{"id":1845,"date":"2016-11-02T21:31:59","date_gmt":"2016-11-02T21:31:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1845"},"modified":"2017-04-19T18:54:02","modified_gmt":"2017-04-19T18:54:02","slug":"linear-factorization-and-descartes-rule-of-signs","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/linear-factorization-and-descartes-rule-of-signs\/","title":{"raw":"Linear Factorization and Descartes Rule of Signs","rendered":"Linear Factorization and Descartes Rule of Signs"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Use linear factorization to find the equation of a polynomial function given it's zeros and a point on it's graph<\/li>\r\n \t<li>Use Descartes rule of signs to determine the maximum \u00a0number of possible real zeros of a polynomial function<\/li>\r\n \t<li>Solve a polynomial function application involving volume<\/li>\r\n<\/ul>\r\n<\/div>\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 that each factor will be in 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 in 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 in 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\" data-number-style=\"arabic\">\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"412896\"]\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 data-effect=\"italics\">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\nor\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<\/em>\u00a0must 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 data-effect=\"bold\">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 data-effect=\"italics\">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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"704164\"][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:\/\/www.myopenmath.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<h2>Descartes\u2019 Rule of Signs<\/h2>\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. For example, the polynomial function below has one sign change.\r\n\r\nThis tells us that the function must have 1 positive real zero.\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\r\nIn this case, [latex]f\\left(\\mathrm{-x}\\right)[\/latex] has 3 sign changes. This tells us that [latex]f\\left(x\\right)[\/latex] could have 3 or 1 negative real zeros.\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\"]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>Positive Real\r\nZeros<\/th>\r\n<th>Negative Real\r\nZeros<\/th>\r\n<th>Complex\r\nZeros<\/th>\r\n<th>Total\r\nZeros<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>2<\/td>\r\n<td>2<\/td>\r\n<td>0<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>0<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0<\/td>\r\n<td>2<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<td>4<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Analysis of the Solution<\/h4>\r\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\" data-media-type=\"image\/jpg\" \/>\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 numbers of positive and negative real zeros for [latex]f\\left(x\\right)=2{x}^{4}-10{x}^{3}+11{x}^{2}-15x+12[\/latex].\u00a0Use a graph to verify the numbers of positive and negative real zeros for the function.\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n\r\n[reveal-answer q=\"941865\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"941865\"]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.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solve real-world applications of polynomial equations<\/h2>\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"801673\"]\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\" data-media-type=\"image\/jpg\" \/>\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\" data-media-type=\"image\/jpg\" \/>\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"24181\"]3 meters by 4 meters by 7 meters[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Use linear factorization to find the equation of a polynomial function given it&#8217;s zeros and a point on it&#8217;s graph<\/li>\n<li>Use Descartes rule of signs to determine the maximum \u00a0number of possible real zeros of a polynomial function<\/li>\n<li>Solve a polynomial function application involving volume<\/li>\n<\/ul>\n<\/div>\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 that each factor will be in the form (<em>x\u00a0\u2013\u00a0c<\/em>), where <em>c<\/em>\u00a0is a complex number.<\/p>\n<p>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 in 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 in the form [latex]a+bi[\/latex],\u00a0then the complex conjugate of the zero, [latex]a-bi[\/latex],\u00a0is also a zero.<\/p>\n<\/div>\n<div 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\" data-number-style=\"arabic\">\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\">Solution<\/span><\/p>\n<div id=\"q412896\" class=\"hidden-answer\" style=\"display: none\">\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.<\/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 data-effect=\"italics\">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>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<\/em>\u00a0must also be a zero of the polynomial because <em>\u2013i<\/em>\u00a0is the complex conjugate of <em>i<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong data-effect=\"bold\">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 data-effect=\"italics\">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\">Solution<\/span><\/p>\n<div id=\"q704164\" class=\"hidden-answer\" style=\"display: none\">[latex]f\\left(x\\right)=-\\frac{1}{2}{x}^{3}+\\frac{5}{2}{x}^{2}-2x+10[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=19266&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Descartes\u2019 Rule of Signs<\/h2>\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. For example, the polynomial function below has one sign change.<\/p>\n<p>This tells us that the function must have 1 positive real zero.<\/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<p>In this case, [latex]f\\left(\\mathrm{-x}\\right)[\/latex] has 3 sign changes. This tells us that [latex]f\\left(x\\right)[\/latex] could have 3 or 1 negative real zeros.<\/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\">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>Positive Real<br \/>\nZeros<\/th>\n<th>Negative Real<br \/>\nZeros<\/th>\n<th>Complex<br \/>\nZeros<\/th>\n<th>Total<br \/>\nZeros<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>2<\/td>\n<td>2<\/td>\n<td>0<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>0<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>2<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>0<\/td>\n<td>4<\/td>\n<td>4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Analysis of the Solution<\/h4>\n<p>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\" data-media-type=\"image\/jpg\" \/><\/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 numbers of positive and negative real zeros for [latex]f\\left(x\\right)=2{x}^{4}-10{x}^{3}+11{x}^{2}-15x+12[\/latex].\u00a0Use a graph to verify the numbers of positive and negative real zeros for the function.<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q941865\">Solution<\/span><\/p>\n<div id=\"q941865\" class=\"hidden-answer\" style=\"display: none\">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.<\/div>\n<\/div>\n<\/div>\n<h2>Solve real-world applications of polynomial equations<\/h2>\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\">Solution<\/span><\/p>\n<div id=\"q801673\" class=\"hidden-answer\" style=\"display: none\">\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.<\/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\" data-media-type=\"image\/jpg\" \/><\/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\" data-media-type=\"image\/jpg\" \/><\/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\">Solution<\/span><\/p>\n<div id=\"q24181\" class=\"hidden-answer\" style=\"display: none\">3 meters by 4 meters by 7 meters<\/div>\n<\/div>\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-1845\">\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>Question ID 19266. <strong>Authored by<\/strong>: Sousa,James, mb Lippman,David. <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>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><\/ul><\/div>\n\t\t\t\t\t\t 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