{"id":223,"date":"2023-06-21T13:22:45","date_gmt":"2023-06-21T13:22:45","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/the-remainder-and-factor-theorems\/"},"modified":"2023-07-04T03:59:55","modified_gmt":"2023-07-04T03:59:55","slug":"the-remainder-and-factor-theorems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/the-remainder-and-factor-theorems\/","title":{"raw":"\u25aa   Theorems Used to Analyze Polynomial Functions","rendered":"\u25aa   Theorems Used to Analyze Polynomial Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li><span class=\"s1\">Evaluate a polynomial using the Remainder Theorem.<\/span><\/li>\r\n \t<li>Use the rational zeros theorem to find rational zeros of a polynomial.<\/li>\r\n \t<li>Use the Factor Theorem to solve a polynomial equation.<\/li>\r\n<\/ul>\r\n<\/div>\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<h2>Using the Rational Zero Theorem to Find Rational Zeros<\/h2>\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<h2>Using the Factor Theorem to Solve a Polynomial Equation<\/h2>\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\nNow use an online graphing calculator to graph [latex]f(x)[\/latex].\r\nOn the next line, enter [latex]g(x) = \\frac{f(x)}{(x-2)}[\/latex].\r\nWhere does [latex]g(x)[\/latex] cross the x-axis? How do those roots compare to the solution we found using the Factor Theorem?\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li><span class=\"s1\">Evaluate a polynomial using the Remainder Theorem.<\/span><\/li>\n<li>Use the rational zeros theorem to find rational zeros of a polynomial.<\/li>\n<li>Use the Factor Theorem to solve a polynomial equation.<\/li>\n<\/ul>\n<\/div>\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<h2>Using the Rational Zero Theorem to Find Rational Zeros<\/h2>\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<h2>Using the Factor Theorem to Solve a Polynomial Equation<\/h2>\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<p>Now use an online graphing calculator to graph [latex]f(x)[\/latex].<br \/>\nOn the next line, enter [latex]g(x) = \\frac{f(x)}{(x-2)}[\/latex].<br \/>\nWhere does [latex]g(x)[\/latex] cross the x-axis? How do those roots compare to the solution we found using the Factor Theorem?<\/p>\n<\/div>\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-223\">\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><\/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":395986,"menu_order":20,"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 free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 2639\",\"author\":\"Anderson,Tophe\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"b5742408-3a4c-4429-b248-4c33fa374f2d","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-223","chapter","type-chapter","status-publish","hentry"],"part":202,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/223","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/223\/revisions"}],"predecessor-version":[{"id":778,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/223\/revisions\/778"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/202"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/223\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=223"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=223"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=223"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}