{"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":"2026-08-07T16:46:40","modified_gmt":"2026-08-07T16:46:40","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 — Precalculus Practice Page Answer Keys"},"content":{"raw":"

Dividing Polynomials Solutions<\/h2>\r\n1.\u00a0The binomial is a factor of the polynomial.\r\n\r\n2.\u00a0[latex]x+6+\\frac{5}{x - 1}\\text{,}\\text{quotient:}x+6\\text{,}\\text{remainder:}\\text{5}[\/latex]\r\n\r\n3.\u00a0[latex]x - 5\\text{,}\\text{quotient: }x - 5\\text{,}\\text{remainder: }\\text{0}[\/latex]\r\n\r\n4.\u00a0[latex]2{x}^{2}-3x+5\\text{,}\\text{quotient:}2{x}^{2}-3x+5\\text{,}\\text{remainder: }\\text{0}[\/latex]\r\n\r\n5.\u00a0[latex]2{x}^{2}+2x+1+\\frac{10}{x - 4}[\/latex]\r\n\r\n6.\u00a0[latex]3{x}^{2}-11x+34-\\frac{106}{x+3}[\/latex]\r\n\r\n7.\u00a0[latex]4{x}^{2}-21x+84-\\frac{323}{x+4}[\/latex]\r\n\r\n8.\u00a0[latex]{x}^{3}-3x+1[\/latex]\r\n\r\n9.\u00a0[latex]\\text{Quotient: }4{x}^{2}+8x+16\\text{,}\\text{remainder: }-1[\/latex]\r\n\r\n10.\u00a0[latex]\\text{Quotient: }3{x}^{2}+3x+5\\text{,}\\text{remainder: }0[\/latex]\r\n\r\n11.\u00a0[latex]2x+3[\/latex]\r\n\r\n12.\u00a0[latex]x+2[\/latex]\r\n

Complex Numbers Solutions<\/h2>\r\n1.\u00a0Add the real parts together and the imaginary parts together.\r\n\r\n2.\u00a0i<\/em>\u00a0times i<\/em>\u00a0equals \u20131, which is not imaginary. (answers vary)\r\n\r\n3.\u00a0[latex]14+7i[\/latex]\r\n\r\n4.\u00a0[latex]-\\frac{23}{29}+\\frac{15}{29}i[\/latex]\r\n\r\n5.\r\n\"Graph\r\n\r\n6.\r\n\"Graph\r\n\r\n7.\u00a0[latex]8-i[\/latex]\r\n\r\n8.\u00a0[latex]-11+4i[\/latex]\r\n\r\n9.\u00a0[latex]6+15i[\/latex]\r\n\r\n10.\u00a0[latex]-16+32i[\/latex]\r\n\r\n11.\u00a0[latex]-4 - 7i[\/latex]\r\n\r\n12.\u00a025\r\n\r\n13.\u00a0[latex]4 - 6i[\/latex]\r\n\r\n14.\u00a0[latex]\\frac{2}{5}+\\frac{11}{5}i[\/latex]\r\n\r\n15. 15i<\/em>\r\n\r\n16.\u00a0[latex]1+i\\sqrt{3}[\/latex]\r\n\r\n17. 1\r\n\r\n18. \u20131\r\n

Zeros of Polynomial Functions Solutions<\/h2>\r\n1.\u00a0Rational zeros can be expressed as fractions whereas real zeros include irrational numbers.\r\n\r\n2.\u00a0Polynomial functions can have repeated zeros, so the fact that number is a zero doesn't preclude it being a zero again.\r\n\r\n3. \u2013106\r\n\r\n[unmatched: no problem 9 shown in this exercise set \u2014 key answer is \"0\"]\r\n\r\n4.\u00a0255\r\n\r\n5. \u20131\r\n\r\n6. \u20132, 1, [latex]\\frac{1}{2}[\/latex]\r\n\r\n7.\u00a0[latex]-\\frac{5}{2}, \\sqrt{6}, -\\sqrt{6}[\/latex]\r\n\r\n8.\u00a0[latex]2, -4, -\\frac{3}{2}[\/latex]\r\n\r\n9. 4, \u20134, \u20135\r\n\r\n10.\u00a0[latex]\\frac{1}{2}, \\frac{1+\\sqrt{5}}{2}, \\frac{1-\\sqrt{5}}{2}[\/latex]\r\n\r\n11.\u00a0[latex]\\frac{3}{2}[\/latex]\r\n\r\n12. 2, 3, \u20131, \u20132\r\n\r\n13.\u00a0[latex]-1, -1, \\sqrt{5}, -\\sqrt{5}[\/latex]\r\n\r\n14.\u00a0[latex]2, 3+2i, 3 - 2i[\/latex]\r\n\r\n15. [latex]-\\frac{1}{2}, 1+4i, 1 - 4i[\/latex]\r\n\r\n16.\u00a0[latex]\\pm 5, \\pm 1, \\pm \\frac{5}{2}[\/latex]\r\n\r\n17.\u00a0[latex]\\pm 1, \\pm \\frac{1}{2}, \\pm \\frac{1}{3}, \\pm \\frac{1}{6}[\/latex]\r\n\r\n18.\u00a0[latex]1, \\frac{1}{2}, -\\frac{1}{3}[\/latex]\r\n\r\n19.\u00a0[latex]2, \\frac{1}{4}, -\\frac{3}{2}[\/latex]\r\n\r\n20.\u00a0[latex]\\frac{5}{4}[\/latex]\r\n\r\n21.\u00a08 by 4 by 6 inches\r\n\r\n22.\u00a08 by 5 by 3 inches\r\n

Inverses and Radical Functions Solutions<\/h2>\r\n1.\u00a0It can be too difficult or impossible to solve for x<\/em>\u00a0in terms of y<\/em>.\r\n\r\n2.\u00a0We will need a restriction on the domain of the answer.\r\n\r\n3.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]\r\n\r\n4.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+3}-1[\/latex]\r\n\r\n5.\u00a0[latex]{f}^{-1}\\left(x\\right)=-\\sqrt{\\frac{x - 5}{3}}[\/latex]\r\n\r\n6.\u00a0[latex]f\\left(x\\right)=\\sqrt{9-x}[\/latex]\r\n\r\n7.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+6}+3[\/latex]\r\n\r\n8.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{x - 5}[\/latex]\r\n\r\n9.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-1}{2},\\left[0,\\infty \\right)[\/latex]\r\n\r\n10.\u00a0[latex]{f}^{-1}\\left(x\\right)={\\left(\\frac{x - 9}{2}\\right)}^{3}[\/latex]\r\n\r\n11.\u00a0[latex]{f}^{-1}\\left(x\\right)={\\frac{2 - 8x}{x}}[\/latex]\r\n\r\n12.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\frac{7x - 3}{1-x}[\/latex]\r\n\r\n13.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]\r\n\"Graph\r\n\r\n14.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{1-x}[\/latex]\r\n\"Graph\r\n\r\n15.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{\\frac{1}{x}}[\/latex]\r\n\"Graph\r\n\r\n16.\u00a0[latex]t\\left(h\\right)=\\sqrt{\\frac{200-h}{4.9}}[\/latex], 5.53 seconds\r\n\r\n17. [latex]r\\left(V\\right)=\\sqrt[3]{\\frac{3V}{4\\pi }}[\/latex], 3.63 feet","rendered":"

Dividing Polynomials Solutions<\/h2>\n

1.\u00a0The binomial is a factor of the polynomial.<\/p>\n

2.\u00a0[latex]x+6+\\frac{5}{x - 1}\\text{,}\\text{quotient:}x+6\\text{,}\\text{remainder:}\\text{5}[\/latex]<\/p>\n

3.\u00a0[latex]x - 5\\text{,}\\text{quotient: }x - 5\\text{,}\\text{remainder: }\\text{0}[\/latex]<\/p>\n

4.\u00a0[latex]2{x}^{2}-3x+5\\text{,}\\text{quotient:}2{x}^{2}-3x+5\\text{,}\\text{remainder: }\\text{0}[\/latex]<\/p>\n

5.\u00a0[latex]2{x}^{2}+2x+1+\\frac{10}{x - 4}[\/latex]<\/p>\n

6.\u00a0[latex]3{x}^{2}-11x+34-\\frac{106}{x+3}[\/latex]<\/p>\n

7.\u00a0[latex]4{x}^{2}-21x+84-\\frac{323}{x+4}[\/latex]<\/p>\n

8.\u00a0[latex]{x}^{3}-3x+1[\/latex]<\/p>\n

9.\u00a0[latex]\\text{Quotient: }4{x}^{2}+8x+16\\text{,}\\text{remainder: }-1[\/latex]<\/p>\n

10.\u00a0[latex]\\text{Quotient: }3{x}^{2}+3x+5\\text{,}\\text{remainder: }0[\/latex]<\/p>\n

11.\u00a0[latex]2x+3[\/latex]<\/p>\n

12.\u00a0[latex]x+2[\/latex]<\/p>\n

Complex Numbers Solutions<\/h2>\n

1.\u00a0Add the real parts together and the imaginary parts together.<\/p>\n

2.\u00a0i<\/em>\u00a0times i<\/em>\u00a0equals \u20131, which is not imaginary. (answers vary)<\/p>\n

3.\u00a0[latex]14+7i[\/latex]<\/p>\n

4.\u00a0[latex]-\\frac{23}{29}+\\frac{15}{29}i[\/latex]<\/p>\n

5.
\n\"Graph<\/p>\n

6.
\n\"Graph<\/p>\n

7.\u00a0[latex]8-i[\/latex]<\/p>\n

8.\u00a0[latex]-11+4i[\/latex]<\/p>\n

9.\u00a0[latex]6+15i[\/latex]<\/p>\n

10.\u00a0[latex]-16+32i[\/latex]<\/p>\n

11.\u00a0[latex]-4 - 7i[\/latex]<\/p>\n

12.\u00a025<\/p>\n

13.\u00a0[latex]4 - 6i[\/latex]<\/p>\n

14.\u00a0[latex]\\frac{2}{5}+\\frac{11}{5}i[\/latex]<\/p>\n

15. 15i<\/em><\/p>\n

16.\u00a0[latex]1+i\\sqrt{3}[\/latex]<\/p>\n

17. 1<\/p>\n

18. \u20131<\/p>\n

Zeros of Polynomial Functions Solutions<\/h2>\n

1.\u00a0Rational zeros can be expressed as fractions whereas real zeros include irrational numbers.<\/p>\n

2.\u00a0Polynomial functions can have repeated zeros, so the fact that number is a zero doesn’t preclude it being a zero again.<\/p>\n

3. \u2013106<\/p>\n

[unmatched: no problem 9 shown in this exercise set \u2014 key answer is “0”]<\/p>\n

4.\u00a0255<\/p>\n

5. \u20131<\/p>\n

6. \u20132, 1, [latex]\\frac{1}{2}[\/latex]<\/p>\n

7.\u00a0[latex]-\\frac{5}{2}, \\sqrt{6}, -\\sqrt{6}[\/latex]<\/p>\n

8.\u00a0[latex]2, -4, -\\frac{3}{2}[\/latex]<\/p>\n

9. 4, \u20134, \u20135<\/p>\n

10.\u00a0[latex]\\frac{1}{2}, \\frac{1+\\sqrt{5}}{2}, \\frac{1-\\sqrt{5}}{2}[\/latex]<\/p>\n

11.\u00a0[latex]\\frac{3}{2}[\/latex]<\/p>\n

12. 2, 3, \u20131, \u20132<\/p>\n

13.\u00a0[latex]-1, -1, \\sqrt{5}, -\\sqrt{5}[\/latex]<\/p>\n

14.\u00a0[latex]2, 3+2i, 3 - 2i[\/latex]<\/p>\n

15. [latex]-\\frac{1}{2}, 1+4i, 1 - 4i[\/latex]<\/p>\n

16.\u00a0[latex]\\pm 5, \\pm 1, \\pm \\frac{5}{2}[\/latex]<\/p>\n

17.\u00a0[latex]\\pm 1, \\pm \\frac{1}{2}, \\pm \\frac{1}{3}, \\pm \\frac{1}{6}[\/latex]<\/p>\n

18.\u00a0[latex]1, \\frac{1}{2}, -\\frac{1}{3}[\/latex]<\/p>\n

19.\u00a0[latex]2, \\frac{1}{4}, -\\frac{3}{2}[\/latex]<\/p>\n

20.\u00a0[latex]\\frac{5}{4}[\/latex]<\/p>\n

21.\u00a08 by 4 by 6 inches<\/p>\n

22.\u00a08 by 5 by 3 inches<\/p>\n

Inverses and Radical Functions Solutions<\/h2>\n

1.\u00a0It can be too difficult or impossible to solve for x<\/em>\u00a0in terms of y<\/em>.<\/p>\n

2.\u00a0We will need a restriction on the domain of the answer.<\/p>\n

3.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]<\/p>\n

4.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+3}-1[\/latex]<\/p>\n

5.\u00a0[latex]{f}^{-1}\\left(x\\right)=-\\sqrt{\\frac{x - 5}{3}}[\/latex]<\/p>\n

6.\u00a0[latex]f\\left(x\\right)=\\sqrt{9-x}[\/latex]<\/p>\n

7.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x+6}+3[\/latex]<\/p>\n

8.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{x - 5}[\/latex]<\/p>\n

9.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\frac{{x}^{2}-1}{2},\\left[0,\\infty \\right)[\/latex]<\/p>\n

10.\u00a0[latex]{f}^{-1}\\left(x\\right)={\\left(\\frac{x - 9}{2}\\right)}^{3}[\/latex]<\/p>\n

11.\u00a0[latex]{f}^{-1}\\left(x\\right)={\\frac{2 - 8x}{x}}[\/latex]<\/p>\n

12.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\frac{7x - 3}{1-x}[\/latex]<\/p>\n

13.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+4[\/latex]
\n\"Graph<\/p>\n

14.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt[3]{1-x}[\/latex]
\n\"Graph<\/p>\n

15.\u00a0[latex]{f}^{-1}\\left(x\\right)=\\sqrt{\\frac{1}{x}}[\/latex]
\n\"Graph<\/p>\n

16.\u00a0[latex]t\\left(h\\right)=\\sqrt{\\frac{200-h}{4.9}}[\/latex], 5.53 seconds<\/p>\n

17. [latex]r\\left(V\\right)=\\sqrt[3]{\\frac{3V}{4\\pi }}[\/latex], 3.63 feet<\/p>\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":"- Select Header -","content_attributions":[],"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":3,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/279\/revisions"}],"predecessor-version":[{"id":394,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/279\/revisions\/394"}],"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}]}}