{"id":279,"date":"2026-01-30T23:00:16","date_gmt":"2026-01-30T23:00:16","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/polynomial-equations-get-stronger-key-precalculus-practice-page-answer-keys\/"},"modified":"2026-01-30T23:02:49","modified_gmt":"2026-01-30T23:02:49","slug":"polynomial-equations-get-stronger-key-precalculus-practice-page-answer-keys","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/polynomial-equations-get-stronger-key-precalculus-practice-page-answer-keys\/","title":{"raw":"Polynomial Equations: Get Stronger Key -- Precalculus Practice Page Answer Keys","rendered":"Polynomial Equations: Get Stronger Key &#8212; Precalculus Practice Page Answer Keys"},"content":{"raw":"\n\n\t\t<div class=\"bc-section section\">\n\t\t\t<div class=\"chapter standard\" id=\"polynomial-equations-get-stronger-key\" title=\"Polynomial Equations: Get Stronger Key\">\n\t<div class=\"chapter-title-wrap\">\n\t\t<p class=\"chapter-number\"><\/p>\n\t\t<h1 class=\"chapter-title\">Polynomial Equations: Get Stronger Key<\/h1>\n\t\t\t\t\t\t\t\t<\/div>\n\t<div class=\"ugc chapter-ugc\">\n\t\t\t\t<h2>Dividing Polynomials Solutions<\/h2> <p>1.&nbsp;The binomial is a factor of the polynomial.<\/p> <p>3.&nbsp;[latex]x+6+\\frac{5}{x - 1}\\text{,}\\text{quotient:}x+6\\text{,}\\text{remainder:}\\text{5}[\/latex]<\/p> <p>7.&nbsp;[latex]x - 5\\text{,}\\text{quotient: }x - 5\\text{,}\\text{remainder: }\\text{0}[\/latex]<\/p> <p>13.&nbsp;[latex]2{x}^{2}-3x+5\\text{,}\\text{quotient:}2{x}^{2}-3x+5\\text{,}\\text{remainder: }\\text{0}[\/latex]<\/p> <p>15.&nbsp;[latex]2{x}^{2}+2x+1+\\frac{10}{x - 4}[\/latex]<\/p> <p>19.&nbsp;[latex]3{x}^{2}-11x+34-\\frac{106}{x+3}[\/latex]<\/p> <p>23.&nbsp;[latex]4{x}^{2}-21x+84-\\frac{323}{x+4}[\/latex]<\/p> <p>29.&nbsp;[latex]{x}^{3}-3x+1[\/latex]<\/p> <p>43.&nbsp;[latex]\\text{Quotient: }4{x}^{2}+8x+16\\text{,}\\text{remainder: }-1[\/latex]<\/p> <p>45.&nbsp;[latex]\\text{Quotient: }3{x}^{2}+3x+5\\text{,}\\text{remainder: }0[\/latex]<\/p> <p>61.&nbsp;[latex]2x+3[\/latex]<\/p> <p>63.&nbsp;[latex]x+2[\/latex]<\/p> <h2>Complex Numbers Solutions<\/h2> <p>1.&nbsp;Add the real parts together and the imaginary parts together.<\/p> <p>3.&nbsp;<em>i<\/em>&nbsp;times <em>i<\/em>&nbsp;equals \u20131, which is not imaginary. (answers vary)<\/p> <p>7.&nbsp;[latex]14+7i[\/latex]<\/p> <p>9.&nbsp;[latex]-\\frac{23}{29}+\\frac{15}{29}i[\/latex]<\/p> <p>13.<br> <img src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230014\/CNX_Precalc_Figure_03_01_2032.jpg\" alt=\"Graph of the plotted point, 1-2i.\"><\/p> <p>15.<br> <img src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230014\/CNX_Precalc_Figure_03_01_2052.jpg\" alt=\"Graph of the plotted point, i.\"><\/p> <p>17.&nbsp;[latex]8-i[\/latex]<\/p> <p>19.&nbsp;[latex]-11+4i[\/latex]<\/p> <p>23.&nbsp;[latex]6+15i[\/latex]<\/p> <p>25.&nbsp;[latex]-16+32i[\/latex]<\/p> <p>27.&nbsp;[latex]-4 - 7i[\/latex]<\/p> <p>29.&nbsp;25<\/p> <p>33.&nbsp;[latex]4 - 6i[\/latex]<\/p> <p>35.&nbsp;[latex]\\frac{2}{5}+\\frac{11}{5}i[\/latex]<\/p> <p>37. 15<em>i<\/em><\/p> <p>39.&nbsp;[latex]1+i\\sqrt{3}[\/latex]<\/p> <p>41. 1<\/p> <p>43. \u20131<\/p> <h2>Zeros of Polynomial Functions Solutions<\/h2> <p>3.&nbsp;Rational zeros can be expressed as fractions whereas real zeros include irrational numbers.<\/p> <p>5.&nbsp;Polynomial functions can have repeated zeros, so the fact that number is a zero doesn\u2019t preclude it being a zero again.<\/p> <p>7. \u2013106<\/p> <p>9.&nbsp;0<\/p> <p>11.&nbsp;255<\/p> <p>13. \u20131<\/p> <p>15. \u20132, 1, [latex]\\frac{1}{2}[\/latex]<\/p> <p>21.&nbsp;[latex]-\\frac{5}{2}, \\sqrt{6}, -\\sqrt{6}[\/latex]<\/p> <p>23.&nbsp;[latex]2, -4, -\\frac{3}{2}[\/latex]<\/p> <p>25. 4, \u20134, \u20135<\/p> <p>29.&nbsp;[latex]\\frac{1}{2}, \\frac{1+\\sqrt{5}}{2}, \\frac{1-\\sqrt{5}}{2}[\/latex]<\/p> <p>31.&nbsp;[latex]\\frac{3}{2}[\/latex]<\/p> <p>33. 2, 3, \u20131, \u20132<\/p> <p>37.&nbsp;[latex]-1, -1, \\sqrt{5}, -\\sqrt{5}[\/latex]<\/p> <p>41.&nbsp;[latex]2, 3+2i, 3 - 2i[\/latex]<\/p> <p>45. [latex]-\\frac{1}{2}, 1+4i, 1 - 4i[\/latex]<\/p> <p>57.&nbsp;[latex]\\pm 5, \\pm 1, \\pm \\frac{5}{2}[\/latex]<\/p> <p>59.&nbsp;[latex]\\pm 1, \\pm \\frac{1}{2}, \\pm \\frac{1}{3}, \\pm \\frac{1}{6}[\/latex]<\/p> <p>61.&nbsp;[latex]1, \\frac{1}{2}, -\\frac{1}{3}[\/latex]<\/p> <p>63.&nbsp;[latex]2, \\frac{1}{4}, -\\frac{3}{2}[\/latex]<\/p> <p>65.&nbsp;[latex]\\frac{5}{4}[\/latex]<\/p> <p>71.&nbsp;8 by 4 by 6 inches<\/p> <p>75.&nbsp;8 by 5 by 3 inches<\/p> <h2>Inverses and Radical Functions Solutions<\/h2> <p>1.&nbsp;It can be too difficult or impossible to solve for <em>x<\/em>&nbsp;in terms of <em>y<\/em>.<\/p> <p>3.&nbsp;We will need a restriction on the domain of the answer.<\/p> <p>5.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<\/p> <p>7.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+3}-1[\/latex]<\/p> <p>9.&nbsp;[latex]{f}^{-1}\\left(x\\right)=-\\sqrt{\\frac{x - 5}{3}}[\/latex]<\/p> <p>11.&nbsp;[latex]f\\left(x\\right)=\\sqrt{9-x}[\/latex]<\/p> <p>31.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+6}+3[\/latex]<\/p> <p>13.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{x - 5}[\/latex]<\/p> <p>17.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-1}{2},\\left[0,\\infty \\right)[\/latex]<\/p> <p>21.&nbsp;[latex]{f}^{-1}\\left(x\\right)={\\left(\\frac{x - 9}{2}\\right)}^{3}[\/latex]<\/p> <p>23.&nbsp;[latex]{f}^{-1}\\left(x\\right)={\\frac{2 - 8x}{x}}[\/latex]<\/p> <p>25.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\frac{7x - 3}{1-x}[\/latex]<\/p> <p>35.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<br> <img src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230014\/CNX_Precalc_Figure_03_08_2042.jpg\" alt=\"Graph of f(x)= (x-4)^2 and its inverse, f^(-1)(x)= sqrt(x)+4.\"><\/p> <p>37.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{1-x}[\/latex]<br> <img src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230015\/CNX_Precalc_Figure_03_08_2062.jpg\" alt=\"Graph of f(x)= 1-x^3 and its inverse, f^(-1)(x)= (1-x)^(1\/3).\"><\/p> <p>41.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{\\frac{1}{x}}[\/latex]<br> <img src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230015\/CNX_Precalc_Figure_03_08_2102.jpg\" alt=\"Graph of f(x)= 1\/x^2 and its inverse, f^(-1)(x)= sqrt(1\/x).\"><\/p> <p>57.&nbsp;[latex]t\\left(h\\right)=\\sqrt{\\frac{200-h}{4.9}}[\/latex], 5.53 seconds<\/p> <p>59. [latex]r\\left(V\\right)=\\sqrt[3]{\\frac{3V}{4\\pi }}[\/latex], 3.63 feet<\/p> \n\t<\/div>\n\t\t\t\n\t\t\t<\/div>\n\n\t\t<\/div>\n\t\n","rendered":"<div class=\"bc-section section\">\n<div class=\"chapter standard\" id=\"polynomial-equations-get-stronger-key\" title=\"Polynomial Equations: Get Stronger Key\">\n<div class=\"chapter-title-wrap\">\n<p class=\"chapter-number\">\n<h1 class=\"chapter-title\">Polynomial Equations: Get Stronger Key<\/h1>\n<\/p><\/div>\n<div class=\"ugc chapter-ugc\">\n<h2>Dividing Polynomials Solutions<\/h2>\n<p>1.&nbsp;The binomial is a factor of the polynomial.<\/p>\n<p>3.&nbsp;[latex]x+6+\\frac{5}{x - 1}\\text{,}\\text{quotient:}x+6\\text{,}\\text{remainder:}\\text{5}[\/latex]<\/p>\n<p>7.&nbsp;[latex]x - 5\\text{,}\\text{quotient: }x - 5\\text{,}\\text{remainder: }\\text{0}[\/latex]<\/p>\n<p>13.&nbsp;[latex]2{x}^{2}-3x+5\\text{,}\\text{quotient:}2{x}^{2}-3x+5\\text{,}\\text{remainder: }\\text{0}[\/latex]<\/p>\n<p>15.&nbsp;[latex]2{x}^{2}+2x+1+\\frac{10}{x - 4}[\/latex]<\/p>\n<p>19.&nbsp;[latex]3{x}^{2}-11x+34-\\frac{106}{x+3}[\/latex]<\/p>\n<p>23.&nbsp;[latex]4{x}^{2}-21x+84-\\frac{323}{x+4}[\/latex]<\/p>\n<p>29.&nbsp;[latex]{x}^{3}-3x+1[\/latex]<\/p>\n<p>43.&nbsp;[latex]\\text{Quotient: }4{x}^{2}+8x+16\\text{,}\\text{remainder: }-1[\/latex]<\/p>\n<p>45.&nbsp;[latex]\\text{Quotient: }3{x}^{2}+3x+5\\text{,}\\text{remainder: }0[\/latex]<\/p>\n<p>61.&nbsp;[latex]2x+3[\/latex]<\/p>\n<p>63.&nbsp;[latex]x+2[\/latex]<\/p>\n<h2>Complex Numbers Solutions<\/h2>\n<p>1.&nbsp;Add the real parts together and the imaginary parts together.<\/p>\n<p>3.&nbsp;<em>i<\/em>&nbsp;times <em>i<\/em>&nbsp;equals \u20131, which is not imaginary. (answers vary)<\/p>\n<p>7.&nbsp;[latex]14+7i[\/latex]<\/p>\n<p>9.&nbsp;[latex]-\\frac{23}{29}+\\frac{15}{29}i[\/latex]<\/p>\n<p>13.<br \/> <img decoding=\"async\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230014\/CNX_Precalc_Figure_03_01_2032.jpg\" alt=\"Graph of the plotted point, 1-2i.\" \/><\/p>\n<p>15.<br \/> <img decoding=\"async\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230014\/CNX_Precalc_Figure_03_01_2052.jpg\" alt=\"Graph of the plotted point, i.\" \/><\/p>\n<p>17.&nbsp;[latex]8-i[\/latex]<\/p>\n<p>19.&nbsp;[latex]-11+4i[\/latex]<\/p>\n<p>23.&nbsp;[latex]6+15i[\/latex]<\/p>\n<p>25.&nbsp;[latex]-16+32i[\/latex]<\/p>\n<p>27.&nbsp;[latex]-4 - 7i[\/latex]<\/p>\n<p>29.&nbsp;25<\/p>\n<p>33.&nbsp;[latex]4 - 6i[\/latex]<\/p>\n<p>35.&nbsp;[latex]\\frac{2}{5}+\\frac{11}{5}i[\/latex]<\/p>\n<p>37. 15<em>i<\/em><\/p>\n<p>39.&nbsp;[latex]1+i\\sqrt{3}[\/latex]<\/p>\n<p>41. 1<\/p>\n<p>43. \u20131<\/p>\n<h2>Zeros of Polynomial Functions Solutions<\/h2>\n<p>3.&nbsp;Rational zeros can be expressed as fractions whereas real zeros include irrational numbers.<\/p>\n<p>5.&nbsp;Polynomial functions can have repeated zeros, so the fact that number is a zero doesn\u2019t preclude it being a zero again.<\/p>\n<p>7. \u2013106<\/p>\n<p>9.&nbsp;0<\/p>\n<p>11.&nbsp;255<\/p>\n<p>13. \u20131<\/p>\n<p>15. \u20132, 1, [latex]\\frac{1}{2}[\/latex]<\/p>\n<p>21.&nbsp;[latex]-\\frac{5}{2}, \\sqrt{6}, -\\sqrt{6}[\/latex]<\/p>\n<p>23.&nbsp;[latex]2, -4, -\\frac{3}{2}[\/latex]<\/p>\n<p>25. 4, \u20134, \u20135<\/p>\n<p>29.&nbsp;[latex]\\frac{1}{2}, \\frac{1+\\sqrt{5}}{2}, \\frac{1-\\sqrt{5}}{2}[\/latex]<\/p>\n<p>31.&nbsp;[latex]\\frac{3}{2}[\/latex]<\/p>\n<p>33. 2, 3, \u20131, \u20132<\/p>\n<p>37.&nbsp;[latex]-1, -1, \\sqrt{5}, -\\sqrt{5}[\/latex]<\/p>\n<p>41.&nbsp;[latex]2, 3+2i, 3 - 2i[\/latex]<\/p>\n<p>45. [latex]-\\frac{1}{2}, 1+4i, 1 - 4i[\/latex]<\/p>\n<p>57.&nbsp;[latex]\\pm 5, \\pm 1, \\pm \\frac{5}{2}[\/latex]<\/p>\n<p>59.&nbsp;[latex]\\pm 1, \\pm \\frac{1}{2}, \\pm \\frac{1}{3}, \\pm \\frac{1}{6}[\/latex]<\/p>\n<p>61.&nbsp;[latex]1, \\frac{1}{2}, -\\frac{1}{3}[\/latex]<\/p>\n<p>63.&nbsp;[latex]2, \\frac{1}{4}, -\\frac{3}{2}[\/latex]<\/p>\n<p>65.&nbsp;[latex]\\frac{5}{4}[\/latex]<\/p>\n<p>71.&nbsp;8 by 4 by 6 inches<\/p>\n<p>75.&nbsp;8 by 5 by 3 inches<\/p>\n<h2>Inverses and Radical Functions Solutions<\/h2>\n<p>1.&nbsp;It can be too difficult or impossible to solve for <em>x<\/em>&nbsp;in terms of <em>y<\/em>.<\/p>\n<p>3.&nbsp;We will need a restriction on the domain of the answer.<\/p>\n<p>5.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<\/p>\n<p>7.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+3}-1[\/latex]<\/p>\n<p>9.&nbsp;[latex]{f}^{-1}\\left(x\\right)=-\\sqrt{\\frac{x - 5}{3}}[\/latex]<\/p>\n<p>11.&nbsp;[latex]f\\left(x\\right)=\\sqrt{9-x}[\/latex]<\/p>\n<p>31.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+6}+3[\/latex]<\/p>\n<p>13.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{x - 5}[\/latex]<\/p>\n<p>17.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-1}{2},\\left[0,\\infty \\right)[\/latex]<\/p>\n<p>21.&nbsp;[latex]{f}^{-1}\\left(x\\right)={\\left(\\frac{x - 9}{2}\\right)}^{3}[\/latex]<\/p>\n<p>23.&nbsp;[latex]{f}^{-1}\\left(x\\right)={\\frac{2 - 8x}{x}}[\/latex]<\/p>\n<p>25.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\frac{7x - 3}{1-x}[\/latex]<\/p>\n<p>35.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<br \/> <img decoding=\"async\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230014\/CNX_Precalc_Figure_03_08_2042.jpg\" alt=\"Graph of f(x)= (x-4)^2 and its inverse, f^(-1)(x)= sqrt(x)+4.\" \/><\/p>\n<p>37.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{1-x}[\/latex]<br \/> <img decoding=\"async\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230015\/CNX_Precalc_Figure_03_08_2062.jpg\" alt=\"Graph of f(x)= 1-x^3 and its inverse, f^(-1)(x)= (1-x)^(1\/3).\" \/><\/p>\n<p>41.&nbsp;[latex]{f}^{-1}\\left(x\\right)=\\sqrt{\\frac{1}{x}}[\/latex]<br \/> <img decoding=\"async\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/60\/2026\/01\/30230015\/CNX_Precalc_Figure_03_08_2102.jpg\" alt=\"Graph of f(x)= 1\/x^2 and its inverse, f^(-1)(x)= sqrt(1\/x).\" \/><\/p>\n<p>57.&nbsp;[latex]t\\left(h\\right)=\\sqrt{\\frac{200-h}{4.9}}[\/latex], 5.53 seconds<\/p>\n<p>59. [latex]r\\left(V\\right)=\\sqrt[3]{\\frac{3V}{4\\pi }}[\/latex], 3.63 feet<\/p>\n<\/p><\/div>\n<\/p><\/div>\n<\/p><\/div>\n","protected":false},"author":13,"menu_order":5,"template":"","meta":{"_candela_citation":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":224,"module-header":"","content_attributions":null,"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/279"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":1,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/279\/revisions"}],"predecessor-version":[{"id":297,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/279\/revisions\/297"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/224"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/279\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=279"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=279"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=279"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}