{"id":17920,"date":"2021-08-23T06:11:43","date_gmt":"2021-08-23T06:11:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/?post_type=chapter&#038;p=17920"},"modified":"2021-09-12T23:24:31","modified_gmt":"2021-09-12T23:24:31","slug":"section-4-7-complex-zeros-fundamental-theorem-of-algebra","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/section-4-7-complex-zeros-fundamental-theorem-of-algebra\/","title":{"raw":"Section 4.7: Complex Zeros; Fundamental Theorem of Algebra","rendered":"Section 4.7: Complex Zeros; Fundamental Theorem of Algebra"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the Conjugate Pairs Theorem.<\/li>\r\n \t<li>Find a polynomial function with specified zeros.<\/li>\r\n \t<li>Find the complex zeros of a polynomial function.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn a previous section we found the real solutions of a quadratic equation. That is, we found the real zeros of a polynomial function of degree 2. Then in a previous section we found the complex solutions of a quadratic equation. That is, we found the complex zeros of a polynomial function of degree 2. In the last section we found the real zeros of a polynomial function of degree 3 or higher. In this section we will find the <i>complex zeros<\/i> of polynomial functions of degree 3 or higher.\r\n<h2>Use the Fundamental Theorem of Algebra<\/h2>\r\n<p id=\"fs-id1165135547255\">Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The <strong>Fundamental Theorem of Algebra <\/strong>tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.<\/p>\r\n<p id=\"fs-id1165137771429\">Suppose <em>f<\/em> is a polynomial function of degree four, and [latex]f\\left(x\\right)=0[\/latex]. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it [latex]{c}_{1}[\/latex]. By the Factor Theorem, we can write [latex]f\\left(x\\right)[\/latex] as a product of [latex]x-{c}_{\\text{1}}[\/latex] and a polynomial quotient. Since [latex]x-{c}_{\\text{1}}[\/latex] is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it [latex]{c}_{\\text{2}}[\/latex]. So we can write the polynomial quotient as a product of [latex]x-{c}_{\\text{2}}[\/latex] and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of [latex]f\\left(x\\right)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165135693782\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The <strong>Fundamental Theorem of Algebra<\/strong> states that, if <em>f(x)<\/em> is a polynomial of degree <em>n &gt; 0<\/em>, then <em>f(x)<\/em> has at least one complex zero.<\/h3>\r\n<p id=\"fs-id1165135409342\">We can use this theorem to argue that, if [latex]f\\left(x\\right)[\/latex] is a polynomial of degree [latex]n&gt;0[\/latex], and <em>a<\/em> is a non-zero real number, then [latex]f\\left(x\\right)[\/latex] has exactly <em>n<\/em> linear factors<\/p>\r\n\r\n<div id=\"eip-750\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\left(x-{c}_{1}\\right)\\left(x-{c}_{2}\\right)\u2026\\left(x-{c}_{n}\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165137889849\">where [latex]{c}_{1},{c}_{2},\u2026,{c}_{n}[\/latex] are complex numbers. Therefore, [latex]f\\left(x\\right)[\/latex] has <em>n<\/em> roots if we allow for multiplicities.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135481201\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165135481208\"><strong>Does every polynomial have at least one imaginary zero?<\/strong><\/p>\r\n<p id=\"fs-id1165137883746\"><em>No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_03_06_06\" class=\"example\">\r\n<div id=\"fs-id1165135658206\" class=\"exercise\">\r\n<div id=\"fs-id1165135658208\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Finding the Zeros of a Polynomial Function with Complex Zeros<\/h3>\r\n<p id=\"fs-id1165135658213\">Find the zeros of [latex]f\\left(x\\right)=3{x}^{3}+9{x}^{2}+x+3[\/latex].<\/p>\r\n\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q278128\">Show Solution<\/span>\r\n<div id=\"q278128\" class=\"hidden-answer\" style=\"display: none;\">\r\n<p id=\"fs-id1165137430512\">The Rational Zero Theorem tells us that if [latex]\\frac{p}{q}[\/latex] is a zero of [latex]f\\left(x\\right)[\/latex], then <em>p<\/em> is a factor of 3 and <em>q<\/em> is a factor of 3.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{p}{q}&amp;=\\frac{\\text{factor of constant term}}{\\text{factor of leading coefficient}} \\\\ &amp;=\\frac{\\text{factor of 3}}{\\text{factor of 3}} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165134212112\">The factors of 3 are [latex]\\pm 1[\/latex] and [latex]\\pm 3[\/latex]. The possible values for [latex]\\frac{p}{q}[\/latex], and therefore the possible rational zeros for the function, are [latex]\\pm 3, \\pm 1,\\text{ and }\\pm \\frac{1}{3}[\/latex]. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let\u2019s begin with \u20133.<\/p>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png\"><img class=\"aligncenter size-full wp-image-13116\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png\" sizes=\"(max-width: 175px) 100vw, 175px\" srcset=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png 175w, https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM-65x43.png 65w\" alt=\"Synthetic division with divisor = -3 and quotient = {3, 9, 1, 3}\" width=\"175\" height=\"115\" \/><\/a>Dividing by [latex]\\left(x+3\\right)[\/latex] gives a remainder of 0, so \u20133 is a zero of the function. The polynomial can be written as\r\n<p style=\"text-align: center;\">[latex]\\left(x+3\\right)\\left(3{x}^{2}+1\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165134224528\">We can then set the quadratic equal to 0 and solve to find the other zeros of the function.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3{x}^{2}+1=0 \\\\ {x}^{2}=-\\frac{1}{3} \\\\ x=\\pm \\sqrt{-\\frac{1}{3}}=\\pm \\frac{i\\sqrt{3}}{3} \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137756902\">The zeros of [latex]f\\left(x\\right)[\/latex] are \u20133 and [latex]\\pm \\frac{i\\sqrt{3}}{3}[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135640998\">Look at the graph of the function <em>f<\/em>. Notice that, at [latex]x=-3[\/latex], the graph crosses the <em>x<\/em>-axis, indicating an odd multiplicity (1) for the zero [latex]x=-3[\/latex]. Also note the presence of the two turning points. This means that, since there is a 3<sup>rd<\/sup> degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the <em>x<\/em>-intercepts for the function are shown. So either the multiplicity of [latex]x=-3[\/latex] is 1 and there are two complex solutions, which is what we found, or the multiplicity at [latex]x=-3[\/latex] is three. Either way, our result is correct.<\/p>\r\n\r\n<div class=\"wp-caption aligncenter\" style=\"width: 497px;\">\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010746\/CNX_Precalc_Figure_03_06_0022.jpg\" alt=\"Graph of a polynomial with its x-intercept at (-3, 0) labeled as &quot;Cross&quot;\" width=\"487\" height=\"289\" \/>\r\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135538762\">Find the zeros of [latex]f\\left(x\\right)=2{x}^{3}+5{x}^{2}-11x+4[\/latex].<\/p>\r\n\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q518560\">Show Solution<\/span>\r\n<div id=\"q518560\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nThe zeros are [latex]\\text{-4, }\\frac{1}{2},\\text{ and 1}\\text{.}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"ohm19264\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19264&amp;theme=oea&amp;iframe_resize_id=ohm19264\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Use the Linear Factorization Theorem to find polynomials with given zeros<\/h2>\r\n<p id=\"fs-id1165135502003\">A vital implication of the <strong>Fundamental Theorem of Algebra<\/strong>, as we stated above, is that a polynomial function of degree <em>n<\/em> will have <em>n<\/em> zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into <em>n<\/em> factors. 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 \u2013 c<\/em>), where <em>c<\/em> is a complex number.<\/p>\r\n<p id=\"eip-651\">Let <em>f<\/em> be a polynomial function with real coefficients, and suppose [latex]a+bi\\text{, }b\\ne 0[\/latex], is a zero of [latex]f\\left(x\\right)[\/latex]. Then, by the Factor Theorem, [latex]x-\\left(a+bi\\right)[\/latex] is a factor of [latex]f\\left(x\\right)[\/latex]. For <em>f<\/em> to have real coefficients, [latex]x-\\left(a-bi\\right)[\/latex] must also be a factor of [latex]f\\left(x\\right)[\/latex]. This is true because any factor other than [latex]x-\\left(a-bi\\right)[\/latex], when multiplied by [latex]x-\\left(a+bi\\right)[\/latex], will 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> with real coefficients has a complex zero [latex]a+bi[\/latex], then the complex conjugate [latex]a-bi[\/latex] must also be a zero of [latex]f\\left(x\\right)[\/latex]. This is called the <strong>Complex Conjugate Theorem<\/strong>.<\/p>\r\n\r\n<div id=\"fs-id1165137933095\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Complex Conjugate Theorem<\/h3>\r\n<p id=\"fs-id1165135436621\">According to the <strong>Linear Factorization Theorem<\/strong>, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\\left(x-c\\right)[\/latex], where <em>c<\/em> is a complex number.<\/p>\r\n<p id=\"fs-id1165135443970\">If the polynomial function <em>f<\/em> has real coefficients and a complex zero in the form [latex]a+bi[\/latex], then the complex conjugate of the zero, [latex]a-bi[\/latex], is also a zero.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_03_06_07\" class=\"example\">\r\n<div id=\"fs-id1165137737023\" class=\"exercise\">\r\n<div id=\"fs-id1165137737025\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Using the Conjugate Pairs Theorem<\/h3>\r\n<p id=\"fs-id1165134151154\">A polynomial function f of degree 5 whose coefficients are real numbers has the zeros 1, [latex]5i[\/latex], and [latex]1+i[\/latex]. Find the remaining zeros.<\/p>\r\n\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q348758\">Show Solution<\/span>\r\n<div id=\"q348758\" class=\"hidden-answer\" style=\"display: none;\">\r\n<p id=\"fs-id1165132964597\">Since f has coefficients that are real numbers, complex zeros appear as conjugate pairs. It follows that [latex]-5i[\/latex], the conjugate of [latex]5i[\/latex] and [latex]1-i[\/latex], the conjugate of [latex]1+i[\/latex], are the two remaining zeros.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137832786\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137832792\">How To: Given the zeros of a polynomial function [latex]f[\/latex] and a point [latex]\\left(c\\text{, }f(c)\\right)[\/latex] on the graph of [latex]f[\/latex], use the Linear Factorization Theorem to find the polynomial function.<\/h3>\r\n<ol id=\"fs-id1165135534938\">\r\n \t<li>Use the zeros to construct the linear factors of the polynomial.<\/li>\r\n \t<li>Multiply the linear factors to expand the polynomial.<\/li>\r\n \t<li>Substitute [latex]\\left(c,f\\left(c\\right)\\right)[\/latex] into the function to determine the leading coefficient.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_06_07\" class=\"example\">\r\n<div id=\"fs-id1165137737023\" class=\"exercise\">\r\n<div id=\"fs-id1165137737025\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros<\/h3>\r\n<p id=\"fs-id1165134151154\">Find a fourth degree polynomial with real coefficients that has zeros of \u20133, 2, <em>i<\/em>, such that [latex]f\\left(-2\\right)=100[\/latex].<\/p>\r\n\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q348756\">Show Solution<\/span>\r\n<div id=\"q348756\" class=\"hidden-answer\" style=\"display: none;\">\r\n<p id=\"fs-id1165132964597\">Because [latex]x=i[\/latex] is a zero, by the Complex Conjugate Theorem [latex]x=-i[\/latex] is also a zero. The polynomial must have factors of [latex]\\left(x+3\\right),\\left(x - 2\\right),\\left(x-i\\right)[\/latex], and [latex]\\left(x+i\\right)[\/latex]. Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let\u2019s begin by multiplying these factors.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;f\\left(x\\right)=a\\left(x+3\\right)\\left(x - 2\\right)\\left(x-i\\right)\\left(x+i\\right)\\\\ &amp;f\\left(x\\right)=a\\left({x}^{2}+x - 6\\right)\\left({x}^{2}+1\\right)\\\\ &amp;f\\left(x\\right)=a\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165134065076\">We need to find <em>a<\/em> to ensure [latex]f\\left(-2\\right)=100[\/latex]. Substitute [latex]x=-2[\/latex] and [latex]f\\left(2\\right)=100[\/latex]\r\ninto [latex]f\\left(x\\right)[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}100&amp;=a\\left({\\left(-2\\right)}^{4}+{\\left(-2\\right)}^{3}-5{\\left(-2\\right)}^{2}+\\left(-2\\right)-6\\right) \\\\ 100&amp;=a\\left(-20\\right) \\\\ -5&amp;=a \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137407591\">So the polynomial function is<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=-5\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165135406977\">or<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135154288\">We found that both <em>i<\/em> and \u2013<em>i<\/em> were zeros, but only one of these zeros needed to be given. If <em>i<\/em> is a zero of a polynomial with real coefficients, then <em>\u2013i<\/em> must also be a zero of the polynomial because <em>\u2013i<\/em> is the complex conjugate of <em>i<\/em>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135340588\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165135340596\"><strong>If 2 + 3<em>i<\/em> were given as a zero of a polynomial with real coefficients, would 2 \u2013 3<em>i<\/em> also need to be a zero?<\/strong><\/p>\r\n<p id=\"fs-id1165134170187\"><em>Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.<\/em><\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134130177\">Find a third degree polynomial with real coefficients that has zeros of 5 and \u20132<em>i<\/em> such that [latex]f\\left(1\\right)=10[\/latex].<\/p>\r\n\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q126079\">Show Solution<\/span>\r\n<div id=\"q126079\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\n[latex]f\\left(x\\right)=-\\frac{1}{2}{x}^{3}+\\frac{5}{2}{x}^{2}-2x+10[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"ohm100297\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100297&amp;theme=oea&amp;iframe_resize_id=ohm100297\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\n<div id=\"Example_03_06_07\" class=\"example\">\r\n<div id=\"fs-id1165137737023\" class=\"exercise\">\r\n<div id=\"fs-id1165137737025\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Use the given complex zero to find the remaining Zeros<\/h3>\r\n<p id=\"fs-id1165134151154\">The polynomial [latex]3x^4+5x^3+25x^2+45x-18[\/latex] has a zero of [latex]3i[\/latex]. Use this zero to find the other zeros.<\/p>\r\n\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q348760\">Show Solution<\/span>\r\n<div id=\"q348760\" class=\"hidden-answer\" style=\"display: none;\">\r\n<p id=\"fs-id1165132964597\">Because [latex]x=3i[\/latex] is a zero, by the Complex Conjugate Theorem [latex]x=-3i[\/latex] is also a zero. The polynomial must have factors of [latex]\\left(x+3i\\right)[\/latex] and [latex]\\left(x - 3i\\right)[\/latex]. We can multiply these together to get [latex]x^2+9[\/latex]. Next we will use long division:<\/p>\r\n<img src=\"http:\/\/www.hutchmath.com\/Images\/ComplexZeros.JPG\" alt=\"Long division of 3x^4+5x^3+25x^2+45x-18 divided by x^2+0x+9, with quotient 3x^2+5x-2\" \/>\r\nNext we will factor the quotient and set it equal to zero to find the other zeros.\r\n<p style=\"text-align: left;\">[latex]\\begin{align}&amp;3x^2+5x-2=\\left(x+2\\right)\\left(3x - 1\\right)\\\\ &amp;0=\\left(x+2\\right)\\left(3x - 1\\right)\\\\ &amp;x=-2,\\frac{1}{3}\\end{align}[\/latex]<\/p>\r\nThe other zeros are [latex]-3i,-2[\/latex], and [latex]\\frac{1}{3}[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135538762\">The polynomial [latex]2x^4-5x^3+11x^2-20x+12[\/latex] has a zero of [latex]-2i[\/latex]. Use this zero to find the other zeros.<\/p>\r\n\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n<span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q518561\">Show Solution<\/span>\r\n<div id=\"q518561\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nThe other zeros are [latex]2i,1[\/latex], and [latex]\\frac{3}{2}[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165135380122\">\r\n \t<li>According to the Fundamental Theorem, every polynomial function has at least one complex zero.<\/li>\r\n \t<li>Every polynomial function with degree greater than 0 has at least one complex zero.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\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&nbsp;\r\n<h2 style=\"text-align: center;\">Section 4.7 Homework Exercises<\/h2>\r\nIn the exercises 1-6, information is given about a polynomial function f whose coefficients are real numbers.\r\nFind the remaining zeros of f.\r\n\r\n1. Degree 3; zeros: 3, [latex]4-i[\/latex]\r\n\r\n2. Degree 3; zeros: 4, [latex]3+i[\/latex]\r\n\r\n3. Degree 4; zeros: [latex]i[\/latex], [latex]4-i[\/latex]\r\n\r\n4. Degree 4; zeros: [latex]-i[\/latex], [latex]2+i[\/latex]\r\n\r\n5. Degree 5; zeros: 1,\u00a0[latex]i[\/latex], [latex]5i[\/latex]\r\n\r\n6. Degree: 5; zeros: 0, 1, 2, [latex]i[\/latex]\r\n\r\nIn exercises 7-12, find a polynomial function f with real coefficients (with a leading coefficient of a) having the given degree and zeros.\r\n\r\n7. Degree 4; zeros: 4 (multiplicity 2); [latex]3+i[\/latex]\r\n\r\n8. Degree 4; zeros:\u00a0 [latex]i[\/latex],\u00a0[latex]1+2i[\/latex]\r\n\r\n9. Degree 5; zeros:\u00a0 2;[latex]-i[\/latex]; [latex]1+i[\/latex]\r\n\r\n10. Degree 6; zeros:\u00a0 [latex]i[\/latex], [latex]4-i[\/latex]; [latex]2+i[\/latex]\r\n\r\n11. Degree 4; zeros: 3 (multiplicity 2); [latex]-i[\/latex]\r\n\r\n12. Degree 5; zeros: 1 (multiplicity 3); [latex]1+i[\/latex]\r\n\r\nIn problems 13-18, use the given zero to find all remaining zeros of each polynomial function.\r\n\r\n13. [latex]f(x)=x^3-5x^2+9x-45[\/latex]; zero: [latex]3i[\/latex]\r\n\r\n14. [latex]f(x)=x^3+3x^2+25x+75[\/latex]; zero: [latex]-5i[\/latex]\r\n\r\n15. [latex]f(x)=4x^4+7x^3+62x^2+112x-32[\/latex]; zero: [latex]-4i[\/latex]\r\n\r\n16. [latex]h(x)=3x^4+5x^3+25x^2+45x-18 [\/latex]; zero: [latex]3i[\/latex]\r\n\r\n17. [latex]h(x)=x^4-7x^3+23x^2-15x-522 [\/latex]; zero: [latex]2-5i[\/latex]\r\n\r\n18.\u00a0[latex]f(x)=x^4-7x^3+14x^2-38x-60 [\/latex]; zero: [latex]1+3i[\/latex]\r\n\r\nFor the following exercises, find all complex zeros (real and non-real). Write f in factored form.\r\n\r\n19. [latex]f(x)=x^3-1[\/latex]\r\n\r\n20. [latex]f(x)=x^4-1[\/latex]\r\n\r\n21. [latex]f(x)=x^3-8x^2+25x-26[\/latex]\r\n\r\n22. [latex]f(x)=x^3+13x^2+57x+85[\/latex]\r\n\r\n23. [latex]f(x)=x^4+5x^2+4[\/latex]\r\n\r\n24. [latex]f(x)=x^4+13x^2+36[\/latex]\r\n\r\n&nbsp;\r\n\r\n&nbsp;","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the Conjugate Pairs Theorem.<\/li>\n<li>Find a polynomial function with specified zeros.<\/li>\n<li>Find the complex zeros of a polynomial function.<\/li>\n<\/ul>\n<\/div>\n<p>In a previous section we found the real solutions of a quadratic equation. That is, we found the real zeros of a polynomial function of degree 2. Then in a previous section we found the complex solutions of a quadratic equation. That is, we found the complex zeros of a polynomial function of degree 2. In the last section we found the real zeros of a polynomial function of degree 3 or higher. In this section we will find the <i>complex zeros<\/i> of polynomial functions of degree 3 or higher.<\/p>\n<h2>Use the Fundamental Theorem of Algebra<\/h2>\n<p id=\"fs-id1165135547255\">Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The <strong>Fundamental Theorem of Algebra <\/strong>tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.<\/p>\n<p id=\"fs-id1165137771429\">Suppose <em>f<\/em> is a polynomial function of degree four, and [latex]f\\left(x\\right)=0[\/latex]. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it [latex]{c}_{1}[\/latex]. By the Factor Theorem, we can write [latex]f\\left(x\\right)[\/latex] as a product of [latex]x-{c}_{\\text{1}}[\/latex] and a polynomial quotient. Since [latex]x-{c}_{\\text{1}}[\/latex] is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it [latex]{c}_{\\text{2}}[\/latex]. So we can write the polynomial quotient as a product of [latex]x-{c}_{\\text{2}}[\/latex] and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of [latex]f\\left(x\\right)[\/latex].<\/p>\n<div id=\"fs-id1165135693782\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The <strong>Fundamental Theorem of Algebra<\/strong> states that, if <em>f(x)<\/em> is a polynomial of degree <em>n &gt; 0<\/em>, then <em>f(x)<\/em> has at least one complex zero.<\/h3>\n<p id=\"fs-id1165135409342\">We can use this theorem to argue that, if [latex]f\\left(x\\right)[\/latex] is a polynomial of degree [latex]n>0[\/latex], and <em>a<\/em> is a non-zero real number, then [latex]f\\left(x\\right)[\/latex] has exactly <em>n<\/em> linear factors<\/p>\n<div id=\"eip-750\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a\\left(x-{c}_{1}\\right)\\left(x-{c}_{2}\\right)\u2026\\left(x-{c}_{n}\\right)[\/latex]<\/div>\n<p id=\"fs-id1165137889849\">where [latex]{c}_{1},{c}_{2},\u2026,{c}_{n}[\/latex] are complex numbers. Therefore, [latex]f\\left(x\\right)[\/latex] has <em>n<\/em> roots if we allow for multiplicities.<\/p>\n<\/div>\n<div id=\"fs-id1165135481201\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165135481208\"><strong>Does every polynomial have at least one imaginary zero?<\/strong><\/p>\n<p id=\"fs-id1165137883746\"><em>No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.<\/em><\/p>\n<\/div>\n<div id=\"Example_03_06_06\" class=\"example\">\n<div id=\"fs-id1165135658206\" class=\"exercise\">\n<div id=\"fs-id1165135658208\" class=\"problem textbox shaded\">\n<h3>Example 1: Finding the Zeros of a Polynomial Function with Complex Zeros<\/h3>\n<p id=\"fs-id1165135658213\">Find the zeros of [latex]f\\left(x\\right)=3{x}^{3}+9{x}^{2}+x+3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q278128\">Show Solution<\/span><\/p>\n<div id=\"q278128\" class=\"hidden-answer\" style=\"display: none;\">\n<p id=\"fs-id1165137430512\">The Rational Zero Theorem tells us that if [latex]\\frac{p}{q}[\/latex] is a zero of [latex]f\\left(x\\right)[\/latex], then <em>p<\/em> is a factor of 3 and <em>q<\/em> is a factor of 3.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{p}{q}&=\\frac{\\text{factor of constant term}}{\\text{factor of leading coefficient}} \\\\ &=\\frac{\\text{factor of 3}}{\\text{factor of 3}} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165134212112\">The factors of 3 are [latex]\\pm 1[\/latex] and [latex]\\pm 3[\/latex]. The possible values for [latex]\\frac{p}{q}[\/latex], and therefore the possible rational zeros for the function, are [latex]\\pm 3, \\pm 1,\\text{ and }\\pm \\frac{1}{3}[\/latex]. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let\u2019s begin with \u20133.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-13116\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png\" sizes=\"auto, (max-width: 175px) 100vw, 175px\" srcset=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM.png 175w, https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/09\/03012930\/Screen-Shot-2015-09-11-at-3.07.51-PM-65x43.png 65w\" alt=\"Synthetic division with divisor = -3 and quotient = {3, 9, 1, 3}\" width=\"175\" height=\"115\" \/><\/a>Dividing by [latex]\\left(x+3\\right)[\/latex] gives a remainder of 0, so \u20133 is a zero of the function. The polynomial can be written as<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+3\\right)\\left(3{x}^{2}+1\\right)[\/latex]<\/p>\n<p id=\"fs-id1165134224528\">We can then set the quadratic equal to 0 and solve to find the other zeros of the function.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3{x}^{2}+1=0 \\\\ {x}^{2}=-\\frac{1}{3} \\\\ x=\\pm \\sqrt{-\\frac{1}{3}}=\\pm \\frac{i\\sqrt{3}}{3} \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137756902\">The zeros of [latex]f\\left(x\\right)[\/latex] are \u20133 and [latex]\\pm \\frac{i\\sqrt{3}}{3}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135640998\">Look at the graph of the function <em>f<\/em>. Notice that, at [latex]x=-3[\/latex], the graph crosses the <em>x<\/em>-axis, indicating an odd multiplicity (1) for the zero [latex]x=-3[\/latex]. Also note the presence of the two turning points. This means that, since there is a 3<sup>rd<\/sup> degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the <em>x<\/em>-intercepts for the function are shown. So either the multiplicity of [latex]x=-3[\/latex] is 1 and there are two complex solutions, which is what we found, or the multiplicity at [latex]x=-3[\/latex] is three. Either way, our result is correct.<\/p>\n<div class=\"wp-caption aligncenter\" style=\"width: 497px;\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010746\/CNX_Precalc_Figure_03_06_0022.jpg\" alt=\"Graph of a polynomial with its x-intercept at (-3, 0) labeled as &quot;Cross&quot;\" width=\"487\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135538762\">Find the zeros of [latex]f\\left(x\\right)=2{x}^{3}+5{x}^{2}-11x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q518560\">Show Solution<\/span><\/p>\n<div id=\"q518560\" class=\"hidden-answer\" style=\"display: none;\">\n<p>The zeros are [latex]\\text{-4, }\\frac{1}{2},\\text{ and 1}\\text{.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm19264\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19264&amp;theme=oea&amp;iframe_resize_id=ohm19264\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use the Linear Factorization Theorem to find polynomials with given zeros<\/h2>\n<p id=\"fs-id1165135502003\">A vital implication of the <strong>Fundamental Theorem of Algebra<\/strong>, as we stated above, is that a polynomial function of degree <em>n<\/em> will have <em>n<\/em> zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into <em>n<\/em> factors. 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 \u2013 c<\/em>), where <em>c<\/em> is a complex number.<\/p>\n<p id=\"eip-651\">Let <em>f<\/em> be a polynomial function with real coefficients, and suppose [latex]a+bi\\text{, }b\\ne 0[\/latex], is a zero of [latex]f\\left(x\\right)[\/latex]. Then, by the Factor Theorem, [latex]x-\\left(a+bi\\right)[\/latex] is a factor of [latex]f\\left(x\\right)[\/latex]. For <em>f<\/em> to have real coefficients, [latex]x-\\left(a-bi\\right)[\/latex] must also be a factor of [latex]f\\left(x\\right)[\/latex]. This is true because any factor other than [latex]x-\\left(a-bi\\right)[\/latex], when multiplied by [latex]x-\\left(a+bi\\right)[\/latex], will 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> with real coefficients has a complex zero [latex]a+bi[\/latex], then the complex conjugate [latex]a-bi[\/latex] must also be a zero of [latex]f\\left(x\\right)[\/latex]. This is called the <strong>Complex Conjugate Theorem<\/strong>.<\/p>\n<div id=\"fs-id1165137933095\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Complex Conjugate Theorem<\/h3>\n<p id=\"fs-id1165135436621\">According to the <strong>Linear Factorization Theorem<\/strong>, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\\left(x-c\\right)[\/latex], where <em>c<\/em> is a complex number.<\/p>\n<p id=\"fs-id1165135443970\">If the polynomial function <em>f<\/em> has real coefficients and a complex zero in the form [latex]a+bi[\/latex], then the complex conjugate of the zero, [latex]a-bi[\/latex], is also a zero.<\/p>\n<\/div>\n<div id=\"Example_03_06_07\" class=\"example\">\n<div id=\"fs-id1165137737023\" class=\"exercise\">\n<div id=\"fs-id1165137737025\" class=\"problem textbox shaded\">\n<h3>Example 2: Using the Conjugate Pairs Theorem<\/h3>\n<p id=\"fs-id1165134151154\">A polynomial function f of degree 5 whose coefficients are real numbers has the zeros 1, [latex]5i[\/latex], and [latex]1+i[\/latex]. Find the remaining zeros.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q348758\">Show Solution<\/span><\/p>\n<div id=\"q348758\" class=\"hidden-answer\" style=\"display: none;\">\n<p id=\"fs-id1165132964597\">Since f has coefficients that are real numbers, complex zeros appear as conjugate pairs. It follows that [latex]-5i[\/latex], the conjugate of [latex]5i[\/latex] and [latex]1-i[\/latex], the conjugate of [latex]1+i[\/latex], are the two remaining zeros.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137832786\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137832792\">How To: Given the zeros of a polynomial function [latex]f[\/latex] and a point [latex]\\left(c\\text{, }f(c)\\right)[\/latex] on the graph of [latex]f[\/latex], use the Linear Factorization Theorem to find the polynomial function.<\/h3>\n<ol id=\"fs-id1165135534938\">\n<li>Use the zeros to construct the linear factors of the polynomial.<\/li>\n<li>Multiply the linear factors to expand the polynomial.<\/li>\n<li>Substitute [latex]\\left(c,f\\left(c\\right)\\right)[\/latex] into the function to determine the leading coefficient.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_06_07\" class=\"example\">\n<div id=\"fs-id1165137737023\" class=\"exercise\">\n<div id=\"fs-id1165137737025\" class=\"problem textbox shaded\">\n<h3>Example 3: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros<\/h3>\n<p id=\"fs-id1165134151154\">Find a fourth degree polynomial with real coefficients that has zeros of \u20133, 2, <em>i<\/em>, such that [latex]f\\left(-2\\right)=100[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q348756\">Show Solution<\/span><\/p>\n<div id=\"q348756\" class=\"hidden-answer\" style=\"display: none;\">\n<p id=\"fs-id1165132964597\">Because [latex]x=i[\/latex] is a zero, by the Complex Conjugate Theorem [latex]x=-i[\/latex] is also a zero. The polynomial must have factors of [latex]\\left(x+3\\right),\\left(x - 2\\right),\\left(x-i\\right)[\/latex], and [latex]\\left(x+i\\right)[\/latex]. Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let\u2019s begin by multiplying these factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&f\\left(x\\right)=a\\left(x+3\\right)\\left(x - 2\\right)\\left(x-i\\right)\\left(x+i\\right)\\\\ &f\\left(x\\right)=a\\left({x}^{2}+x - 6\\right)\\left({x}^{2}+1\\right)\\\\ &f\\left(x\\right)=a\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)\\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165134065076\">We need to find <em>a<\/em> to ensure [latex]f\\left(-2\\right)=100[\/latex]. Substitute [latex]x=-2[\/latex] and [latex]f\\left(2\\right)=100[\/latex]<br \/>\ninto [latex]f\\left(x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}100&=a\\left({\\left(-2\\right)}^{4}+{\\left(-2\\right)}^{3}-5{\\left(-2\\right)}^{2}+\\left(-2\\right)-6\\right) \\\\ 100&=a\\left(-20\\right) \\\\ -5&=a \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137407591\">So the polynomial function is<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=-5\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)[\/latex]<\/p>\n<p id=\"fs-id1165135406977\">or<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135154288\">We found that both <em>i<\/em> and \u2013<em>i<\/em> were zeros, but only one of these zeros needed to be given. If <em>i<\/em> is a zero of a polynomial with real coefficients, then <em>\u2013i<\/em> must also be a zero of the polynomial because <em>\u2013i<\/em> is the complex conjugate of <em>i<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135340588\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165135340596\"><strong>If 2 + 3<em>i<\/em> were given as a zero of a polynomial with real coefficients, would 2 \u2013 3<em>i<\/em> also need to be a zero?<\/strong><\/p>\n<p id=\"fs-id1165134170187\"><em>Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.<\/em><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134130177\">Find a third degree polynomial with real coefficients that has zeros of 5 and \u20132<em>i<\/em> such that [latex]f\\left(1\\right)=10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q126079\">Show Solution<\/span><\/p>\n<div id=\"q126079\" class=\"hidden-answer\" style=\"display: none;\">\n<p>[latex]f\\left(x\\right)=-\\frac{1}{2}{x}^{3}+\\frac{5}{2}{x}^{2}-2x+10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm100297\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100297&amp;theme=oea&amp;iframe_resize_id=ohm100297\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"Example_03_06_07\" class=\"example\">\n<div id=\"fs-id1165137737023\" class=\"exercise\">\n<div id=\"fs-id1165137737025\" class=\"problem textbox shaded\">\n<h3>Example 4: Use the given complex zero to find the remaining Zeros<\/h3>\n<p id=\"fs-id1165134151154\">The polynomial [latex]3x^4+5x^3+25x^2+45x-18[\/latex] has a zero of [latex]3i[\/latex]. Use this zero to find the other zeros.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q348760\">Show Solution<\/span><\/p>\n<div id=\"q348760\" class=\"hidden-answer\" style=\"display: none;\">\n<p id=\"fs-id1165132964597\">Because [latex]x=3i[\/latex] is a zero, by the Complex Conjugate Theorem [latex]x=-3i[\/latex] is also a zero. The polynomial must have factors of [latex]\\left(x+3i\\right)[\/latex] and [latex]\\left(x - 3i\\right)[\/latex]. We can multiply these together to get [latex]x^2+9[\/latex]. Next we will use long division:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/www.hutchmath.com\/Images\/ComplexZeros.JPG\" alt=\"Long division of 3x^4+5x^3+25x^2+45x-18 divided by x^2+0x+9, with quotient 3x^2+5x-2\" \/><br \/>\nNext we will factor the quotient and set it equal to zero to find the other zeros.<\/p>\n<p style=\"text-align: left;\">[latex]\\begin{align}&3x^2+5x-2=\\left(x+2\\right)\\left(3x - 1\\right)\\\\ &0=\\left(x+2\\right)\\left(3x - 1\\right)\\\\ &x=-2,\\frac{1}{3}\\end{align}[\/latex]<\/p>\n<p>The other zeros are [latex]-3i,-2[\/latex], and [latex]\\frac{1}{3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135538762\">The polynomial [latex]2x^4-5x^3+11x^2-20x+12[\/latex] has a zero of [latex]-2i[\/latex]. Use this zero to find the other zeros.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p><span class=\"show-answer collapsed\" style=\"cursor: pointer;\" data-target=\"q518561\">Show Solution<\/span><\/p>\n<div id=\"q518561\" class=\"hidden-answer\" style=\"display: none;\">\n<p>The other zeros are [latex]2i,1[\/latex], and [latex]\\frac{3}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165135380122\">\n<li>According to the Fundamental Theorem, every polynomial function has at least one complex zero.<\/li>\n<li>Every polynomial function with degree greater than 0 has at least one complex zero.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\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<p>&nbsp;<\/p>\n<h2 style=\"text-align: center;\">Section 4.7 Homework Exercises<\/h2>\n<p>In the exercises 1-6, information is given about a polynomial function f whose coefficients are real numbers.<br \/>\nFind the remaining zeros of f.<\/p>\n<p>1. Degree 3; zeros: 3, [latex]4-i[\/latex]<\/p>\n<p>2. Degree 3; zeros: 4, [latex]3+i[\/latex]<\/p>\n<p>3. Degree 4; zeros: [latex]i[\/latex], [latex]4-i[\/latex]<\/p>\n<p>4. Degree 4; zeros: [latex]-i[\/latex], [latex]2+i[\/latex]<\/p>\n<p>5. Degree 5; zeros: 1,\u00a0[latex]i[\/latex], [latex]5i[\/latex]<\/p>\n<p>6. Degree: 5; zeros: 0, 1, 2, [latex]i[\/latex]<\/p>\n<p>In exercises 7-12, find a polynomial function f with real coefficients (with a leading coefficient of a) having the given degree and zeros.<\/p>\n<p>7. Degree 4; zeros: 4 (multiplicity 2); [latex]3+i[\/latex]<\/p>\n<p>8. Degree 4; zeros:\u00a0 [latex]i[\/latex],\u00a0[latex]1+2i[\/latex]<\/p>\n<p>9. Degree 5; zeros:\u00a0 2;[latex]-i[\/latex]; [latex]1+i[\/latex]<\/p>\n<p>10. Degree 6; zeros:\u00a0 [latex]i[\/latex], [latex]4-i[\/latex]; [latex]2+i[\/latex]<\/p>\n<p>11. Degree 4; zeros: 3 (multiplicity 2); [latex]-i[\/latex]<\/p>\n<p>12. Degree 5; zeros: 1 (multiplicity 3); [latex]1+i[\/latex]<\/p>\n<p>In problems 13-18, use the given zero to find all remaining zeros of each polynomial function.<\/p>\n<p>13. [latex]f(x)=x^3-5x^2+9x-45[\/latex]; zero: [latex]3i[\/latex]<\/p>\n<p>14. [latex]f(x)=x^3+3x^2+25x+75[\/latex]; zero: [latex]-5i[\/latex]<\/p>\n<p>15. [latex]f(x)=4x^4+7x^3+62x^2+112x-32[\/latex]; zero: [latex]-4i[\/latex]<\/p>\n<p>16. [latex]h(x)=3x^4+5x^3+25x^2+45x-18[\/latex]; zero: [latex]3i[\/latex]<\/p>\n<p>17. [latex]h(x)=x^4-7x^3+23x^2-15x-522[\/latex]; zero: [latex]2-5i[\/latex]<\/p>\n<p>18.\u00a0[latex]f(x)=x^4-7x^3+14x^2-38x-60[\/latex]; zero: [latex]1+3i[\/latex]<\/p>\n<p>For the following exercises, find all complex zeros (real and non-real). Write f in factored form.<\/p>\n<p>19. [latex]f(x)=x^3-1[\/latex]<\/p>\n<p>20. [latex]f(x)=x^4-1[\/latex]<\/p>\n<p>21. [latex]f(x)=x^3-8x^2+25x-26[\/latex]<\/p>\n<p>22. [latex]f(x)=x^3+13x^2+57x+85[\/latex]<\/p>\n<p>23. [latex]f(x)=x^4+5x^2+4[\/latex]<\/p>\n<p>24. [latex]f(x)=x^4+13x^2+36[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"author":264444,"menu_order":7,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-17920","chapter","type-chapter","status-publish","hentry"],"part":17771,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17920","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":56,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17920\/revisions"}],"predecessor-version":[{"id":18141,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17920\/revisions\/18141"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/17771"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17920\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=17920"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=17920"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=17920"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=17920"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}