{"id":1833,"date":"2016-11-02T20:56:46","date_gmt":"2016-11-02T20:56:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1833"},"modified":"2021-01-14T21:14:35","modified_gmt":"2021-01-14T21:14:35","slug":"introduction-zeros-of-polynomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/chapter\/introduction-zeros-of-polynomials\/","title":{"raw":"Methods for Finding Zeros of Polynomials","rendered":"Methods for Finding Zeros of Polynomials"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li class=\"li2\"><span class=\"s1\">Evaluate a polynomial using the Remainder Theorem.<\/span><\/li>\r\n \t<li><span class=\"s1\">Use the Rational Zero Theorem to find rational zeros.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Use the Factor Theorem to solve a polynomial equation.<\/span><\/li>\r\n \t<li>Use synthetic division to find the zeros of a polynomial function.<\/li>\r\n \t<li>Use the Fundamental Theorem of Algebra to find complex zeros of a polynomial function.<\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Use the Linear Factorization Theorem to find polynomials with given zeros.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Use Descartes\u2019 Rule of Signs\u00a0to determine the maximum number of possible real zeros of a polynomial function.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Solve real-world applications of polynomial equations.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\nA 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?\r\n\r\nThis 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.\r\n<h2>Theorems Used to Analyze Polynomial Functions<\/h2>\r\nIn 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>). Let\u2019s walk through the proof of the theorem.\r\n\r\nRecall 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\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=d\\left(x\\right)q\\left(x\\right)+r\\left(x\\right)[\/latex]<\/p>\r\nIf the divisor, <em>d<\/em>(<em>x<\/em>), is <em>x<\/em> \u2013\u00a0<em>k<\/em>, this takes the form\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex]<\/p>\r\nSince 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\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\left(k\\right)=\\left(k-k\\right)q\\left(k\\right)+r\\hfill \\\\ \\text{}f\\left(k\\right)=0\\cdot q\\left(k\\right)+r\\hfill \\\\ \\text{}f\\left(k\\right)=r\\hfill \\end{array}[\/latex]<\/p>\r\nIn 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>.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Remainder Theorem<\/h3>\r\nIf 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].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>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>\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 class=\"textbox exercises\">\r\n<h3>Example: Using the Remainder Theorem to Evaluate a Polynomial<\/h3>\r\nUse 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].\r\n\r\n[reveal-answer q=\"830571\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"830571\"]\r\n\r\nTo find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ 2\\overline{)\\begin{array}{lllllllll}6\\hfill &amp; -1\\hfill &amp; -15\\hfill &amp; 2\\hfill &amp; -7\\hfill \\\\ \\hfill &amp; \\text{ }12\\hfill &amp; \\text{ }\\text{ }\\text{ }22\\hfill &amp; 14\\hfill &amp; \\text{ }\\text{ }32\\hfill \\end{array}}\\\\ \\begin{array}{llllll}\\hfill &amp; \\text{}6\\hfill &amp; 11\\hfill &amp; \\text{ }\\text{ }\\text{ }7\\hfill &amp; \\text{ }\\text{ }16\\hfill &amp; \\text{ }\\text{ }25\\hfill \\end{array}\\end{array}[\/latex]<\/p>\r\nThe remainder is [latex]25[\/latex]. Therefore, [latex]f\\left(2\\right)=25[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can check our answer by evaluating [latex]f\\left(2\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}f\\left(x\\right) &amp; =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\\\ f\\left(2\\right) &amp; =6{\\left(2\\right)}^{4}-{\\left(2\\right)}^{3}-15{\\left(2\\right)}^{2}+2\\left(2\\right)-7 \\\\ f\\left(2\\right) &amp; =25\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the Remainder Theorem to evaluate [latex]f\\left(x\\right)=2{x}^{5}+4{x}^{4}-3{x}^{3}+8{x}^{2}+7[\/latex]\r\nat [latex]x=-3[\/latex].\r\n\r\n[reveal-answer q=\"276041\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"276041\"]\r\n\r\n[latex]f(-3)=-2[\/latex]\r\n\r\nThis is what your synthetic division should have looked like:\r\n\r\n<img class=\"alignnone size-medium wp-image-921\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4695\/2019\/07\/17090130\/Screen-Shot-2020-11-17-at-12.57.59-AM-300x116.png\" alt=\"\" width=\"300\" height=\"116\" \/>\r\n\r\nNote: there was no [latex]x[\/latex] term, so a zero was needed\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question]214138[\/ohm_question]\r\n\r\n<\/div>\r\n<h3>Using the Rational Zero Theorem to Find Rational Zeros<\/h3>\r\nAnother 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 coefficient of the polynomial\r\n\r\nConsider a quadratic function with two zeros, [latex]x=\\frac{2}{5}[\/latex]\u00a0and [latex]x=\\frac{3}{4}[\/latex].\r\n\r\nThese zeros have factors associated with them. Let us set each factor equal to 0 and then construct the original quadratic function.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201550\/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=\"818\" height=\"147\" \/>\r\n\r\nNotice 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.\r\n\r\nWe can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Rational Zero Theorem<\/h3>\r\nThe <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].\r\n\r\nWhen the leading coefficient is 1, the possible rational zeros are the factors of the constant term.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>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>\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 class=\"textbox exercises\">\r\n<h3>Example: Listing All Possible Rational Zeros<\/h3>\r\nList all possible rational zeros of [latex]f\\left(x\\right)=2{x}^{4}-5{x}^{3}+{x}^{2}-4[\/latex].\r\n\r\n[reveal-answer q=\"746924\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"746924\"]\r\n\r\nThe 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.\r\n\r\nThe constant term is \u20134; the factors of \u20134 are [latex]p=\\pm 1,\\pm 2,\\pm 4[\/latex].\r\n\r\nThe leading coefficient is 2; the factors of 2 are [latex]q=\\pm 1,\\pm 2[\/latex].\r\n\r\nIf 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{p}{q}=\\pm \\frac{1}{1},\\pm \\frac{1}{2}\\text{ }&amp; \\frac{p}{q}=\\pm \\frac{2}{1},\\pm \\frac{2}{2}\\text{ }&amp; \\frac{p}{q}=\\pm \\frac{4}{1},\\pm \\frac{4}{2}\\end{array}[\/latex]<\/p>\r\nNote 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.\r\n<p style=\"text-align: center;\">[latex]\\frac{p}{q}=\\frac{\\text{Factors of the constant term}}{\\text{Factors of the leading coefficient}}=\\pm 1,\\pm 2,\\pm 4,\\pm \\frac{1}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Rational Zero Theorem to Find Rational Zeros<\/h3>\r\nUse the Rational Zero Theorem to find the rational zeros of [latex]f\\left(x\\right)=2{x}^{3}+{x}^{2}-4x+1[\/latex].\r\n[reveal-answer q=\"637649\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"637649\"]\r\n\r\nThe 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{p}{q}=\\frac{\\text{Factors of the constant term}}{\\text{Factors of the leading coefficient}}\\hfill \\\\ \\text{}\\frac{p}{q}=\\frac{\\text{Factors of 1}}{\\text{Factors of 2}}\\hfill \\end{array}[\/latex]<\/p>\r\nThe 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].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }f\\left(-1\\right)=2{\\left(-1\\right)}^{3}+{\\left(-1\\right)}^{2}-4\\left(-1\\right)+1=4\\hfill \\\\ \\text{ }f\\left(1\\right)=2{\\left(1\\right)}^{3}+{\\left(1\\right)}^{2}-4\\left(1\\right)+1=0\\hfill \\\\ \\text{ }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\\hfill \\\\ \\text{ }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}\\hfill \\end{array}[\/latex]<\/p>\r\nOf 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].\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\nUse the Rational Zero Theorem to find the rational zeros of [latex]f\\left(x\\right)={x}^{3}-3{x}^{2}-6x+8[\/latex].\r\n\r\n[reveal-answer q=\"38787\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"38787\"]\r\n\r\n[latex]-2, 1, \\text{and } 4[\/latex] are zeros of the polynomial. [\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Using the Factor Theorem to Solve a Polynomial Equation<\/h3>\r\nThe <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 [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex].\r\n\r\nIf <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].\r\n\r\nNotice, 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].\r\n\r\nSimilarly, 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.\r\n\r\nThis 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.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Factor Theorem<\/h3>\r\nAccording 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].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>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>\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 class=\"textbox exercises\">\r\n<h3>Example: Using the Factor Theorem to Solve a Polynomial Equation<\/h3>\r\nShow 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 polynomial.\r\n\r\n[reveal-answer q=\"976488\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"976488\"]\r\n\r\nWe can use synthetic division to show that [latex]\\left(x+2\\right)[\/latex] is a factor of the polynomial.\r\n\r\n<img class=\"aligncenter wp-image-13110 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205544\/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=\"192\" height=\"120\" \/>\r\n\r\nThe 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:\r\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left({x}^{2}-8x+15\\right)[\/latex]<\/p>\r\nWe can factor the quadratic factor to write the polynomial as\r\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x - 3\\right)\\left(x - 5\\right)[\/latex]<\/p>\r\nBy the Factor Theorem, the zeros of [latex]{x}^{3}-6{x}^{2}-x+30[\/latex] are \u20132, 3, and 5.\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\nUse 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.\r\n\r\n[reveal-answer q=\"50598\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"50598\"]\r\n\r\nThe zeros are 2, \u20132, and \u20134.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Zeros of a Polynomial Functions<\/h2>\r\nThe 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 <strong>synthetic division<\/strong> repeatedly to determine all of the <strong>zeros<\/strong> of a polynomial function.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polynomial function [latex]f[\/latex], use synthetic division to find its zeros<\/h3>\r\n<ol>\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 from 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 class=\"textbox exercises\">\r\n<h3>Example: Finding the Zeros of a Polynomial Function with Repeated Real Zeros<\/h3>\r\nFind the zeros of [latex]f\\left(x\\right)=4{x}^{3}-3x - 1[\/latex].\r\n\r\n[reveal-answer q=\"571513\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"571513\"]\r\n\r\nThe 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{p}{q}=\\frac{\\text{Factors of the constant term}}{\\text{Factors of the leading coefficient}}\\hfill \\\\ \\text{}\\frac{p}{q}=\\frac{\\text{Factors of -1}}{\\text{Factors of 4}}\\hfill \\end{array}[\/latex]<\/p>\r\nThe 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].\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\r\n<a href=\"https:\/\/courses.candelalearning.com\/precalcone1xmommaster\/wp-content\/uploads\/sites\/1226\/2015\/09\/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\/wp-content\/uploads\/sites\/896\/2016\/11\/02205550\/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>\r\n\r\nDividing 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 [latex]\\left(x - 1\\right)\\left(4{x}^{2}+4x+1\\right)[\/latex].\r\n\r\nThe quadratic is a perfect square. [latex]f\\left(x\\right)[\/latex]\u00a0can be written as [latex]\\left(x - 1\\right){\\left(2x+1\\right)}^{2}[\/latex].\r\n\r\nWe 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+1=0\\hfill \\\\ \\text{ }x=-\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThe zeros of the function are 1 and [latex]-\\frac{1}{2}[\/latex] with multiplicity 2.\r\n<h4>Analysis of the Solution<\/h4>\r\nLook at the graph of the function <em>f<\/em>. 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<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205551\/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\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>The Fundamental Theorem of Algebra<\/h3>\r\nNow 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.\r\n\r\nSuppose <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]. 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].\r\n<div class=\"textbox\">\r\n<h3>A General Note: The <strong>Fundamental Theorem of Algebra<\/strong><\/h3>\r\nThe Fundamental Theorem of Algebra states that, if [latex]f(x)[\/latex] is a polynomial of degree [latex]n&gt;0[\/latex], then [latex]f(x)[\/latex] has at least one complex zero.\r\n\r\nWe 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.\r\n\r\nThe polynomial can be written as\r\n<p 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]<\/p>\r\nwhere [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.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Does every polynomial have at least one imaginary zero?<\/strong>\r\n\r\n<em>No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Zeros of a Polynomial Function with Complex Zeros<\/h3>\r\nFind the zeros of [latex]f\\left(x\\right)=3{x}^{3}+9{x}^{2}+x+3[\/latex].\r\n\r\n[reveal-answer q=\"791291\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"791291\"]\r\n\r\nThe 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{p}{q}=\\frac{\\text{Factors of the constant term}}{\\text{Factor of the leading coefficient}}\\hfill \\\\ \\text{}\\frac{p}{q}=\\frac{\\text{Factors of 3}}{\\text{Factors of 3}}\\hfill \\end{array}[\/latex]<\/p>\r\nThe 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.\r\n\r\n<a href=\"https:\/\/courses.candelalearning.com\/precalcone1xmommaster\/wp-content\/uploads\/sites\/1226\/2015\/09\/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\/wp-content\/uploads\/sites\/896\/2016\/11\/02205554\/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 [latex]\\left(x+3\\right)\\left(3{x}^{2}+1\\right)[\/latex].\r\n\r\nWe can then set the quadratic equal to 0 and solve to find the other zeros of the function.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3{x}^{2}+1=0\\hfill \\\\ \\text{ }{x}^{2}=-\\frac{1}{3}\\hfill \\\\ \\text{ }x=\\pm \\sqrt{-\\frac{1}{3}}=\\pm \\frac{i\\sqrt{3}}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe zeros of [latex]f\\left(x\\right)[\/latex]\u00a0are \u20133 and [latex]\\pm \\frac{i\\sqrt{3}}{3}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nLook 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.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205555\/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\" \/>\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\nFind the zeros of [latex]f\\left(x\\right)=2{x}^{3}+5{x}^{2}-11x+4[\/latex].\r\n\r\n[reveal-answer q=\"696690\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"696690\"]\r\n\r\nThe zeros are [latex]\\text{-4, }\\frac{1}{2},\\text{ and 1}\\text{.}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=103644&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Linear Factorization and Descartes Rule of Signs<\/h2>\r\nA vital implication of the <strong>Fundamental Theorem of Algebra\u00a0<\/strong>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 each factor will be of the form (<em>x\u00a0\u2013\u00a0c<\/em>) where <em>c<\/em>\u00a0is a complex number.\r\n\r\nLet <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>.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Complex Conjugate Theorem<\/h3>\r\nAccording 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 of the form [latex]\\left(x-c\\right)[\/latex] where <em>c<\/em>\u00a0is a complex number.\r\n\r\nIf the polynomial function <em>f<\/em>\u00a0has real coefficients and a complex zero of the form [latex]a+bi[\/latex],\u00a0then the complex conjugate of the zero, [latex]a-bi[\/latex],\u00a0is also a zero.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>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 class=\"textbox exercises\">\r\n<h3>Example: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros<\/h3>\r\nFind 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].\r\n\r\n[reveal-answer q=\"412896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"412896\"]\r\n\r\nBecause [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.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}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{array}[\/latex]<\/p>\r\nWe 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].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}100=a\\left({\\left(-2\\right)}^{4}+{\\left(-2\\right)}^{3}-5{\\left(-2\\right)}^{2}+\\left(-2\\right)-6\\right)\\hfill \\\\ 100=a\\left(-20\\right)\\hfill \\\\ -5=a\\hfill \\end{array}[\/latex]<\/p>\r\nSo the polynomial function is:\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 style=\"text-align: center;\">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\nWe 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\u00a0<\/em>must also be a zero of the polynomial because <em>\u2013i<\/em>\u00a0is the complex conjugate of <em>i<\/em>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<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>\r\n\r\n<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>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind 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].\r\n\r\n[reveal-answer q=\"704164\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"704164\"]\r\n\r\n[latex]f\\left(x\\right)=-\\frac{1}{2}{x}^{3}+\\frac{5}{2}{x}^{2}-2x+10[\/latex][\/hidden-answer]\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19266&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Descartes\u2019 Rule of Signs<\/h3>\r\nThere 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.\r\n\r\nThere is a similar relationship between the number of sign changes in [latex]f\\left(-x\\right)[\/latex] and the number of negative real zeros.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Descartes\u2019 Rule of Signs<\/h3>\r\nAccording 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:\r\n<ul>\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 class=\"textbox exercises\">\r\n<h3>Example: Using Descartes\u2019 Rule of Signs<\/h3>\r\nUse 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].\r\n[reveal-answer q=\"143065\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"143065\"]\r\n\r\nBegin by determining the number of sign changes.\r\n\r\n<img class=\"aligncenter size-full wp-image-11813\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205558\/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\" \/>\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 style=\"text-align: center;\">[latex]\\begin{array}{l}f\\left(-x\\right)=-{\\left(-x\\right)}^{4}-3{\\left(-x\\right)}^{3}+6{\\left(-x\\right)}^{2}-4\\left(-x\\right)-12\\hfill \\\\ f\\left(-x\\right)=-{x}^{4}+3{x}^{3}+6{x}^{2}+4x - 12\\hfill \\end{array}[\/latex]<img class=\"aligncenter size-full wp-image-11814\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205600\/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\" \/><\/p>\r\nAgain, there are two sign changes, so there are either 2 or 0 negative real roots.\r\n\r\nThere are four possibilities, as we can see below.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th style=\"width: 148px; text-align: center;\">Positive Real Zeros<\/th>\r\n<th style=\"width: 109px; text-align: center;\">Negative Real Zeros<\/th>\r\n<th style=\"width: 74px; text-align: center;\">Complex Zeros<\/th>\r\n<th style=\"width: 54px; text-align: center;\">Total Zeros<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 148px;\">2<\/td>\r\n<td style=\"width: 109px;\">2<\/td>\r\n<td style=\"width: 74px;\">0<\/td>\r\n<td style=\"width: 54px;\">4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 148px;\">2<\/td>\r\n<td style=\"width: 109px;\">0<\/td>\r\n<td style=\"width: 74px;\">2<\/td>\r\n<td style=\"width: 54px;\">4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 148px;\">0<\/td>\r\n<td style=\"width: 109px;\">2<\/td>\r\n<td style=\"width: 74px;\">2<\/td>\r\n<td style=\"width: 54px;\">4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 148px;\">0<\/td>\r\n<td style=\"width: 109px;\">0<\/td>\r\n<td style=\"width: 74px;\">4<\/td>\r\n<td style=\"width: 54px;\">4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can confirm the numbers of positive and negative real roots by examining a graph of the function.\u00a0We can see from the graph that the function has 0 positive real roots and 2 negative real roots.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205602\/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\" \/>\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\nUse Descartes\u2019 Rule of Signs to determine the maximum possible number 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 number of positive and negative real zeros for the function.\r\n\r\n[reveal-answer q=\"941865\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"941865\"]\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.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Solving Real-world Applications of Polynomial Equations<\/h3>\r\nWe 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.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Polynomial Equations<\/h3>\r\nA 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?\r\n\r\n[reveal-answer q=\"801673\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"801673\"]\r\n\r\nBegin 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.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}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{array}[\/latex]<\/p>\r\nSubstitute the given volume into this equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }351=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2}\\hfill &amp; \\text{Substitute 351 for }V.\\hfill \\\\ 1053={w}^{3}+4{w}^{2}\\hfill &amp; \\text{Multiply both sides by 3}.\\hfill \\\\ \\text{ }0={w}^{3}+4{w}^{2}-1053 \\hfill &amp; \\text{Subtract 1053 from both sides}.\\hfill \\end{array}[\/latex]<\/p>\r\nDescartes' 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].\r\n\r\n<img class=\"aligncenter size-full wp-image-13122\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205604\/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=\"155\" height=\"100\" \/>\r\n\r\nSince 1 is not a solution, we will check [latex]x=3[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205606\/CNX_Precalc_revised_eq12.png\" alt=\".\" width=\"100\" \/>\r\n\r\nSince 3 is not a solution either, we will test [latex]x=9[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205608\/CNX_Precalc_revised_eq22.png\" alt=\".\" width=\"120\" \/>\r\n\r\nSynthetic 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.\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\nThe sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.\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\nA 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?\r\n\r\n[reveal-answer q=\"24181\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"24181\"]\r\n\r\n3 meters by 4 meters by 7 meters[\/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] if 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 of Algebra, 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 easiest 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] where <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=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li class=\"li2\"><span class=\"s1\">Evaluate a polynomial using the Remainder Theorem.<\/span><\/li>\n<li><span class=\"s1\">Use the Rational Zero Theorem to find rational zeros.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Use the Factor Theorem to solve a polynomial equation.<\/span><\/li>\n<li>Use synthetic division to find the zeros of a polynomial function.<\/li>\n<li>Use the Fundamental Theorem of Algebra to find complex zeros of a polynomial function.<\/li>\n<li class=\"li2\"><span class=\"s1\">Use the Linear Factorization Theorem to find polynomials with given zeros.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Use Descartes\u2019 Rule of Signs\u00a0to determine the maximum number of possible real zeros of a polynomial function.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Solve real-world applications of polynomial equations.<\/span><\/li>\n<\/ul>\n<\/div>\n<p>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>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>Theorems Used to Analyze Polynomial Functions<\/h2>\n<p>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>). Let\u2019s walk through the proof of the theorem.<\/p>\n<p>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<p style=\"text-align: center;\">[latex]f\\left(x\\right)=d\\left(x\\right)q\\left(x\\right)+r\\left(x\\right)[\/latex]<\/p>\n<p>If the divisor, <em>d<\/em>(<em>x<\/em>), is <em>x<\/em> \u2013\u00a0<em>k<\/em>, this takes the form<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex]<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\left(k\\right)=\\left(k-k\\right)q\\left(k\\right)+r\\hfill \\\\ \\text{}f\\left(k\\right)=0\\cdot q\\left(k\\right)+r\\hfill \\\\ \\text{}f\\left(k\\right)=r\\hfill \\end{array}[\/latex]<\/p>\n<p>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 class=\"textbox\">\n<h3>A General Note: The Remainder Theorem<\/h3>\n<p>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 class=\"textbox\">\n<h3>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>\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 class=\"textbox exercises\">\n<h3>Example: Using the Remainder Theorem to Evaluate a Polynomial<\/h3>\n<p>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=\"q830571\">Show Solution<\/span><\/p>\n<div id=\"q830571\" class=\"hidden-answer\" style=\"display: none\">\n<p>To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ 2\\overline{)\\begin{array}{lllllllll}6\\hfill & -1\\hfill & -15\\hfill & 2\\hfill & -7\\hfill \\\\ \\hfill & \\text{ }12\\hfill & \\text{ }\\text{ }\\text{ }22\\hfill & 14\\hfill & \\text{ }\\text{ }32\\hfill \\end{array}}\\\\ \\begin{array}{llllll}\\hfill & \\text{}6\\hfill & 11\\hfill & \\text{ }\\text{ }\\text{ }7\\hfill & \\text{ }\\text{ }16\\hfill & \\text{ }\\text{ }25\\hfill \\end{array}\\end{array}[\/latex]<\/p>\n<p>The remainder is [latex]25[\/latex]. Therefore, [latex]f\\left(2\\right)=25[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our answer by evaluating [latex]f\\left(2\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}f\\left(x\\right) & =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\\\ f\\left(2\\right) & =6{\\left(2\\right)}^{4}-{\\left(2\\right)}^{3}-15{\\left(2\\right)}^{2}+2\\left(2\\right)-7 \\\\ f\\left(2\\right) & =25\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=2{x}^{5}+4{x}^{4}-3{x}^{3}+8{x}^{2}+7[\/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=\"q276041\">Show Solution<\/span><\/p>\n<div id=\"q276041\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(-3)=-2[\/latex]<\/p>\n<p>This is what your synthetic division should have looked like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-921\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4695\/2019\/07\/17090130\/Screen-Shot-2020-11-17-at-12.57.59-AM-300x116.png\" alt=\"\" width=\"300\" height=\"116\" \/><\/p>\n<p>Note: there was no [latex]x[\/latex] term, so a zero was needed<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm214138\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=214138&theme=oea&iframe_resize_id=ohm214138&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3>Using the Rational Zero Theorem to Find Rational Zeros<\/h3>\n<p>Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial, but first we need a pool of rational numbers to test. The <strong>Rational Zero Theorem<\/strong> helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial<\/p>\n<p>Consider a quadratic function with two zeros, [latex]x=\\frac{2}{5}[\/latex]\u00a0and [latex]x=\\frac{3}{4}[\/latex].<\/p>\n<p>These zeros have factors associated with them. Let us set each factor equal to 0 and then construct the original quadratic function.<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\/924\/2015\/11\/25201550\/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=\"818\" height=\"147\" \/><\/p>\n<p>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>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 class=\"textbox\">\n<h3>A General Note: The Rational Zero Theorem<\/h3>\n<p>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>When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a polynomial function [latex]f\\left(x\\right)[\/latex], use the Rational Zero Theorem to find rational zeros<\/h3>\n<ol>\n<li>Determine all factors of the constant term and all factors of the leading coefficient.<\/li>\n<li>Determine all possible values of [latex]\\frac{p}{q}[\/latex], where <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 class=\"textbox exercises\">\n<h3>Example: Listing All Possible Rational Zeros<\/h3>\n<p>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=\"q746924\">Show Solution<\/span><\/p>\n<div id=\"q746924\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>The constant term is \u20134; the factors of \u20134 are [latex]p=\\pm 1,\\pm 2,\\pm 4[\/latex].<\/p>\n<p>The leading coefficient is 2; the factors of 2 are [latex]q=\\pm 1,\\pm 2[\/latex].<\/p>\n<p>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{array}{l}\\frac{p}{q}=\\pm \\frac{1}{1},\\pm \\frac{1}{2}\\text{ }& \\frac{p}{q}=\\pm \\frac{2}{1},\\pm \\frac{2}{2}\\text{ }& \\frac{p}{q}=\\pm \\frac{4}{1},\\pm \\frac{4}{2}\\end{array}[\/latex]<\/p>\n<p>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 constant term}}{\\text{Factors of the leading coefficient}}=\\pm 1,\\pm 2,\\pm 4,\\pm \\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Rational Zero Theorem to Find Rational Zeros<\/h3>\n<p>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=\"q637649\">Show Solution<\/span><\/p>\n<div id=\"q637649\" class=\"hidden-answer\" style=\"display: none\">\n<p>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{array}{l}\\frac{p}{q}=\\frac{\\text{Factors of the constant term}}{\\text{Factors of the leading coefficient}}\\hfill \\\\ \\text{}\\frac{p}{q}=\\frac{\\text{Factors of 1}}{\\text{Factors of 2}}\\hfill \\end{array}[\/latex]<\/p>\n<p>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]\\begin{array}{l}\\text{ }f\\left(-1\\right)=2{\\left(-1\\right)}^{3}+{\\left(-1\\right)}^{2}-4\\left(-1\\right)+1=4\\hfill \\\\ \\text{ }f\\left(1\\right)=2{\\left(1\\right)}^{3}+{\\left(1\\right)}^{2}-4\\left(1\\right)+1=0\\hfill \\\\ \\text{ }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\\hfill \\\\ \\text{ }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}\\hfill \\end{array}[\/latex]<\/p>\n<p>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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the Rational Zero Theorem to find the rational zeros of [latex]f\\left(x\\right)={x}^{3}-3{x}^{2}-6x+8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q38787\">Show Solution<\/span><\/p>\n<div id=\"q38787\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-2, 1, \\text{and } 4[\/latex] are zeros of the polynomial. <\/p><\/div>\n<\/div>\n<\/div>\n<h3>Using the Factor Theorem to Solve a Polynomial Equation<\/h3>\n<p>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 [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex].<\/p>\n<p>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>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>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>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 class=\"textbox\">\n<h3>A General Note: The Factor Theorem<\/h3>\n<p>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 class=\"textbox\">\n<h3>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>\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 class=\"textbox exercises\">\n<h3>Example: Using the Factor Theorem to Solve a Polynomial Equation<\/h3>\n<p>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 polynomial.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q976488\">Show Solution<\/span><\/p>\n<div id=\"q976488\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can use synthetic division to show that [latex]\\left(x+2\\right)[\/latex] is a factor of the polynomial.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-13110 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205544\/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=\"192\" height=\"120\" \/><\/p>\n<p>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>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>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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q50598\">Show Solution<\/span><\/p>\n<div id=\"q50598\" class=\"hidden-answer\" style=\"display: none\">\n<p>The zeros are 2, \u20132, and \u20134.<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Finding Zeros of a Polynomial Functions<\/h2>\n<p>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 <strong>synthetic division<\/strong> repeatedly to determine all of the <strong>zeros<\/strong> of a polynomial function.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a polynomial function [latex]f[\/latex], use synthetic division to find its zeros<\/h3>\n<ol>\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 from 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 class=\"textbox exercises\">\n<h3>Example: Finding the Zeros of a Polynomial Function with Repeated Real Zeros<\/h3>\n<p>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=\"q571513\">Show Solution<\/span><\/p>\n<div id=\"q571513\" class=\"hidden-answer\" style=\"display: none\">\n<p>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{array}{l}\\frac{p}{q}=\\frac{\\text{Factors of the constant term}}{\\text{Factors of the leading coefficient}}\\hfill \\\\ \\text{}\\frac{p}{q}=\\frac{\\text{Factors of -1}}{\\text{Factors of 4}}\\hfill \\end{array}[\/latex]<\/p>\n<p>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].<br \/>\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.<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalcone1xmommaster\/wp-content\/uploads\/sites\/1226\/2015\/09\/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\/wp-content\/uploads\/sites\/896\/2016\/11\/02205550\/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><\/p>\n<p>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 [latex]\\left(x - 1\\right)\\left(4{x}^{2}+4x+1\\right)[\/latex].<\/p>\n<p>The quadratic is a perfect square. [latex]f\\left(x\\right)[\/latex]\u00a0can be written as [latex]\\left(x - 1\\right){\\left(2x+1\\right)}^{2}[\/latex].<\/p>\n<p>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{array}{l}2x+1=0\\hfill \\\\ \\text{ }x=-\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>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>. 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<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205551\/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<\/div>\n<\/div>\n<\/div>\n<h3>The Fundamental Theorem of Algebra<\/h3>\n<p>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>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]. 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 class=\"textbox\">\n<h3>A General Note: The <strong>Fundamental Theorem of Algebra<\/strong><\/h3>\n<p>The Fundamental Theorem of Algebra states that, if [latex]f(x)[\/latex] is a polynomial of degree [latex]n>0[\/latex], then [latex]f(x)[\/latex] has at least one complex zero.<\/p>\n<p>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<p>The polynomial can be written as<\/p>\n<p 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]<\/p>\n<p>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 class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Does every polynomial have at least one imaginary zero?<\/strong><\/p>\n<p><em>No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.<\/em><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Zeros of a Polynomial Function with Complex Zeros<\/h3>\n<p>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=\"q791291\">Show Solution<\/span><\/p>\n<div id=\"q791291\" class=\"hidden-answer\" style=\"display: none\">\n<p>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{array}{l}\\frac{p}{q}=\\frac{\\text{Factors of the constant term}}{\\text{Factor of the leading coefficient}}\\hfill \\\\ \\text{}\\frac{p}{q}=\\frac{\\text{Factors of 3}}{\\text{Factors of 3}}\\hfill \\end{array}[\/latex]<\/p>\n<p>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:\/\/courses.candelalearning.com\/precalcone1xmommaster\/wp-content\/uploads\/sites\/1226\/2015\/09\/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\/wp-content\/uploads\/sites\/896\/2016\/11\/02205554\/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 [latex]\\left(x+3\\right)\\left(3{x}^{2}+1\\right)[\/latex].<\/p>\n<p>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{array}{l}3{x}^{2}+1=0\\hfill \\\\ \\text{ }{x}^{2}=-\\frac{1}{3}\\hfill \\\\ \\text{ }x=\\pm \\sqrt{-\\frac{1}{3}}=\\pm \\frac{i\\sqrt{3}}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>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>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<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205555\/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<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q696690\">Show Solution<\/span><\/p>\n<div id=\"q696690\" class=\"hidden-answer\" style=\"display: none\">\n<p>The zeros are [latex]\\text{-4, }\\frac{1}{2},\\text{ and 1}\\text{.}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=103644&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Linear Factorization and Descartes Rule of Signs<\/h2>\n<p>A vital implication of the <strong>Fundamental Theorem of Algebra\u00a0<\/strong>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 each factor will be of the form (<em>x\u00a0\u2013\u00a0c<\/em>) where <em>c<\/em>\u00a0is a complex number.<\/p>\n<p>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 class=\"textbox\">\n<h3>A General Note: Complex Conjugate Theorem<\/h3>\n<p>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 of the form [latex]\\left(x-c\\right)[\/latex] where <em>c<\/em>\u00a0is a complex number.<\/p>\n<p>If the polynomial function <em>f<\/em>\u00a0has real coefficients and a complex zero of 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 class=\"textbox\">\n<h3>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 class=\"textbox exercises\">\n<h3>Example: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros<\/h3>\n<p>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=\"q412896\">Show Solution<\/span><\/p>\n<div id=\"q412896\" class=\"hidden-answer\" style=\"display: none\">\n<p>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{array}{l}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{array}[\/latex]<\/p>\n<p>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{array}{l}100=a\\left({\\left(-2\\right)}^{4}+{\\left(-2\\right)}^{3}-5{\\left(-2\\right)}^{2}+\\left(-2\\right)-6\\right)\\hfill \\\\ 100=a\\left(-20\\right)\\hfill \\\\ -5=a\\hfill \\end{array}[\/latex]<\/p>\n<p>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 style=\"text-align: center;\">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>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\u00a0<\/em>must 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 class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><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><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=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q704164\">Show Solution<\/span><\/p>\n<div id=\"q704164\" 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><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19266&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h3>Descartes\u2019 Rule of Signs<\/h3>\n<p>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.<\/p>\n<p>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<div class=\"textbox\">\n<h3>A General Note: Descartes\u2019 Rule of Signs<\/h3>\n<p>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>\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 class=\"textbox exercises\">\n<h3>Example: Using Descartes\u2019 Rule of Signs<\/h3>\n<p>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=\"q143065\">Show Solution<\/span><\/p>\n<div id=\"q143065\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by determining the number of sign changes.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-11813\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205558\/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\" \/><\/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 style=\"text-align: center;\">[latex]\\begin{array}{l}f\\left(-x\\right)=-{\\left(-x\\right)}^{4}-3{\\left(-x\\right)}^{3}+6{\\left(-x\\right)}^{2}-4\\left(-x\\right)-12\\hfill \\\\ f\\left(-x\\right)=-{x}^{4}+3{x}^{3}+6{x}^{2}+4x - 12\\hfill \\end{array}[\/latex]<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-11814\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205600\/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\" \/><\/p>\n<p>Again, there are two sign changes, so there are either 2 or 0 negative real roots.<\/p>\n<p>There are four possibilities, as we can see below.<\/p>\n<table>\n<thead>\n<tr>\n<th style=\"width: 148px; text-align: center;\">Positive Real Zeros<\/th>\n<th style=\"width: 109px; text-align: center;\">Negative Real Zeros<\/th>\n<th style=\"width: 74px; text-align: center;\">Complex Zeros<\/th>\n<th style=\"width: 54px; text-align: center;\">Total Zeros<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"width: 148px;\">2<\/td>\n<td style=\"width: 109px;\">2<\/td>\n<td style=\"width: 74px;\">0<\/td>\n<td style=\"width: 54px;\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 148px;\">2<\/td>\n<td style=\"width: 109px;\">0<\/td>\n<td style=\"width: 74px;\">2<\/td>\n<td style=\"width: 54px;\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 148px;\">0<\/td>\n<td style=\"width: 109px;\">2<\/td>\n<td style=\"width: 74px;\">2<\/td>\n<td style=\"width: 54px;\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 148px;\">0<\/td>\n<td style=\"width: 109px;\">0<\/td>\n<td style=\"width: 74px;\">4<\/td>\n<td style=\"width: 54px;\">4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Analysis of the Solution<\/h4>\n<p>We can confirm the numbers of positive and negative real roots by examining a graph of the function.\u00a0We can see from the graph that the function has 0 positive real roots and 2 negative real roots.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205602\/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<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use Descartes\u2019 Rule of Signs to determine the maximum possible number 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 number 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=\"q941865\">Show Solution<\/span><\/p>\n<div id=\"q941865\" 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><\/div>\n<\/div>\n<\/div>\n<h3>Solving Real-world Applications of Polynomial Equations<\/h3>\n<p>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 class=\"textbox exercises\">\n<h3>Example: Solving Polynomial Equations<\/h3>\n<p>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=\"q801673\">Show Solution<\/span><\/p>\n<div id=\"q801673\" class=\"hidden-answer\" style=\"display: none\">\n<p>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{array}{l}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{array}[\/latex]<\/p>\n<p>Substitute the given volume into this equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }351=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2}\\hfill & \\text{Substitute 351 for }V.\\hfill \\\\ 1053={w}^{3}+4{w}^{2}\\hfill & \\text{Multiply both sides by 3}.\\hfill \\\\ \\text{ }0={w}^{3}+4{w}^{2}-1053 \\hfill & \\text{Subtract 1053 from both sides}.\\hfill \\end{array}[\/latex]<\/p>\n<p>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<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-13122\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205604\/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=\"155\" height=\"100\" \/><\/p>\n<p>Since 1 is not a solution, we will check [latex]x=3[\/latex].<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205606\/CNX_Precalc_revised_eq12.png\" alt=\".\" width=\"100\" \/><\/p>\n<p>Since 3 is not a solution either, we will test [latex]x=9[\/latex].<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205608\/CNX_Precalc_revised_eq22.png\" alt=\".\" width=\"120\" \/><\/p>\n<p>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>The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q24181\">Show Solution<\/span><\/p>\n<div id=\"q24181\" class=\"hidden-answer\" style=\"display: none\">\n<p>3 meters by 4 meters by 7 meters<\/p><\/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] if 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 of Algebra, 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 easiest 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] where <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-1833\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 2639. <strong>Authored by<\/strong>: Anderson, Tophe. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 103644. <strong>Authored by<\/strong>: Jim Smart. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 19266. <strong>Authored by<\/strong>: Sousa, James. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/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":21,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for 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