137660817\">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 to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading <strong>coefficient<\/strong> of the polynomial<\/p>\r\n<p id=\"fs-id1165135508309\">Consider a quadratic function with two zeros, [latex]x=\\frac{2}{5}[\/latex]\u00a0and [latex]x=\\frac{3}{4}[\/latex].<\/p>\r\nBy the Factor Theorem, these zeros have factors associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching factor.<span id=\"eip-id1165135315549\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010746\/CNX_Precalc_Figure_03_03_0225.jpg\" alt=\"This image shows x minus two fifths equals 0 or x minus three fourths equals 0. Beside this math is the sentence, 'Set each factor equal to 0.' Next it shows that five x minus 2 equals 0 or 4 x minus 3 equals 0. Beside this math is the sentence, 'Multiply both sides of the equation to eliminate fractions.' Next it shows that f of x is equal to (5 x minus 2) times (4 x minus 3). Beside this math is the sentence, 'Create the quadratic function, multiplying the factors.' Next it shows f of x equals 20 x squared minus 23 x plus 6. Beside this math is the sentence, 'Expand the polynomial.' The last equation shows f of x equals (5 times 4) times x squared minus 23 x plus (2 times 3). Set each factor equal to zero. Multiply both sides of the equation to eliminate fractions. Create the quadratic function, multiplying the factors. Expand the polynomial.\" width=\"975\" height=\"175\" \/><\/span>\r\n<p id=\"fs-id1165135485170\">Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3. Similarly, two of the factors from the leading coefficient, 20, are the two denominators from the original rational roots: 5 and 4.<\/p>\r\n<p id=\"fs-id1165137761317\">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 the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros.<\/p>\r\n\r\n<div id=\"fs-id1165137737069\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Rational Zero Theorem<\/h3>\r\n<p id=\"fs-id1165135530393\">The <strong>Rational Zero Theorem<\/strong> states that, if the polynomial [latex]f\\left(x\\right)={a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+...+{a}_{1}x+{a}_{0}[\/latex] has integer coefficients, then every rational zero of [latex]f\\left(x\\right)[\/latex]\u00a0has the form [latex]\\frac{p}{q}[\/latex] where <em>p<\/em>\u00a0is a factor of the constant term [latex]{a}_{0}[\/latex] and <em>q<\/em>\u00a0is a factor of the leading coefficient [latex]{a}_{n}[\/latex].<\/p>\r\n<p id=\"fs-id1165137736282\">When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137736287\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137446561\">How To: Given a polynomial function [latex]f\\left(x\\right)[\/latex], use the Rational Zero Theorem to find rational zeros.<\/h3>\r\n<ol id=\"fs-id1165137662157\">\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 <em>p<\/em>\u00a0is a factor of the constant term and <em>q<\/em>\u00a0is 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<\/div>\r\n<div id=\"Example_03_06_03\" class=\"example\">\r\n<div id=\"fs-id1165135319520\" class=\"exercise\">\r\n<div id=\"fs-id1165135319523\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Listing All Possible Rational Zeros<\/h3>\r\n<p id=\"fs-id1165135640968\">List all possible rational zeros of [latex]f\\left(x\\right)=2{x}^{4}-5{x}^{3}+{x}^{2}-4[\/latex].<\/p>\r\n[reveal-answer q=\"655905\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"655905\"]\r\n<p id=\"fs-id1165134179675\">The only possible rational zeros of [latex]f\\left(x\\right)[\/latex]\u00a0are the quotients of the factors of the last term, \u20134, and the factors of the leading coefficient, 2.<\/p>\r\n<p id=\"fs-id1165132972963\">The constant term is \u20134; the factors of \u20134 are [latex]p=\\pm 1,\\pm 2,\\pm 4[\/latex].<\/p>\r\n<p id=\"fs-id1165137571546\">The leading coefficient is 2; the factors of 2 are [latex]q=\\pm 1,\\pm 2[\/latex].<\/p>\r\n<p id=\"fs-id1165135557830\">If any of the four real zeros are rational zeros, then they will be of one of the following factors of \u20134 divided by one of the factors of 2.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\frac{p}{q}=\\pm \\frac{1}{1},\\pm \\frac{1}{2} &amp;&amp; \\frac{p}{q}=\\pm \\frac{2}{1},\\pm \\frac{2}{2} &amp;&amp; \\frac{p}{q}=\\pm \\frac{4}{1},\\pm \\frac{4}{2}\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135351678\">Note that [latex]\\frac{2}{2}=1[\/latex]\u00a0and [latex]\\frac{4}{2}=2[\/latex], which have already been listed. So we can shorten our list.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{p}{q}=\\frac{\\text{Factors of the last}}{\\text{Factors of the first}}=\\pm 1,\\pm 2,\\pm 4,\\pm \\frac{1}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_03_06_04\" class=\"example\">\r\n<div id=\"fs-id1165137611543\" class=\"exercise\">\r\n<div id=\"fs-id1165137611546\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Using the Rational Zero Theorem to Find Rational Zeros<\/h3>\r\n<p id=\"fs-id1165137459844\">Use the Rational Zero Theorem to find the rational zeros of [latex]f\\left(x\\right)=2{x}^{3}+{x}^{2}-4x+1[\/latex].<\/p>\r\n[reveal-answer q=\"565085\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"565085\"]\r\n<p id=\"fs-id1165135154382\">The Rational Zero Theorem tells us that if [latex]\\frac{p}{q}[\/latex] is a zero of [latex]f\\left(x\\right)[\/latex],\u00a0then <em>p<\/em>\u00a0is a factor of 1 and <em>q<\/em>\u00a0is a factor of 2.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{p}{q}&amp;=\\frac{\\text{factor of constant term}}{\\text{factor of leading coefficient}} \\\\[1mm] &amp;=\\frac{\\text{factor of 1}}{\\text{factor of 2}} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135209688\">The factors of 1 are [latex]\\pm 1[\/latex] and the factors of 2 are [latex]\\pm 1[\/latex] and [latex]\\pm 2[\/latex]. The possible values for [latex]\\frac{p}{q}[\/latex] are [latex]\\pm 1[\/latex] and [latex]\\pm \\frac{1}{2}[\/latex]. These are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for <em>x<\/em>\u00a0in [latex]f\\left(x\\right)[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(-1\\right)=2{\\left(-1\\right)}^{3}+{\\left(-1\\right)}^{2}-4\\left(-1\\right)+1=4[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(1\\right)=2{\\left(1\\right)}^{3}+{\\left(1\\right)}^{2}-4\\left(1\\right)+1=0[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(-\\frac{1}{2}\\right)=2{\\left(-\\frac{1}{2}\\right)}^{3}+{\\left(-\\frac{1}{2}\\right)}^{2}-4\\left(-\\frac{1}{2}\\right)+1=3[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(\\frac{1}{2}\\right)=2{\\left(\\frac{1}{2}\\right)}^{3}+{\\left(\\frac{1}{2}\\right)}^{2}-4\\left(\\frac{1}{2}\\right)+1=-\\frac{1}{2}[\/latex]<\/p>\r\n<p id=\"fs-id1165137731309\">Of those, [latex]-1,-\\frac{1}{2},\\text{ and }\\frac{1}{2}[\/latex] are not zeros of [latex]f\\left(x\\right)[\/latex]. 1 is the only rational zero of [latex]f\\left(x\\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134198689\">Use the Rational Zero Theorem to find the rational zeros of [latex]f\\left(x\\right)={x}^{3}-5{x}^{2}+2x+1[\/latex].<\/p>\r\n[reveal-answer q=\"544964\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"544964\"]\r\n\r\nThere are no rational zeros.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]43500[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Find zeros of a polynomial function<\/h2>\r\n<p id=\"fs-id1165135530405\">The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use <span class=\"no-emphasis\">synthetic division<\/span> repeatedly to determine all of the zeros of a polynomial function.<\/p>\r\n\r\n<div id=\"fs-id1165134155131\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165133364811\">How To: Given a polynomial function [latex]f[\/latex], use synthetic division to find its zeros.<\/h3>\r\n<ol id=\"fs-id1165135400178\">\r\n \t<li>Use the Rational Zero Theorem to list all possible rational zeros of the function.<\/li>\r\n \t<li>Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.<\/li>\r\n \t<li>Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.<\/li>\r\n \t<li>Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_06_05\" class=\"example\">\r\n<div id=\"fs-id1165137761977\" class=\"exercise\">\r\n<div id=\"fs-id1165134198663\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Finding the Zeros of a Polynomial Function with Repeated Real Zeros<\/h3>\r\n<p id=\"fs-id1165134198668\">Find the zeros of [latex]f\\left(x\\right)=4{x}^{3}-3x - 1[\/latex].<\/p>\r\n[reveal-answer q=\"663920\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"663920\"]\r\n<p id=\"fs-id1165135393366\">The Rational Zero Theorem tells us that if [latex]\\frac{p}{q}[\/latex] is a zero of [latex]f\\left(x\\right)[\/latex], then <em>p\u00a0<\/em>is a factor of \u20131 and\u00a0<em>q<\/em>\u00a0is a factor of 4.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{p}{q}&amp;=\\frac{\\text{factor of constant term}}{\\text{factor of leading coefficient}} \\\\ &amp;=\\frac{\\text{factor of -1}}{\\text{factor of 4}} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135501059\">The factors of \u20131 are [latex]\\pm 1[\/latex]\u00a0and the factors of 4 are [latex]\\pm 1,\\pm 2[\/latex], and [latex]\\pm 4[\/latex]. The possible values for [latex]\\frac{p}{q}[\/latex] are [latex]\\pm 1,\\pm \\frac{1}{2}[\/latex], and [latex]\\pm \\frac{1}{4}[\/latex].<\/p>\r\nThese are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let\u2019s begin with 1.\r\n<div id=\"eip-id1165137397701\" class=\"equation unnumbered\" style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.05.49-PM.png\"><img class=\"aligncenter size-full wp-image-13113\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.05.49-PM.png\" alt=\"Synthetic division with 1 as the divisor and {4, 0, -3, -1} as the quotient. Solution is {4, 4, 1, 0}\" width=\"166\" height=\"122\" \/><\/a><\/div>\r\n<p id=\"fs-id1165137934395\">Dividing by [latex]\\left(x - 1\\right)[\/latex]\u00a0gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(x - 1\\right)\\left(4{x}^{2}+4x+1\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137605971\">The quadratic is a perfect square. [latex]f\\left(x\\right)[\/latex]\u00a0can be written as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(x - 1\\right){\\left(2x+1\\right)}^{2}[\/latex].<\/p>\r\n<p id=\"fs-id1165135564194\">We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2x+1=0 \\\\ x=-\\frac{1}{2} \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165135651485\">The zeros of the function are 1 and [latex]-\\frac{1}{2}[\/latex] with multiplicity 2.<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nLook at the graph of the function <em>f<\/em>\u00a0in Figure 1. Notice, at [latex]x=-0.5[\/latex], the graph bounces off the <em>x<\/em>-axis, indicating the even multiplicity (2,4,6\u2026) for the zero \u20130.5.\u00a0At [latex]x=1[\/latex], the graph crosses the <em>x<\/em>-axis, indicating the odd multiplicity (1,3,5\u2026) for the zero [latex]x=1[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010746\/CNX_Precalc_Figure_03_06_0012.jpg\" alt=\"Graph of a polynomial that have its local maximum at (-0.5, 0) labeled as \" width=\"487\" height=\"289\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Use the Fundamental Theorem of Algebra<\/h2>\r\n<p id=\"fs-id1165135547255\">Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The <strong>Fundamental Theorem of Algebra <\/strong>tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.<\/p>\r\n<p id=\"fs-id1165137771429\">Suppose <em>f<\/em>\u00a0is a polynomial function of degree four, and [latex]f\\left(x\\right)=0[\/latex]. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it [latex]{c}_{1}[\/latex]. By the Factor Theorem, we can write [latex]f\\left(x\\right)[\/latex] as a product of [latex]x-{c}_{\\text{1}}[\/latex] and a polynomial quotient. Since [latex]x-{c}_{\\text{1}}[\/latex] is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it [latex]{c}_{\\text{2}}[\/latex]. So we can write the polynomial quotient as a product of [latex]x-{c}_{\\text{2}}[\/latex] and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of [latex]f\\left(x\\right)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165135693782\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The <strong>Fundamental Theorem of Algebra<\/strong> states that, if <em>f(x)<\/em> is a polynomial of degree <em>n &gt; 0<\/em>, then <em>f(x)<\/em> has at least one complex zero.<\/h3>\r\n<p id=\"fs-id1165135409342\">We can use this theorem to argue that, if [latex]f\\left(x\\right)[\/latex] is a polynomial of degree [latex]n&gt;0[\/latex], and <em>a<\/em>\u00a0is a non-zero real number, then [latex]f\\left(x\\right)[\/latex] has exactly <em>n<\/em>\u00a0linear factors<\/p>\r\n\r\n<div id=\"eip-750\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\left(x-{c}_{1}\\right)\\left(x-{c}_{2}\\right)...\\left(x-{c}_{n}\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165137889849\">where [latex]{c}_{1},{c}_{2},...,{c}_{n}[\/latex] are complex numbers. Therefore, [latex]f\\left(x\\right)[\/latex] has <em>n<\/em>\u00a0roots if we allow for multiplicities.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135481201\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165135481208\"><strong>Does every polynomial have at least one imaginary zero?<\/strong><\/p>\r\n<p id=\"fs-id1165137883746\"><em>No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_03_06_06\" class=\"example\">\r\n<div id=\"fs-id1165135658206\" class=\"exercise\">\r\n<div id=\"fs-id1165135658208\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Finding the Zeros of a Polynomial Function with Complex Zeros<\/h3>\r\n<p id=\"fs-id1165135658213\">Find the zeros of [latex]f\\left(x\\right)=3{x}^{3}+9{x}^{2}+x+3[\/latex].<\/p>\r\n[reveal-answer q=\"278128\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"278128\"]\r\n<p id=\"fs-id1165137430512\">The Rational Zero Theorem tells us that if [latex]\\frac{p}{q}[\/latex] is a zero of [latex]f\\left(x\\right)[\/latex], then <em>p<\/em>\u00a0is a factor of 3 and\u00a0<em>q<\/em>\u00a0is a factor of 3.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{p}{q}&amp;=\\frac{\\text{factor of constant term}}{\\text{factor of leading coefficient}} \\\\ &amp;=\\frac{\\text{factor of 3}}{\\text{factor of 3}} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165134212112\">The factors of 3 are [latex]\\pm 1[\/latex] and [latex]\\pm 3[\/latex]. The possible values for [latex]\\frac{p}{q}[\/latex], and therefore the possible rational zeros for the function, are [latex]\\pm 3, \\pm 1,\\text{ and }\\pm \\frac{1}{3}[\/latex]. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let\u2019s begin with \u20133.<\/p>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png\"><img class=\"aligncenter size-full wp-image-13116\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png\" alt=\"Synthetic division with divisor = -3 and quotient = {3, 9, 1, 3}\" width=\"175\" height=\"115\" \/><\/a>Dividing by [latex]\\left(x+3\\right)[\/latex] gives a remainder of 0, so \u20133 is a zero of the function. The polynomial can be written as\r\n<p style=\"text-align: center;\">[latex]\\left(x+3\\right)\\left(3{x}^{2}+1\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165134224528\">We can then set the quadratic equal to 0 and solve to find the other zeros of the function.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3{x}^{2}+1=0 \\\\ {x}^{2}=-\\frac{1}{3} \\\\ x=\\pm \\sqrt{-\\frac{1}{3}}=\\pm \\frac{i\\sqrt{3}}{3} \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137756902\">The zeros of [latex]f\\left(x\\right)[\/latex]\u00a0are \u20133 and [latex]\\pm \\frac{i\\sqrt{3}}{3}[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135640998\">Look at the graph of the function <em>f<\/em>. Notice that, at [latex]x=-3[\/latex], the graph crosses the <em>x<\/em>-axis, indicating an odd multiplicity (1) for the zero [latex]x=-3[\/latex]. Also note the presence of the two turning points. This means that, since there is a 3<sup>rd<\/sup> degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the <em>x<\/em>-intercepts for the function are shown. So either the multiplicity of [latex]x=-3[\/latex] is 1 and there are two complex solutions, which is what we found, or the multiplicity at [latex]x=-3[\/latex] is three. Either way, our result is correct.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010746\/CNX_Precalc_Figure_03_06_0022.jpg\" alt=\"Graph of a polynomial with its x-intercept at (-3, 0) labeled as &quot;Cross&quot;\" width=\"487\" height=\"289\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135538762\">Find the zeros of [latex]f\\left(x\\right)=2{x}^{3}+5{x}^{2}-11x+4[\/latex].<\/p>\r\n[reveal-answer q=\"518560\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"518560\"]\r\n\r\nThe zeros are [latex]\\text{-4, }\\frac{1}{2},\\text{ and 1}\\text{.}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]19264[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use the Linear Factorization Theorem to find polynomials with given zeros<\/h2>\r\n<p id=\"fs-id1165135502003\">A vital implication of the <strong>Fundamental Theorem of Algebra<\/strong>, as we stated above, is that a polynomial function of degree <em>n<\/em>\u00a0will have <em>n<\/em>\u00a0zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into <em>n<\/em>\u00a0factors. The <strong>Linear Factorization Theorem<\/strong> tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form (<em>x\u00a0\u2013\u00a0c<\/em>), where <em>c<\/em>\u00a0is a complex number.<\/p>\r\n<p id=\"eip-651\">Let <em>f<\/em>\u00a0be a polynomial function with real coefficients, and suppose [latex]a+bi\\text{, }b\\ne 0[\/latex],\u00a0is a zero of [latex]f\\left(x\\right)[\/latex].\u00a0Then, by the Factor Theorem, [latex]x-\\left(a+bi\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex].\u00a0For <em>f<\/em>\u00a0to have real coefficients, [latex]x-\\left(a-bi\\right)[\/latex]\u00a0must also be a factor of [latex]f\\left(x\\right)[\/latex].\u00a0This is true because any factor other than [latex]x-\\left(a-bi\\right)[\/latex],\u00a0when multiplied by [latex]x-\\left(a+bi\\right)[\/latex],\u00a0will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function <em>f<\/em>\u00a0with real coefficients has a complex zero [latex]a+bi[\/latex],\u00a0then the complex conjugate [latex]a-bi[\/latex]\u00a0must also be a zero of [latex]f\\left(x\\right)[\/latex]. This is called the <strong>Complex Conjugate Theorem<\/strong>.<\/p>\r\n\r\n<div id=\"fs-id1165137933095\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Complex Conjugate Theorem<\/h3>\r\n<p id=\"fs-id1165135436621\">According to the <strong>Linear Factorization Theorem<\/strong>, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\\left(x-c\\right)[\/latex], where <em>c<\/em>\u00a0is a complex number.<\/p>\r\n<p id=\"fs-id1165135443970\">If the polynomial function <em>f<\/em>\u00a0has real coefficients and a complex zero in the form [latex]a+bi[\/latex],\u00a0then the complex conjugate of the zero, [latex]a-bi[\/latex],\u00a0is also a zero.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137832786\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137832792\">How To: Given the zeros of a polynomial function [latex]f[\/latex] and a point [latex]\\left(c\\text{, }f(c)\\right)[\/latex]\u00a0on the graph of [latex]f[\/latex], use the Linear Factorization Theorem to find the polynomial function.<\/h3>\r\n<ol id=\"fs-id1165135534938\">\r\n \t<li>Use the zeros to construct the linear factors of the polynomial.<\/li>\r\n \t<li>Multiply the linear factors to expand the polynomial.<\/li>\r\n \t<li>Substitute [latex]\\left(c,f\\left(c\\right)\\right)[\/latex] into the function to determine the leading coefficient.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_06_07\" class=\"example\">\r\n<div id=\"fs-id1165137737023\" class=\"exercise\">\r\n<div id=\"fs-id1165137737025\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros<\/h3>\r\n<p id=\"fs-id1165134151154\">Find a fourth degree polynomial with real coefficients that has zeros of \u20133, 2, <em>i<\/em>, such that [latex]f\\left(-2\\right)=100[\/latex].<\/p>\r\n[reveal-answer q=\"348756\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348756\"]\r\n<p id=\"fs-id1165132964597\">Because [latex]x=i[\/latex]\u00a0is a zero, by the Complex Conjugate Theorem [latex]x=-i[\/latex]\u00a0is also a zero. The polynomial must have factors of [latex]\\left(x+3\\right),\\left(x - 2\\right),\\left(x-i\\right)[\/latex], and [latex]\\left(x+i\\right)[\/latex]. Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let\u2019s begin by multiplying these factors.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;f\\left(x\\right)=a\\left(x+3\\right)\\left(x - 2\\right)\\left(x-i\\right)\\left(x+i\\right)\\\\ &amp;f\\left(x\\right)=a\\left({x}^{2}+x - 6\\right)\\left({x}^{2}+1\\right)\\\\ &amp;f\\left(x\\right)=a\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165134065076\">We need to find <em>a<\/em> to ensure [latex]f\\left(-2\\right)=100[\/latex]. Substitute [latex]x=-2[\/latex] and [latex]f\\left(2\\right)=100[\/latex]\r\ninto [latex]f\\left(x\\right)[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}100&amp;=a\\left({\\left(-2\\right)}^{4}+{\\left(-2\\right)}^{3}-5{\\left(-2\\right)}^{2}+\\left(-2\\right)-6\\right) \\\\ 100&amp;=a\\left(-20\\right) \\\\ -5&amp;=a \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137407591\">So the polynomial function is<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=-5\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135406977\">or<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135154288\">We found that both <em>i<\/em>\u00a0and \u2013<em>i<\/em> were zeros, but only one of these zeros needed to be given. If <em>i<\/em>\u00a0is a zero of a polynomial with real coefficients, then <em>\u2013i<\/em>\u00a0must also be a zero of the polynomial because <em>\u2013i<\/em>\u00a0is the complex conjugate of <em>i<\/em>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135340588\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165135340596\"><strong>If 2 + 3<em>i<\/em>\u00a0were given as a zero of a polynomial with real coefficients, would 2 \u2013\u00a03<em>i<\/em>\u00a0also need to be a zero?<\/strong><\/p>\r\n<p id=\"fs-id1165134170187\"><em>Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.<\/em><\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134130177\">Find a third degree polynomial with real coefficients that has zeros of 5 and \u20132<em>i<\/em>\u00a0such that [latex]f\\left(1\\right)=10[\/latex].<\/p>\r\n[reveal-answer q=\"126079\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"126079\"]\r\n\r\n[latex]f\\left(x\\right)=-\\frac{1}{2}{x}^{3}+\\frac{5}{2}{x}^{2}-2x+10[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]100297[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use Descartes\u2019 Rule of Signs<\/h2>\r\n<p id=\"fs-id1165135177655\">There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order,<strong> Descartes\u2019 Rule of Signs<\/strong> tells us of a relationship between the number of sign changes in [latex]f\\left(x\\right)[\/latex] and the number of positive real zeros. For example, the polynomial function below has one sign change.<\/p>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02015816\/Descartes-fx.jpg\"><img class=\"alignnone size-full wp-image-15127\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02015816\/Descartes-fx.jpg\" alt=\"Polynomial function f(x)=x^4+x^3+x^2+x-2 showing one sign change between +x and -2\" width=\"487\" height=\"45\" \/><\/a>\r\n\r\n<span id=\"fs-id1165134378690\">This tells us that the function must have 1 positive real zero.<\/span>\r\n<p id=\"fs-id1165135206084\">There is a similar relationship between the number of sign changes in [latex]f\\left(-x\\right)[\/latex] and the number of negative real zeros.<\/p>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02020146\/Descartes-f-x.jpg\"><img class=\"alignnone size-full wp-image-15128\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02020146\/Descartes-f-x.jpg\" alt=\"f(-x)=^4-x^3+x^2-x-1 showing 3 sign changes\" width=\"487\" height=\"76\" \/><\/a>\r\n\r\n<span id=\"fs-id1165135152070\">In this case, [latex]f\\left(\\mathrm{-x}\\right)[\/latex] has 3 sign changes. This tells us that [latex]f\\left(x\\right)[\/latex] could have 3 or 1 negative real zeros.<\/span>\r\n<div id=\"fs-id1165137844271\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Descartes\u2019 Rule of Signs<\/h3>\r\n<p id=\"fs-id1165137844280\">According to <strong>Descartes\u2019 Rule of Signs<\/strong>, if we let [latex]f\\left(x\\right)={a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+...+{a}_{1}x+{a}_{0}[\/latex]\u00a0be a polynomial function with real coefficients:<\/p>\r\n\r\n<ul id=\"fs-id1165134351104\">\r\n \t<li>The number of positive real zeros is either equal to the number of sign changes of [latex]f\\left(x\\right)[\/latex] or is less than the number of sign changes by an even integer.<\/li>\r\n \t<li>The number of negative real zeros is either equal to the number of sign changes of [latex]f\\left(-x\\right)[\/latex] or is less than the number of sign changes by an even integer.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"Example_03_06_08\" class=\"example\">\r\n<div id=\"fs-id1165134149118\" class=\"exercise\">\r\n<div id=\"fs-id1165134149120\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Using Descartes\u2019 Rule of Signs<\/h3>\r\n<p id=\"fs-id1165134149125\">Use Descartes\u2019 Rule of Signs to determine the possible numbers of positive and negative real zeros for [latex]f\\left(x\\right)=-{x}^{4}-3{x}^{3}+6{x}^{2}-4x - 12[\/latex].<\/p>\r\n[reveal-answer q=\"858296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"858296\"]\r\n<p id=\"fs-id1165137932381\">Begin by determining the number of sign changes.<\/p>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/08\/03012922\/Screen-Shot-2015-08-04-at-12.31.54-PM.png\"><img class=\"aligncenter size-full wp-image-11813\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/08\/03012922\/Screen-Shot-2015-08-04-at-12.31.54-PM.png\" alt=\"Screen Shot 2015-08-04 at 12.31.54 PM\" width=\"534\" height=\"57\" \/><\/a>\r\n\r\nThere are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine [latex]f\\left(-x\\right)[\/latex] to determine the number of negative real roots.\r\n<p id=\"fs-id1165137696389\" style=\"text-align: center;\">[latex]\\begin{align}&amp;f\\left(-x\\right)=-{\\left(-x\\right)}^{4}-3{\\left(-x\\right)}^{3}+6{\\left(-x\\right)}^{2}-4\\left(-x\\right)-12 \\\\ &amp;f\\left(-x\\right)=-{x}^{4}+3{x}^{3}+6{x}^{2}+4x - 12 \\end{align}[\/latex]<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/08\/03012922\/Screen-Shot-2015-08-04-at-12.32.40-PM.png\"><img class=\"aligncenter size-full wp-image-11814\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/08\/03012922\/Screen-Shot-2015-08-04-at-12.32.40-PM.png\" alt=\"Screen Shot 2015-08-04 at 12.32.40 PM\" width=\"536\" height=\"52\" \/><\/a><\/p>\r\nAgain, there are two sign changes, so there are either 2 or 0 negative real roots.\r\n<p id=\"fs-id1165135383102\">There are four possibilities, as we can see below.<\/p>\r\n\r\n<table id=\"Table_03_06_01\" summary=\"..\"><colgroup> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th>Positive Real\r\nZeros<\/th>\r\n<th>Negative Real\r\nZeros<\/th>\r\n<th>Complex\r\nZeros<\/th>\r\n<th>Total\r\nZeros<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>2<\/td>\r\n<td>2<\/td>\r\n<td>0<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>0<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0<\/td>\r\n<td>2<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<td>4<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135628642\">We can confirm the numbers of positive and negative real roots by examining a graph of the function.\u00a0We can see from the graph in Figure 3 that the function has 0 positive real roots and 2 negative real roots.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010747\/CNX_Precalc_Figure_03_06_0072.jpg\" alt=\"Graph of f(x)=-x^4-3x^3+6x^2-4x-12 with x-intercepts at -4.42 and -1.\" width=\"487\" height=\"403\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n[\/hidden-answer]<b><\/b>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135177622\">Use Descartes\u2019 Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for [latex]f\\left(x\\right)=2{x}^{4}-10{x}^{3}+11{x}^{2}-15x+12[\/latex].\u00a0Use a graph to verify the numbers of positive and negative real zeros for the function.<\/p>\r\n[reveal-answer q=\"682370\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"682370\"]\r\n\r\nThere must be 4, 2, or 0 positive real roots and 0 negative real roots. The graph shows that there are 2 positive real zeros and 0 negative real zeros.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Solve real-world applications of polynomial equations<\/h2>\r\n<p id=\"fs-id1165135440219\">We have now introduced a variety of tools for solving polynomial equations. Let\u2019s use these tools to solve the bakery problem from the beginning of the section.<\/p>\r\n\r\n<div id=\"Example_03_06_09\" class=\"example\">\r\n<div id=\"fs-id1165135440229\" class=\"exercise\">\r\n<div id=\"fs-id1165135440231\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Solving Polynomial Equations<\/h3>\r\n<p id=\"fs-id1165135440236\">A new bakery offers decorated sheet cakes for children\u2019s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?<\/p>\r\n[reveal-answer q=\"448563\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"448563\"]\r\n<p id=\"fs-id1165135549014\">Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[\/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[\/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\\frac{1}{3}w[\/latex]. Let\u2019s write the volume of the cake in terms of width of the cake.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;V=\\left(w+4\\right)\\left(w\\right)\\left(\\frac{1}{3}w\\right)\\\\ &amp;V=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2}\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135173357\">Substitute the given volume into this equation.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;351=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2} &amp;&amp; \\text{Substitute 351 for }V. \\\\ &amp;1053={w}^{3}+4{w}^{2} &amp;&amp; \\text{Multiply both sides by 3}. \\\\ &amp;0={w}^{3}+4{w}^{2}-1053 &amp;&amp; \\text{Subtract 1053 from both sides}. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165134070008\">Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\\pm 3,\\pm 9,\\pm 13,\\pm 27,\\pm 39,\\pm 81,\\pm 117,\\pm 351[\/latex],\u00a0and [latex]\\pm 1053[\/latex].\u00a0We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let\u2019s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[\/latex].<\/p>\r\n\r\n<div id=\"eip-id1165133027598\" class=\"equation unnumbered\" style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012931\/Screen-Shot-2015-09-11-at-3.12.09-PM.png\"><img class=\"aligncenter wp-image-13122\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012931\/Screen-Shot-2015-09-11-at-3.12.09-PM.png\" alt=\"Synthetic Division with divisor = 1, and quotient = {1, 4, 0, -1053}. Solution is {1, 5, 5, -1048}\" width=\"163\" height=\"105\" \/><\/a><\/div>\r\n<p id=\"fs-id1165135693719\">Since 1 is not a solution, we will check [latex]x=3[\/latex].<\/p>\r\n<p id=\"eip-id7072398\"><span id=\"eip-id1169268826692\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010747\/CNX_Precalc_revised_eq12.png\" alt=\".\" width=\"164\" height=\"74\" \/><\/span><\/p>\r\n<p id=\"fs-id1165134037580\">Since 3 is not a solution either, we will test [latex]x=9[\/latex].<\/p>\r\n<p id=\"eip-id1168304942510\"><span id=\"eip-id1165198399599\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010747\/CNX_Precalc_revised_eq22.png\" alt=\".\" width=\"179\" height=\"67\" \/><\/span><\/p>\r\n<p id=\"fs-id1165135238389\">Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.<\/p>\r\n<p style=\"text-align: center;\">[latex]l=w+4=9+4=13\\text{ and }h=\\frac{1}{3}w=\\frac{1}{3}\\left(9\\right)=3[\/latex]<\/p>\r\n<p id=\"fs-id1165137843234\">The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137843251\">A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?<\/p>\r\n[reveal-answer q=\"342195\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342195\"]\r\n\r\n3 meters by 4 meters by 7 meters\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165135380122\">\r\n \t<li>To find [latex]f\\left(k\\right)[\/latex], determine the remainder of the polynomial [latex]f\\left(x\\right)[\/latex] when it is divided by [latex]x-k[\/latex].<\/li>\r\n \t<li><em>k<\/em>\u00a0is a zero of [latex]f\\left(x\\right)[\/latex]\u00a0if and only if [latex]\\left(x-k\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex].<\/li>\r\n \t<li>Each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading coefficient.<\/li>\r\n \t<li>When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.<\/li>\r\n \t<li>Synthetic division can be used to find the zeros of a polynomial function.<\/li>\r\n \t<li>According to the Fundamental Theorem, every polynomial function has at least one complex zero.<\/li>\r\n \t<li>Every polynomial function with degree greater than 0 has at least one complex zero.<\/li>\r\n \t<li>Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form [latex]\\left(x-c\\right)[\/latex], where <em>c<\/em>\u00a0is a complex number.<\/li>\r\n \t<li>The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer.<\/li>\r\n \t<li>The number of negative real zeros of a polynomial function is either the number of sign changes of [latex]f\\left(-x\\right)[\/latex]\u00a0or less than the number of sign changes by an even integer.<\/li>\r\n \t<li>Polynomial equations model many real-world scenarios. Solving the equations is most easily done by synthetic division.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165133281424\" class=\"definition\">\r\n \t<dt><strong>Descartes\u2019 Rule of Signs<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165133281430\">a rule that determines the maximum possible numbers of positive and negative real zeros based on the number of sign changes of [latex]f\\left(x\\right)[\/latex] and [latex]f\\left(-x\\right)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135459801\" class=\"definition\">\r\n \t<dt><strong>Factor Theorem<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135459806\"><em>k<\/em>\u00a0is a zero of polynomial function [latex]f\\left(x\\right)[\/latex] if and only if [latex]\\left(x-k\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133045332\" class=\"definition\">\r\n \t<dt><strong>Fundamental Theorem of Algebra<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165133045337\">a polynomial function with degree greater than 0 has at least one complex zero<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133045341\" class=\"definition\">\r\n \t<dt><strong>Linear Factorization Theorem<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165133045347\">allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\\left(x-c\\right)[\/latex],\u00a0where <em>c<\/em>\u00a0is a complex number<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135456904\" class=\"definition\">\r\n \t<dt><strong>Rational Zero Theorem<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135456910\">the possible rational zeros of a polynomial function have the form [latex]\\frac{p}{q}[\/latex] where <em>p<\/em>\u00a0is a factor of the constant term and <em>q<\/em>\u00a0is a factor of the leading coefficient.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137938597\" class=\"definition\">\r\n \t<dt><strong>Remainder Theorem<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137938602\">if a polynomial [latex]f\\left(x\\right)[\/latex] is divided by [latex]x-k[\/latex], then the remainder is equal to the value [latex]f\\left(k\\right)[\/latex]<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate a polynomial using the Remainder Theorem.<\/li>\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>Use Descartes\u2019 Rule of Signs.<\/li>\n<li>Solve real-world applications of polynomial equations<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137758829\">A new bakery offers decorated sheet cakes for children\u2019s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?<\/p>\n<p id=\"eip-151\">This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations.<\/p>\n<h2>Evaluate a polynomial using the Remainder Theorem<\/h2>\n<p id=\"fs-id1165135471230\">In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the <strong>Remainder Theorem<\/strong>. If the polynomial is divided by <em>x<\/em> \u2013\u00a0<em>k<\/em>, the remainder may be found quickly by evaluating the polynomial function at <em>k<\/em>, that is, <em>f<\/em>(<em>k<\/em>)\u00a0Let\u2019s walk through the proof of the theorem.<\/p>\n<p id=\"fs-id1165134085965\">Recall that the <strong>Division Algorithm<\/strong> states that, given a polynomial dividend <em>f<\/em>(<em>x<\/em>)\u00a0and a non-zero polynomial divisor <em>d<\/em>(<em>x<\/em>)\u00a0where the degree of\u00a0<em>d<\/em>(<em>x<\/em>) is less than or equal to the degree of <em>f<\/em>(<em>x<\/em>), there exist unique polynomials <em>q<\/em>(<em>x<\/em>)\u00a0and <em>r<\/em>(<em>x<\/em>)\u00a0such that<\/p>\n<div id=\"eip-753\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=d\\left(x\\right)q\\left(x\\right)+r\\left(x\\right)[\/latex]<\/div>\n<p id=\"fs-id1165134094600\">If the divisor, <em>d<\/em>(<em>x<\/em>), is <em>x<\/em> \u2013\u00a0<em>k<\/em>, this takes the form<\/p>\n<div id=\"eip-567\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex]<\/div>\n<p id=\"fs-id1165137447771\">Since the divisor <em>x<\/em> \u2013\u00a0<em>k<\/em>\u00a0is linear, the remainder will be a constant, <em>r<\/em>. And, if we evaluate this for <em>x<\/em> =\u00a0<em>k<\/em>, we have<\/p>\n<div id=\"eip-791\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}f\\left(k\\right)&=\\left(k-k\\right)q\\left(k\\right)+r \\\\ &=0\\cdot q\\left(k\\right)+r \\\\ &=r \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135572088\">In other words, <em>f<\/em>(<em>k<\/em>)\u00a0is the remainder obtained by dividing <em>f<\/em>(<em>x<\/em>)\u00a0by <em>x<\/em> \u2013\u00a0<em>k<\/em>.<\/p>\n<div id=\"fs-id1165134042705\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Remainder Theorem<\/h3>\n<p id=\"fs-id1165137725230\">If a polynomial [latex]f\\left(x\\right)[\/latex] is divided by <em>x<\/em> \u2013\u00a0<em>k<\/em>, then the remainder is the value [latex]f\\left(k\\right)[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165137823272\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165134130150\">How To: Given a polynomial function [latex]f[\/latex], evaluate [latex]f\\left(x\\right)[\/latex] at [latex]x=k[\/latex] using the Remainder Theorem.<\/h3>\n<ol id=\"fs-id1165137870981\">\n<li>Use synthetic division to divide the polynomial by [latex]x-k[\/latex].<\/li>\n<li>The remainder is the value [latex]f\\left(k\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_06_01\" class=\"example\">\n<div id=\"fs-id1165135199549\" class=\"exercise\">\n<div id=\"fs-id1165137727226\" class=\"problem textbox shaded\">\n<h3>Example 1: Using the Remainder Theorem to Evaluate a Polynomial<\/h3>\n<p id=\"fs-id1165137855032\">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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q540399\">Show Solution<\/span><\/p>\n<div id=\"q540399\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137771250\">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;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02010113\/Synthetic-Division-Example-1.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-15126\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02010113\/Synthetic-Division-Example-1.jpg\" alt=\"Synthetic division with divisor 2 and quotient {6,-1, -15, 2, -7}. Solution is {6,11,7, 16,25}\" width=\"255\" height=\"113\" \/><\/a><\/p>\n<p id=\"fs-id1165137932598\">The remainder is 25. Therefore, [latex]f\\left(2\\right)=25[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135397290\">We can check our answer by evaluating [latex]f\\left(2\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}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 \\\\ &=25 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137892503\">Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=2{x}^{5}-3{x}^{4}-9{x}^{3}+8{x}^{2}+2[\/latex]<br \/>\nat [latex]x=-3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q947133\">Show Solution<\/span><\/p>\n<div id=\"q947133\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(-3\\right)=-412[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Use the Factor Theorem to solve a polynomial equation<\/h2>\n<p id=\"fs-id1165137459796\">The <strong>Factor Theorem <\/strong>is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm tells us<\/p>\n<div id=\"eip-10\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex].<\/div>\n<p id=\"fs-id1165137463592\">If <em>k<\/em>\u00a0is a zero, then the remainder <em>r<\/em>\u00a0is [latex]f\\left(k\\right)=0[\/latex]\u00a0and [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+0[\/latex]\u00a0or [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)[\/latex].<\/p>\n<p id=\"fs-id1165135176357\">Notice, written in this form, <em>x<\/em>\u00a0\u2013\u00a0<em>k<\/em> is a factor of [latex]f\\left(x\\right)[\/latex]. We can conclude if <em>k\u00a0<\/em>is a zero of [latex]f\\left(x\\right)[\/latex], then [latex]x-k[\/latex] is a factor of [latex]f\\left(x\\right)[\/latex].<\/p>\n<p id=\"fs-id1165135684373\">Similarly, if [latex]x-k[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex],\u00a0then the remainder of the Division Algorithm [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex]\u00a0is 0. This tells us that <em>k<\/em>\u00a0is a zero.<\/p>\n<p id=\"fs-id1165132943504\">This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree <em>n<\/em>\u00a0in the complex number system will have <em>n<\/em>\u00a0zeros. We can use the Factor Theorem to completely factor a polynomial into the product of <em>n<\/em>\u00a0factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.<\/p>\n<div id=\"fs-id1165135173601\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Factor Theorem<\/h3>\n<p id=\"fs-id1165135575975\">According to the <strong>Factor Theorem<\/strong>, <em>k<\/em>\u00a0is a zero of [latex]f\\left(x\\right)[\/latex]\u00a0if and only if [latex]\\left(x-k\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165137809910\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165133356017\">How To: Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial.<strong><br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id1165135383467\">\n<li>Use synthetic division to divide the polynomial by [latex]\\left(x-k\\right)[\/latex].<\/li>\n<li>Confirm that the remainder is 0.<\/li>\n<li>Write the polynomial as the product of [latex]\\left(x-k\\right)[\/latex] and the quadratic quotient.<\/li>\n<li>If possible, factor the quadratic.<\/li>\n<li>Write the polynomial as the product of factors.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_06_02\" class=\"example\">\n<div id=\"fs-id1165134273636\" class=\"exercise\">\n<div id=\"fs-id1165134273638\" class=\"problem textbox shaded\">\n<h3>Example 2: Using the Factor Theorem to Solve a Polynomial Equation<\/h3>\n<p id=\"fs-id1165134198723\">Show that [latex]\\left(x+2\\right)[\/latex]\u00a0is a factor of [latex]{x}^{3}-6{x}^{2}-x+30[\/latex]. Find the remaining factors. Use the factors to determine the zeros of the <strong>polynomial<\/strong>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q884250\">Show Solution<\/span><\/p>\n<div id=\"q884250\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134109661\">We can use synthetic division to show that [latex]\\left(x+2\\right)[\/latex] is a factor of the polynomial.<\/p>\n<div id=\"eip-id1165134503086\" class=\"equation unnumbered\" style=\"text-align: center;\">\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.03.13-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-13110\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.03.13-PM.png\" alt=\"Synthetic division with divisor -2 and quotient {1, 6, -1, 30}. Solution is {1, -8, 15, 0}\" width=\"229\" height=\"143\" \/><\/a><\/p>\n<\/div>\n<p id=\"fs-id1165134261634\">The remainder is zero, so [latex]\\left(x+2\\right)[\/latex] is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left({x}^{2}-8x+15\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135440193\">We can factor the quadratic factor to write the polynomial as<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x - 3\\right)\\left(x - 5\\right)[\/latex]<\/p>\n<p id=\"fs-id1165137605065\">By the Factor Theorem, the zeros of [latex]{x}^{3}-6{x}^{2}-x+30[\/latex] are \u20132, 3, and 5.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134569514\">Use the Factor Theorem to find the zeros of [latex]f\\left(x\\right)={x}^{3}+4{x}^{2}-4x - 16[\/latex]\u00a0given that [latex]\\left(x - 2\\right)[\/latex]\u00a0is a factor of the polynomial.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q536719\">Show Solution<\/span><\/p>\n<div id=\"q536719\" class=\"hidden-answer\" style=\"display: none\">\n<p>The zeros are 2, \u20132, and \u20134.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm2367\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2367&theme=oea&iframe_resize_id=ohm2367\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use the Rational Zero Theorem to find rational zeros<\/h2>\n<p id=\"fs-id1165137660817\">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 to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading <strong>coefficient<\/strong> of the polynomial<\/p>\n<p id=\"fs-id1165135508309\">Consider a quadratic function with two zeros, [latex]x=\\frac{2}{5}[\/latex]\u00a0and [latex]x=\\frac{3}{4}[\/latex].<\/p>\n<p>By the Factor Theorem, these zeros have factors associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching factor.<span id=\"eip-id1165135315549\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010746\/CNX_Precalc_Figure_03_03_0225.jpg\" alt=\"This image shows x minus two fifths equals 0 or x minus three fourths equals 0. Beside this math is the sentence, 'Set each factor equal to 0.' Next it shows that five x minus 2 equals 0 or 4 x minus 3 equals 0. Beside this math is the sentence, 'Multiply both sides of the equation to eliminate fractions.' Next it shows that f of x is equal to (5 x minus 2) times (4 x minus 3). Beside this math is the sentence, 'Create the quadratic function, multiplying the factors.' Next it shows f of x equals 20 x squared minus 23 x plus 6. Beside this math is the sentence, 'Expand the polynomial.' The last equation shows f of x equals (5 times 4) times x squared minus 23 x plus (2 times 3). Set each factor equal to zero. Multiply both sides of the equation to eliminate fractions. Create the quadratic function, multiplying the factors. Expand the polynomial.\" width=\"975\" height=\"175\" \/><\/span><\/p>\n<p id=\"fs-id1165135485170\">Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3. Similarly, two of the factors from the leading coefficient, 20, are the two denominators from the original rational roots: 5 and 4.<\/p>\n<p id=\"fs-id1165137761317\">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 the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros.<\/p>\n<div id=\"fs-id1165137737069\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Rational Zero Theorem<\/h3>\n<p id=\"fs-id1165135530393\">The <strong>Rational Zero Theorem<\/strong> states that, if the polynomial [latex]f\\left(x\\right)={a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+...+{a}_{1}x+{a}_{0}[\/latex] has integer coefficients, then every rational zero of [latex]f\\left(x\\right)[\/latex]\u00a0has the form [latex]\\frac{p}{q}[\/latex] where <em>p<\/em>\u00a0is a factor of the constant term [latex]{a}_{0}[\/latex] and <em>q<\/em>\u00a0is a factor of the leading coefficient [latex]{a}_{n}[\/latex].<\/p>\n<p id=\"fs-id1165137736282\">When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.<\/p>\n<\/div>\n<div id=\"fs-id1165137736287\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137446561\">How To: Given a polynomial function [latex]f\\left(x\\right)[\/latex], use the Rational Zero Theorem to find rational zeros.<\/h3>\n<ol id=\"fs-id1165137662157\">\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 <em>p<\/em>\u00a0is a factor of the constant term and <em>q<\/em>\u00a0is 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<\/div>\n<div id=\"Example_03_06_03\" class=\"example\">\n<div id=\"fs-id1165135319520\" class=\"exercise\">\n<div id=\"fs-id1165135319523\" class=\"problem textbox shaded\">\n<h3>Example 3: Listing All Possible Rational Zeros<\/h3>\n<p id=\"fs-id1165135640968\">List all possible rational zeros of [latex]f\\left(x\\right)=2{x}^{4}-5{x}^{3}+{x}^{2}-4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q655905\">Show Solution<\/span><\/p>\n<div id=\"q655905\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134179675\">The only possible rational zeros of [latex]f\\left(x\\right)[\/latex]\u00a0are the quotients of the factors of the last term, \u20134, and the factors of the leading coefficient, 2.<\/p>\n<p id=\"fs-id1165132972963\">The constant term is \u20134; the factors of \u20134 are [latex]p=\\pm 1,\\pm 2,\\pm 4[\/latex].<\/p>\n<p id=\"fs-id1165137571546\">The leading coefficient is 2; the factors of 2 are [latex]q=\\pm 1,\\pm 2[\/latex].<\/p>\n<p id=\"fs-id1165135557830\">If any of the four real zeros are rational zeros, then they will be of one of the following factors of \u20134 divided by one of the factors of 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\frac{p}{q}=\\pm \\frac{1}{1},\\pm \\frac{1}{2} && \\frac{p}{q}=\\pm \\frac{2}{1},\\pm \\frac{2}{2} && \\frac{p}{q}=\\pm \\frac{4}{1},\\pm \\frac{4}{2}\\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135351678\">Note that [latex]\\frac{2}{2}=1[\/latex]\u00a0and [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}}=\\pm 1,\\pm 2,\\pm 4,\\pm \\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_03_06_04\" class=\"example\">\n<div id=\"fs-id1165137611543\" class=\"exercise\">\n<div id=\"fs-id1165137611546\" class=\"problem textbox shaded\">\n<h3>Example 4: Using the Rational Zero Theorem to Find Rational Zeros<\/h3>\n<p id=\"fs-id1165137459844\">Use the Rational Zero Theorem to find the rational zeros of [latex]f\\left(x\\right)=2{x}^{3}+{x}^{2}-4x+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q565085\">Show Solution<\/span><\/p>\n<div id=\"q565085\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135154382\">The Rational Zero Theorem tells us that if [latex]\\frac{p}{q}[\/latex] is a zero of [latex]f\\left(x\\right)[\/latex],\u00a0then <em>p<\/em>\u00a0is a factor of 1 and <em>q<\/em>\u00a0is a factor of 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{p}{q}&=\\frac{\\text{factor of constant term}}{\\text{factor of leading coefficient}} \\\\[1mm] &=\\frac{\\text{factor of 1}}{\\text{factor of 2}} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135209688\">The factors of 1 are [latex]\\pm 1[\/latex] and the factors of 2 are [latex]\\pm 1[\/latex] and [latex]\\pm 2[\/latex]. The possible values for [latex]\\frac{p}{q}[\/latex] are [latex]\\pm 1[\/latex] and [latex]\\pm \\frac{1}{2}[\/latex]. These are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for <em>x<\/em>\u00a0in [latex]f\\left(x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(-1\\right)=2{\\left(-1\\right)}^{3}+{\\left(-1\\right)}^{2}-4\\left(-1\\right)+1=4[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(1\\right)=2{\\left(1\\right)}^{3}+{\\left(1\\right)}^{2}-4\\left(1\\right)+1=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(-\\frac{1}{2}\\right)=2{\\left(-\\frac{1}{2}\\right)}^{3}+{\\left(-\\frac{1}{2}\\right)}^{2}-4\\left(-\\frac{1}{2}\\right)+1=3[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(\\frac{1}{2}\\right)=2{\\left(\\frac{1}{2}\\right)}^{3}+{\\left(\\frac{1}{2}\\right)}^{2}-4\\left(\\frac{1}{2}\\right)+1=-\\frac{1}{2}[\/latex]<\/p>\n<p id=\"fs-id1165137731309\">Of those, [latex]-1,-\\frac{1}{2},\\text{ and }\\frac{1}{2}[\/latex] are not zeros of [latex]f\\left(x\\right)[\/latex]. 1 is the only rational zero of [latex]f\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134198689\">Use the Rational Zero Theorem to find the rational zeros of [latex]f\\left(x\\right)={x}^{3}-5{x}^{2}+2x+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q544964\">Show Solution<\/span><\/p>\n<div id=\"q544964\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are no rational zeros.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm43500\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=43500&theme=oea&iframe_resize_id=ohm43500\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Find zeros of a polynomial function<\/h2>\n<p id=\"fs-id1165135530405\">The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use <span class=\"no-emphasis\">synthetic division<\/span> repeatedly to determine all of the zeros of a polynomial function.<\/p>\n<div id=\"fs-id1165134155131\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165133364811\">How To: Given a polynomial function [latex]f[\/latex], use synthetic division to find its zeros.<\/h3>\n<ol id=\"fs-id1165135400178\">\n<li>Use the Rational Zero Theorem to list all possible rational zeros of the function.<\/li>\n<li>Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.<\/li>\n<li>Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.<\/li>\n<li>Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_06_05\" class=\"example\">\n<div id=\"fs-id1165137761977\" class=\"exercise\">\n<div id=\"fs-id1165134198663\" class=\"problem textbox shaded\">\n<h3>Example 5: Finding the Zeros of a Polynomial Function with Repeated Real Zeros<\/h3>\n<p id=\"fs-id1165134198668\">Find the zeros of [latex]f\\left(x\\right)=4{x}^{3}-3x - 1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q663920\">Show Solution<\/span><\/p>\n<div id=\"q663920\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135393366\">The Rational Zero Theorem tells us that if [latex]\\frac{p}{q}[\/latex] is a zero of [latex]f\\left(x\\right)[\/latex], then <em>p\u00a0<\/em>is a factor of \u20131 and\u00a0<em>q<\/em>\u00a0is a factor of 4.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{p}{q}&=\\frac{\\text{factor of constant term}}{\\text{factor of leading coefficient}} \\\\ &=\\frac{\\text{factor of -1}}{\\text{factor of 4}} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135501059\">The factors of \u20131 are [latex]\\pm 1[\/latex]\u00a0and the factors of 4 are [latex]\\pm 1,\\pm 2[\/latex], and [latex]\\pm 4[\/latex]. The possible values for [latex]\\frac{p}{q}[\/latex] are [latex]\\pm 1,\\pm \\frac{1}{2}[\/latex], and [latex]\\pm \\frac{1}{4}[\/latex].<\/p>\n<p>These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let\u2019s begin with 1.<\/p>\n<div id=\"eip-id1165137397701\" class=\"equation unnumbered\" style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.05.49-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-13113\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.05.49-PM.png\" alt=\"Synthetic division with 1 as the divisor and {4, 0, -3, -1} as the quotient. Solution is {4, 4, 1, 0}\" width=\"166\" height=\"122\" \/><\/a><\/div>\n<p id=\"fs-id1165137934395\">Dividing by [latex]\\left(x - 1\\right)[\/latex]\u00a0gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x - 1\\right)\\left(4{x}^{2}+4x+1\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137605971\">The quadratic is a perfect square. [latex]f\\left(x\\right)[\/latex]\u00a0can be written as<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x - 1\\right){\\left(2x+1\\right)}^{2}[\/latex].<\/p>\n<p id=\"fs-id1165135564194\">We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2x+1=0 \\\\ x=-\\frac{1}{2} \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165135651485\">The zeros of the function are 1 and [latex]-\\frac{1}{2}[\/latex] with multiplicity 2.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Look at the graph of the function <em>f<\/em>\u00a0in Figure 1. Notice, at [latex]x=-0.5[\/latex], the graph bounces off the <em>x<\/em>-axis, indicating the even multiplicity (2,4,6\u2026) for the zero \u20130.5.\u00a0At [latex]x=1[\/latex], the graph crosses the <em>x<\/em>-axis, indicating the odd multiplicity (1,3,5\u2026) for the zero [latex]x=1[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010746\/CNX_Precalc_Figure_03_06_0012.jpg\" alt=\"Graph of a polynomial that have its local maximum at (-0.5, 0) labeled as\" width=\"487\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use the Fundamental Theorem of Algebra<\/h2>\n<p id=\"fs-id1165135547255\">Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The <strong>Fundamental Theorem of Algebra <\/strong>tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.<\/p>\n<p id=\"fs-id1165137771429\">Suppose <em>f<\/em>\u00a0is a polynomial function of degree four, and [latex]f\\left(x\\right)=0[\/latex]. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it [latex]{c}_{1}[\/latex]. By the Factor Theorem, we can write [latex]f\\left(x\\right)[\/latex] as a product of [latex]x-{c}_{\\text{1}}[\/latex] and a polynomial quotient. Since [latex]x-{c}_{\\text{1}}[\/latex] is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it [latex]{c}_{\\text{2}}[\/latex]. So we can write the polynomial quotient as a product of [latex]x-{c}_{\\text{2}}[\/latex] and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of [latex]f\\left(x\\right)[\/latex].<\/p>\n<div id=\"fs-id1165135693782\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The <strong>Fundamental Theorem of Algebra<\/strong> states that, if <em>f(x)<\/em> is a polynomial of degree <em>n &gt; 0<\/em>, then <em>f(x)<\/em> has at least one complex zero.<\/h3>\n<p id=\"fs-id1165135409342\">We can use this theorem to argue that, if [latex]f\\left(x\\right)[\/latex] is a polynomial of degree [latex]n>0[\/latex], and <em>a<\/em>\u00a0is a non-zero real number, then [latex]f\\left(x\\right)[\/latex] has exactly <em>n<\/em>\u00a0linear factors<\/p>\n<div id=\"eip-750\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\left(x-{c}_{1}\\right)\\left(x-{c}_{2}\\right)...\\left(x-{c}_{n}\\right)[\/latex]<\/div>\n<p id=\"fs-id1165137889849\">where [latex]{c}_{1},{c}_{2},...,{c}_{n}[\/latex] are complex numbers. Therefore, [latex]f\\left(x\\right)[\/latex] has <em>n<\/em>\u00a0roots if we allow for multiplicities.<\/p>\n<\/div>\n<div id=\"fs-id1165135481201\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165135481208\"><strong>Does every polynomial have at least one imaginary zero?<\/strong><\/p>\n<p id=\"fs-id1165137883746\"><em>No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.<\/em><\/p>\n<\/div>\n<div id=\"Example_03_06_06\" class=\"example\">\n<div id=\"fs-id1165135658206\" class=\"exercise\">\n<div id=\"fs-id1165135658208\" class=\"problem textbox shaded\">\n<h3>Example 6: Finding the Zeros of a Polynomial Function with Complex Zeros<\/h3>\n<p id=\"fs-id1165135658213\">Find the zeros of [latex]f\\left(x\\right)=3{x}^{3}+9{x}^{2}+x+3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q278128\">Show Solution<\/span><\/p>\n<div id=\"q278128\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137430512\">The Rational Zero Theorem tells us that if [latex]\\frac{p}{q}[\/latex] is a zero of [latex]f\\left(x\\right)[\/latex], then <em>p<\/em>\u00a0is a factor of 3 and\u00a0<em>q<\/em>\u00a0is a factor of 3.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{p}{q}&=\\frac{\\text{factor of constant term}}{\\text{factor of leading coefficient}} \\\\ &=\\frac{\\text{factor of 3}}{\\text{factor of 3}} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165134212112\">The factors of 3 are [latex]\\pm 1[\/latex] and [latex]\\pm 3[\/latex]. The possible values for [latex]\\frac{p}{q}[\/latex], and therefore the possible rational zeros for the function, are [latex]\\pm 3, \\pm 1,\\text{ and }\\pm \\frac{1}{3}[\/latex]. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let\u2019s begin with \u20133.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-13116\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png\" alt=\"Synthetic division with divisor = -3 and quotient = {3, 9, 1, 3}\" width=\"175\" height=\"115\" \/><\/a>Dividing by [latex]\\left(x+3\\right)[\/latex] gives a remainder of 0, so \u20133 is a zero of the function. The polynomial can be written as<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+3\\right)\\left(3{x}^{2}+1\\right)[\/latex]<\/p>\n<p id=\"fs-id1165134224528\">We can then set the quadratic equal to 0 and solve to find the other zeros of the function.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3{x}^{2}+1=0 \\\\ {x}^{2}=-\\frac{1}{3} \\\\ x=\\pm \\sqrt{-\\frac{1}{3}}=\\pm \\frac{i\\sqrt{3}}{3} \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137756902\">The zeros of [latex]f\\left(x\\right)[\/latex]\u00a0are \u20133 and [latex]\\pm \\frac{i\\sqrt{3}}{3}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135640998\">Look at the graph of the function <em>f<\/em>. Notice that, at [latex]x=-3[\/latex], the graph crosses the <em>x<\/em>-axis, indicating an odd multiplicity (1) for the zero [latex]x=-3[\/latex]. Also note the presence of the two turning points. This means that, since there is a 3<sup>rd<\/sup> degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the <em>x<\/em>-intercepts for the function are shown. So either the multiplicity of [latex]x=-3[\/latex] is 1 and there are two complex solutions, which is what we found, or the multiplicity at [latex]x=-3[\/latex] is three. Either way, our result is correct.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010746\/CNX_Precalc_Figure_03_06_0022.jpg\" alt=\"Graph of a polynomial with its x-intercept at (-3, 0) labeled as &quot;Cross&quot;\" width=\"487\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135538762\">Find the zeros of [latex]f\\left(x\\right)=2{x}^{3}+5{x}^{2}-11x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q518560\">Show Solution<\/span><\/p>\n<div id=\"q518560\" class=\"hidden-answer\" style=\"display: none\">\n<p>The zeros are [latex]\\text{-4, }\\frac{1}{2},\\text{ and 1}\\text{.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm19264\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19264&theme=oea&iframe_resize_id=ohm19264\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use the Linear Factorization Theorem to find polynomials with given zeros<\/h2>\n<p id=\"fs-id1165135502003\">A vital implication of the <strong>Fundamental Theorem of Algebra<\/strong>, as we stated above, is that a polynomial function of degree <em>n<\/em>\u00a0will have <em>n<\/em>\u00a0zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into <em>n<\/em>\u00a0factors. The <strong>Linear Factorization Theorem<\/strong> tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form (<em>x\u00a0\u2013\u00a0c<\/em>), where <em>c<\/em>\u00a0is a complex number.<\/p>\n<p id=\"eip-651\">Let <em>f<\/em>\u00a0be a polynomial function with real coefficients, and suppose [latex]a+bi\\text{, }b\\ne 0[\/latex],\u00a0is a zero of [latex]f\\left(x\\right)[\/latex].\u00a0Then, by the Factor Theorem, [latex]x-\\left(a+bi\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex].\u00a0For <em>f<\/em>\u00a0to have real coefficients, [latex]x-\\left(a-bi\\right)[\/latex]\u00a0must also be a factor of [latex]f\\left(x\\right)[\/latex].\u00a0This is true because any factor other than [latex]x-\\left(a-bi\\right)[\/latex],\u00a0when multiplied by [latex]x-\\left(a+bi\\right)[\/latex],\u00a0will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function <em>f<\/em>\u00a0with real coefficients has a complex zero [latex]a+bi[\/latex],\u00a0then the complex conjugate [latex]a-bi[\/latex]\u00a0must also be a zero of [latex]f\\left(x\\right)[\/latex]. This is called the <strong>Complex Conjugate Theorem<\/strong>.<\/p>\n<div id=\"fs-id1165137933095\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Complex Conjugate Theorem<\/h3>\n<p id=\"fs-id1165135436621\">According to the <strong>Linear Factorization Theorem<\/strong>, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\\left(x-c\\right)[\/latex], where <em>c<\/em>\u00a0is a complex number.<\/p>\n<p id=\"fs-id1165135443970\">If the polynomial function <em>f<\/em>\u00a0has real coefficients and a complex zero in the form [latex]a+bi[\/latex],\u00a0then the complex conjugate of the zero, [latex]a-bi[\/latex],\u00a0is also a zero.<\/p>\n<\/div>\n<div id=\"fs-id1165137832786\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137832792\">How To: Given the zeros of a polynomial function [latex]f[\/latex] and a point [latex]\\left(c\\text{, }f(c)\\right)[\/latex]\u00a0on the graph of [latex]f[\/latex], use the Linear Factorization Theorem to find the polynomial function.<\/h3>\n<ol id=\"fs-id1165135534938\">\n<li>Use the zeros to construct the linear factors of the polynomial.<\/li>\n<li>Multiply the linear factors to expand the polynomial.<\/li>\n<li>Substitute [latex]\\left(c,f\\left(c\\right)\\right)[\/latex] into the function to determine the leading coefficient.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_06_07\" class=\"example\">\n<div id=\"fs-id1165137737023\" class=\"exercise\">\n<div id=\"fs-id1165137737025\" class=\"problem textbox shaded\">\n<h3>Example 7: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros<\/h3>\n<p id=\"fs-id1165134151154\">Find a fourth degree polynomial with real coefficients that has zeros of \u20133, 2, <em>i<\/em>, such that [latex]f\\left(-2\\right)=100[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348756\">Show Solution<\/span><\/p>\n<div id=\"q348756\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165132964597\">Because [latex]x=i[\/latex]\u00a0is a zero, by the Complex Conjugate Theorem [latex]x=-i[\/latex]\u00a0is also a zero. The polynomial must have factors of [latex]\\left(x+3\\right),\\left(x - 2\\right),\\left(x-i\\right)[\/latex], and [latex]\\left(x+i\\right)[\/latex]. Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let\u2019s begin by multiplying these factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&f\\left(x\\right)=a\\left(x+3\\right)\\left(x - 2\\right)\\left(x-i\\right)\\left(x+i\\right)\\\\ &f\\left(x\\right)=a\\left({x}^{2}+x - 6\\right)\\left({x}^{2}+1\\right)\\\\ &f\\left(x\\right)=a\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)\\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165134065076\">We need to find <em>a<\/em> to ensure [latex]f\\left(-2\\right)=100[\/latex]. Substitute [latex]x=-2[\/latex] and [latex]f\\left(2\\right)=100[\/latex]<br \/>\ninto [latex]f\\left(x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}100&=a\\left({\\left(-2\\right)}^{4}+{\\left(-2\\right)}^{3}-5{\\left(-2\\right)}^{2}+\\left(-2\\right)-6\\right) \\\\ 100&=a\\left(-20\\right) \\\\ -5&=a \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137407591\">So the polynomial function is<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=-5\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135406977\">or<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135154288\">We found that both <em>i<\/em>\u00a0and \u2013<em>i<\/em> were zeros, but only one of these zeros needed to be given. If <em>i<\/em>\u00a0is a zero of a polynomial with real coefficients, then <em>\u2013i<\/em>\u00a0must also be a zero of the polynomial because <em>\u2013i<\/em>\u00a0is the complex conjugate of <em>i<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135340588\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165135340596\"><strong>If 2 + 3<em>i<\/em>\u00a0were given as a zero of a polynomial with real coefficients, would 2 \u2013\u00a03<em>i<\/em>\u00a0also need to be a zero?<\/strong><\/p>\n<p id=\"fs-id1165134170187\"><em>Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.<\/em><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134130177\">Find a third degree polynomial with real coefficients that has zeros of 5 and \u20132<em>i<\/em>\u00a0such that [latex]f\\left(1\\right)=10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q126079\">Show Solution<\/span><\/p>\n<div id=\"q126079\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(x\\right)=-\\frac{1}{2}{x}^{3}+\\frac{5}{2}{x}^{2}-2x+10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm100297\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100297&theme=oea&iframe_resize_id=ohm100297\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use Descartes\u2019 Rule of Signs<\/h2>\n<p id=\"fs-id1165135177655\">There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order,<strong> Descartes\u2019 Rule of Signs<\/strong> tells us of a relationship between the number of sign changes in [latex]f\\left(x\\right)[\/latex] and the number of positive real zeros. For example, the polynomial function below has one sign change.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02015816\/Descartes-fx.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-15127\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02015816\/Descartes-fx.jpg\" alt=\"Polynomial function f(x)=x^4+x^3+x^2+x-2 showing one sign change between +x and -2\" width=\"487\" height=\"45\" \/><\/a><\/p>\n<p><span id=\"fs-id1165134378690\">This tells us that the function must have 1 positive real zero.<\/span><\/p>\n<p id=\"fs-id1165135206084\">There is a similar relationship between the number of sign changes in [latex]f\\left(-x\\right)[\/latex] and the number of negative real zeros.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02020146\/Descartes-f-x.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-15128\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/08\/02020146\/Descartes-f-x.jpg\" alt=\"f(-x)=^4-x^3+x^2-x-1 showing 3 sign changes\" width=\"487\" height=\"76\" \/><\/a><\/p>\n<p><span id=\"fs-id1165135152070\">In this case, [latex]f\\left(\\mathrm{-x}\\right)[\/latex] has 3 sign changes. This tells us that [latex]f\\left(x\\right)[\/latex] could have 3 or 1 negative real zeros.<\/span><\/p>\n<div id=\"fs-id1165137844271\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Descartes\u2019 Rule of Signs<\/h3>\n<p id=\"fs-id1165137844280\">According to <strong>Descartes\u2019 Rule of Signs<\/strong>, if we let [latex]f\\left(x\\right)={a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+...+{a}_{1}x+{a}_{0}[\/latex]\u00a0be a polynomial function with real coefficients:<\/p>\n<ul id=\"fs-id1165134351104\">\n<li>The number of positive real zeros is either equal to the number of sign changes of [latex]f\\left(x\\right)[\/latex] or is less than the number of sign changes by an even integer.<\/li>\n<li>The number of negative real zeros is either equal to the number of sign changes of [latex]f\\left(-x\\right)[\/latex] or is less than the number of sign changes by an even integer.<\/li>\n<\/ul>\n<\/div>\n<div id=\"Example_03_06_08\" class=\"example\">\n<div id=\"fs-id1165134149118\" class=\"exercise\">\n<div id=\"fs-id1165134149120\" class=\"problem textbox shaded\">\n<h3>Example 7: Using Descartes\u2019 Rule of Signs<\/h3>\n<p id=\"fs-id1165134149125\">Use Descartes\u2019 Rule of Signs to determine the possible numbers of positive and negative real zeros for [latex]f\\left(x\\right)=-{x}^{4}-3{x}^{3}+6{x}^{2}-4x - 12[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q858296\">Show Solution<\/span><\/p>\n<div id=\"q858296\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137932381\">Begin by determining the number of sign changes.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/08\/03012922\/Screen-Shot-2015-08-04-at-12.31.54-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-11813\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/08\/03012922\/Screen-Shot-2015-08-04-at-12.31.54-PM.png\" alt=\"Screen Shot 2015-08-04 at 12.31.54 PM\" width=\"534\" height=\"57\" \/><\/a><\/p>\n<p>There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine [latex]f\\left(-x\\right)[\/latex] to determine the number of negative real roots.<\/p>\n<p id=\"fs-id1165137696389\" style=\"text-align: center;\">[latex]\\begin{align}&f\\left(-x\\right)=-{\\left(-x\\right)}^{4}-3{\\left(-x\\right)}^{3}+6{\\left(-x\\right)}^{2}-4\\left(-x\\right)-12 \\\\ &f\\left(-x\\right)=-{x}^{4}+3{x}^{3}+6{x}^{2}+4x - 12 \\end{align}[\/latex]<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/08\/03012922\/Screen-Shot-2015-08-04-at-12.32.40-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-11814\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/08\/03012922\/Screen-Shot-2015-08-04-at-12.32.40-PM.png\" alt=\"Screen Shot 2015-08-04 at 12.32.40 PM\" width=\"536\" height=\"52\" \/><\/a><\/p>\n<p>Again, there are two sign changes, so there are either 2 or 0 negative real roots.<\/p>\n<p id=\"fs-id1165135383102\">There are four possibilities, as we can see below.<\/p>\n<table id=\"Table_03_06_01\" summary=\"..\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th>Positive Real<br \/>\nZeros<\/th>\n<th>Negative Real<br \/>\nZeros<\/th>\n<th>Complex<br \/>\nZeros<\/th>\n<th>Total<br \/>\nZeros<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>2<\/td>\n<td>2<\/td>\n<td>0<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>0<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>2<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>0<\/td>\n<td>4<\/td>\n<td>4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135628642\">We can confirm the numbers of positive and negative real roots by examining a graph of the function.\u00a0We can see from the graph in Figure 3 that the function has 0 positive real roots and 2 negative real roots.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010747\/CNX_Precalc_Figure_03_06_0072.jpg\" alt=\"Graph of f(x)=-x^4-3x^3+6x^2-4x-12 with x-intercepts at -4.42 and -1.\" width=\"487\" height=\"403\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135177622\">Use Descartes\u2019 Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for [latex]f\\left(x\\right)=2{x}^{4}-10{x}^{3}+11{x}^{2}-15x+12[\/latex].\u00a0Use a graph to verify the numbers of positive and negative real zeros for the function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q682370\">Show Solution<\/span><\/p>\n<div id=\"q682370\" class=\"hidden-answer\" style=\"display: none\">\n<p>There must be 4, 2, or 0 positive real roots and 0 negative real roots. The graph shows that there are 2 positive real zeros and 0 negative real zeros.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Solve real-world applications of polynomial equations<\/h2>\n<p id=\"fs-id1165135440219\">We have now introduced a variety of tools for solving polynomial equations. Let\u2019s use these tools to solve the bakery problem from the beginning of the section.<\/p>\n<div id=\"Example_03_06_09\" class=\"example\">\n<div id=\"fs-id1165135440229\" class=\"exercise\">\n<div id=\"fs-id1165135440231\" class=\"problem textbox shaded\">\n<h3>Example 8: Solving Polynomial Equations<\/h3>\n<p id=\"fs-id1165135440236\">A new bakery offers decorated sheet cakes for children\u2019s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q448563\">Show Solution<\/span><\/p>\n<div id=\"q448563\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135549014\">Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[\/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[\/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\\frac{1}{3}w[\/latex]. Let\u2019s write the volume of the cake in terms of width of the cake.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&V=\\left(w+4\\right)\\left(w\\right)\\left(\\frac{1}{3}w\\right)\\\\ &V=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2}\\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135173357\">Substitute the given volume into this equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&351=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2} && \\text{Substitute 351 for }V. \\\\ &1053={w}^{3}+4{w}^{2} && \\text{Multiply both sides by 3}. \\\\ &0={w}^{3}+4{w}^{2}-1053 && \\text{Subtract 1053 from both sides}. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165134070008\">Descartes&#8217; rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\\pm 3,\\pm 9,\\pm 13,\\pm 27,\\pm 39,\\pm 81,\\pm 117,\\pm 351[\/latex],\u00a0and [latex]\\pm 1053[\/latex].\u00a0We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let\u2019s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[\/latex].<\/p>\n<div id=\"eip-id1165133027598\" class=\"equation unnumbered\" style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012931\/Screen-Shot-2015-09-11-at-3.12.09-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-13122\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012931\/Screen-Shot-2015-09-11-at-3.12.09-PM.png\" alt=\"Synthetic Division with divisor = 1, and quotient = {1, 4, 0, -1053}. Solution is {1, 5, 5, -1048}\" width=\"163\" height=\"105\" \/><\/a><\/div>\n<p id=\"fs-id1165135693719\">Since 1 is not a solution, we will check [latex]x=3[\/latex].<\/p>\n<p id=\"eip-id7072398\"><span id=\"eip-id1169268826692\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010747\/CNX_Precalc_revised_eq12.png\" alt=\".\" width=\"164\" height=\"74\" \/><\/span><\/p>\n<p id=\"fs-id1165134037580\">Since 3 is not a solution either, we will test [latex]x=9[\/latex].<\/p>\n<p id=\"eip-id1168304942510\"><span id=\"eip-id1165198399599\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010747\/CNX_Precalc_revised_eq22.png\" alt=\".\" width=\"179\" height=\"67\" \/><\/span><\/p>\n<p id=\"fs-id1165135238389\">Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.<\/p>\n<p style=\"text-align: center;\">[latex]l=w+4=9+4=13\\text{ and }h=\\frac{1}{3}w=\\frac{1}{3}\\left(9\\right)=3[\/latex]<\/p>\n<p id=\"fs-id1165137843234\">The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137843251\">A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q342195\">Show Solution<\/span><\/p>\n<div id=\"q342195\" class=\"hidden-answer\" style=\"display: none\">\n<p>3 meters by 4 meters by 7 meters<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165135380122\">\n<li>To find [latex]f\\left(k\\right)[\/latex], determine the remainder of the polynomial [latex]f\\left(x\\right)[\/latex] when it is divided by [latex]x-k[\/latex].<\/li>\n<li><em>k<\/em>\u00a0is a zero of [latex]f\\left(x\\right)[\/latex]\u00a0if and only if [latex]\\left(x-k\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex].<\/li>\n<li>Each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading coefficient.<\/li>\n<li>When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.<\/li>\n<li>Synthetic division can be used to find the zeros of a polynomial function.<\/li>\n<li>According to the Fundamental Theorem, every polynomial function has at least one complex zero.<\/li>\n<li>Every polynomial function with degree greater than 0 has at least one complex zero.<\/li>\n<li>Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form [latex]\\left(x-c\\right)[\/latex], where <em>c<\/em>\u00a0is a complex number.<\/li>\n<li>The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer.<\/li>\n<li>The number of negative real zeros of a polynomial function is either the number of sign changes of [latex]f\\left(-x\\right)[\/latex]\u00a0or less than the number of sign changes by an even integer.<\/li>\n<li>Polynomial equations model many real-world scenarios. Solving the equations is most easily done by synthetic division.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165133281424\" class=\"definition\">\n<dt><strong>Descartes\u2019 Rule of Signs<\/strong><\/dt>\n<dd id=\"fs-id1165133281430\">a rule that determines the maximum possible numbers of positive and negative real zeros based on the number of sign changes of [latex]f\\left(x\\right)[\/latex] and [latex]f\\left(-x\\right)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135459801\" class=\"definition\">\n<dt><strong>Factor Theorem<\/strong><\/dt>\n<dd id=\"fs-id1165135459806\"><em>k<\/em>\u00a0is a zero of polynomial function [latex]f\\left(x\\right)[\/latex] if and only if [latex]\\left(x-k\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133045332\" class=\"definition\">\n<dt><strong>Fundamental Theorem of Algebra<\/strong><\/dt>\n<dd id=\"fs-id1165133045337\">a polynomial function with degree greater than 0 has at least one complex zero<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133045341\" class=\"definition\">\n<dt><strong>Linear Factorization Theorem<\/strong><\/dt>\n<dd id=\"fs-id1165133045347\">allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\\left(x-c\\right)[\/latex],\u00a0where <em>c<\/em>\u00a0is a complex number<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135456904\" class=\"definition\">\n<dt><strong>Rational Zero Theorem<\/strong><\/dt>\n<dd id=\"fs-id1165135456910\">the possible rational zeros of a polynomial function have the form [latex]\\frac{p}{q}[\/latex] where <em>p<\/em>\u00a0is a factor of the constant term and <em>q<\/em>\u00a0is a factor of the leading coefficient.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137938597\" class=\"definition\">\n<dt><strong>Remainder Theorem<\/strong><\/dt>\n<dd id=\"fs-id1165137938602\">if a polynomial [latex]f\\left(x\\right)[\/latex] is divided by [latex]x-k[\/latex], then the remainder is equal to the value [latex]f\\left(k\\right)[\/latex]<\/dd>\n<\/dl>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-13872\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":97803,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax 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