{"id":1783,"date":"2015-11-12T18:30:45","date_gmt":"2015-11-12T18:30:45","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1783"},"modified":"2015-11-12T18:30:45","modified_gmt":"2015-11-12T18:30:45","slug":"solutions-25","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/solutions-25\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"<h2>Solutions to Try Its<\/h2>\n1.\u00a0[latex]\\frac{3}{x - 3}-\\frac{2}{x - 2}[\/latex]\n\n2.\u00a0[latex]\\frac{6}{x - 1}-\\frac{5}{{\\left(x - 1\\right)}^{2}}[\/latex]\n\n3.\u00a0[latex]\\frac{3}{x - 1}+\\frac{2x - 4}{{x}^{2}+1}[\/latex]\n\n4.\u00a0[latex]\\frac{x - 2}{{x}^{2}-2x+3}+\\frac{2x+1}{{\\left({x}^{2}-2x+3\\right)}^{2}}[\/latex]\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\n1.\u00a0No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, [latex]\\frac{1}{{x}^{2}+1}[\/latex] cannot be decomposed because the denominator cannot be factored.\n\n3.\u00a0Graph both sides and ensure they are equal.\n\n5.\u00a0If we choose [latex]x=-1[\/latex], then the <em>B<\/em>-term disappears, letting us immediately know that [latex]A=3[\/latex]. We could alternatively plug in [latex]x=-\\frac{5}{3}[\/latex], giving us a <em>B<\/em>-value of [latex]-2[\/latex].\n\n7.\u00a0[latex]\\frac{8}{x+3}-\\frac{5}{x - 8}[\/latex]\n\n9.\u00a0[latex]\\frac{1}{x+5}+\\frac{9}{x+2}[\/latex]\n\n11.\u00a0[latex]\\frac{3}{5x - 2}+\\frac{4}{4x - 1}[\/latex]\n\n13.\u00a0[latex]\\frac{5}{2\\left(x+3\\right)}+\\frac{5}{2\\left(x - 3\\right)}[\/latex]\n\n15.\u00a0[latex]\\frac{3}{x+2}+\\frac{3}{x - 2}[\/latex]\n\n17.\u00a0[latex]\\frac{9}{5\\left(x+2\\right)}+\\frac{11}{5\\left(x - 3\\right)}[\/latex]\n\n19.\u00a0[latex]\\frac{8}{x - 3}-\\frac{5}{x - 2}[\/latex]\n\n21.\u00a0[latex]\\frac{1}{x - 2}+\\frac{2}{{\\left(x - 2\\right)}^{2}}[\/latex]\n\n23.\u00a0[latex]-\\frac{6}{4x+5}+\\frac{3}{{\\left(4x+5\\right)}^{2}}[\/latex]\n\n25.\u00a0[latex]-\\frac{1}{x - 7}-\\frac{2}{{\\left(x - 7\\right)}^{2}}[\/latex]\n\n27.\u00a0[latex]\\frac{4}{x}-\\frac{3}{2\\left(x+1\\right)}+\\frac{7}{2{\\left(x+1\\right)}^{2}}[\/latex]\n\n29.\u00a0[latex]\\frac{4}{x}+\\frac{2}{{x}^{2}}-\\frac{3}{3x+2}+\\frac{7}{2{\\left(3x+2\\right)}^{2}}[\/latex]\n\n31.\u00a0[latex]\\frac{x+1}{{x}^{2}+x+3}+\\frac{3}{x+2}[\/latex]\n\n33.\u00a0[latex]\\frac{4 - 3x}{{x}^{2}+3x+8}+\\frac{1}{x - 1}[\/latex]\n\n35.\u00a0[latex]\\frac{2x - 1}{{x}^{2}+6x+1}+\\frac{2}{x+3}[\/latex]\n\n37.\u00a0[latex]\\frac{1}{{x}^{2}+x+1}+\\frac{4}{x - 1}[\/latex]\n\n39.\u00a0[latex]\\frac{2}{{x}^{2}-3x+9}+\\frac{3}{x+3}[\/latex]\n\n41.\u00a0[latex]-\\frac{1}{4{x}^{2}+6x+9}+\\frac{1}{2x - 3}[\/latex]\n\n43.\u00a0[latex]\\frac{1}{x}+\\frac{1}{x+6}-\\frac{4x}{{x}^{2}-6x+36}[\/latex]\n\n45.\u00a0[latex]\\frac{x+6}{{x}^{2}+1}+\\frac{4x+3}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex]\n\n47.\u00a0[latex]\\frac{x+1}{x+2}+\\frac{2x+3}{{\\left(x+2\\right)}^{2}}[\/latex]\n\n49.\u00a0[latex]\\frac{1}{{x}^{2}+3x+25}-\\frac{3x}{{\\left({x}^{2}+3x+25\\right)}^{2}}[\/latex]\n\n51.\u00a0[latex]\\frac{1}{8x}-\\frac{x}{8\\left({x}^{2}+4\\right)}+\\frac{10-x}{2{\\left({x}^{2}+4\\right)}^{2}}[\/latex]\n\n53.\u00a0[latex]-\\frac{16}{x}-\\frac{9}{{x}^{2}}+\\frac{16}{x - 1}-\\frac{7}{{\\left(x - 1\\right)}^{2}}[\/latex]\n\n55.\u00a0[latex]\\frac{1}{x+1}-\\frac{2}{{\\left(x+1\\right)}^{2}}+\\frac{5}{{\\left(x+1\\right)}^{3}}[\/latex]\n\n57.\u00a0[latex]\\frac{5}{x - 2}-\\frac{3}{10\\left(x+2\\right)}+\\frac{7}{x+8}-\\frac{7}{10\\left(x - 8\\right)}[\/latex]\n\n59.\u00a0[latex]-\\frac{5}{4x}-\\frac{5}{2\\left(x+2\\right)}+\\frac{11}{2\\left(x+4\\right)}+\\frac{5}{4\\left(x+4\\right)}[\/latex]","rendered":"<h2>Solutions to Try Its<\/h2>\n<p>1.\u00a0[latex]\\frac{3}{x - 3}-\\frac{2}{x - 2}[\/latex]<\/p>\n<p>2.\u00a0[latex]\\frac{6}{x - 1}-\\frac{5}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\n<p>3.\u00a0[latex]\\frac{3}{x - 1}+\\frac{2x - 4}{{x}^{2}+1}[\/latex]<\/p>\n<p>4.\u00a0[latex]\\frac{x - 2}{{x}^{2}-2x+3}+\\frac{2x+1}{{\\left({x}^{2}-2x+3\\right)}^{2}}[\/latex]<\/p>\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\n<p>1.\u00a0No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, [latex]\\frac{1}{{x}^{2}+1}[\/latex] cannot be decomposed because the denominator cannot be factored.<\/p>\n<p>3.\u00a0Graph both sides and ensure they are equal.<\/p>\n<p>5.\u00a0If we choose [latex]x=-1[\/latex], then the <em>B<\/em>-term disappears, letting us immediately know that [latex]A=3[\/latex]. We could alternatively plug in [latex]x=-\\frac{5}{3}[\/latex], giving us a <em>B<\/em>-value of [latex]-2[\/latex].<\/p>\n<p>7.\u00a0[latex]\\frac{8}{x+3}-\\frac{5}{x - 8}[\/latex]<\/p>\n<p>9.\u00a0[latex]\\frac{1}{x+5}+\\frac{9}{x+2}[\/latex]<\/p>\n<p>11.\u00a0[latex]\\frac{3}{5x - 2}+\\frac{4}{4x - 1}[\/latex]<\/p>\n<p>13.\u00a0[latex]\\frac{5}{2\\left(x+3\\right)}+\\frac{5}{2\\left(x - 3\\right)}[\/latex]<\/p>\n<p>15.\u00a0[latex]\\frac{3}{x+2}+\\frac{3}{x - 2}[\/latex]<\/p>\n<p>17.\u00a0[latex]\\frac{9}{5\\left(x+2\\right)}+\\frac{11}{5\\left(x - 3\\right)}[\/latex]<\/p>\n<p>19.\u00a0[latex]\\frac{8}{x - 3}-\\frac{5}{x - 2}[\/latex]<\/p>\n<p>21.\u00a0[latex]\\frac{1}{x - 2}+\\frac{2}{{\\left(x - 2\\right)}^{2}}[\/latex]<\/p>\n<p>23.\u00a0[latex]-\\frac{6}{4x+5}+\\frac{3}{{\\left(4x+5\\right)}^{2}}[\/latex]<\/p>\n<p>25.\u00a0[latex]-\\frac{1}{x - 7}-\\frac{2}{{\\left(x - 7\\right)}^{2}}[\/latex]<\/p>\n<p>27.\u00a0[latex]\\frac{4}{x}-\\frac{3}{2\\left(x+1\\right)}+\\frac{7}{2{\\left(x+1\\right)}^{2}}[\/latex]<\/p>\n<p>29.\u00a0[latex]\\frac{4}{x}+\\frac{2}{{x}^{2}}-\\frac{3}{3x+2}+\\frac{7}{2{\\left(3x+2\\right)}^{2}}[\/latex]<\/p>\n<p>31.\u00a0[latex]\\frac{x+1}{{x}^{2}+x+3}+\\frac{3}{x+2}[\/latex]<\/p>\n<p>33.\u00a0[latex]\\frac{4 - 3x}{{x}^{2}+3x+8}+\\frac{1}{x - 1}[\/latex]<\/p>\n<p>35.\u00a0[latex]\\frac{2x - 1}{{x}^{2}+6x+1}+\\frac{2}{x+3}[\/latex]<\/p>\n<p>37.\u00a0[latex]\\frac{1}{{x}^{2}+x+1}+\\frac{4}{x - 1}[\/latex]<\/p>\n<p>39.\u00a0[latex]\\frac{2}{{x}^{2}-3x+9}+\\frac{3}{x+3}[\/latex]<\/p>\n<p>41.\u00a0[latex]-\\frac{1}{4{x}^{2}+6x+9}+\\frac{1}{2x - 3}[\/latex]<\/p>\n<p>43.\u00a0[latex]\\frac{1}{x}+\\frac{1}{x+6}-\\frac{4x}{{x}^{2}-6x+36}[\/latex]<\/p>\n<p>45.\u00a0[latex]\\frac{x+6}{{x}^{2}+1}+\\frac{4x+3}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex]<\/p>\n<p>47.\u00a0[latex]\\frac{x+1}{x+2}+\\frac{2x+3}{{\\left(x+2\\right)}^{2}}[\/latex]<\/p>\n<p>49.\u00a0[latex]\\frac{1}{{x}^{2}+3x+25}-\\frac{3x}{{\\left({x}^{2}+3x+25\\right)}^{2}}[\/latex]<\/p>\n<p>51.\u00a0[latex]\\frac{1}{8x}-\\frac{x}{8\\left({x}^{2}+4\\right)}+\\frac{10-x}{2{\\left({x}^{2}+4\\right)}^{2}}[\/latex]<\/p>\n<p>53.\u00a0[latex]-\\frac{16}{x}-\\frac{9}{{x}^{2}}+\\frac{16}{x - 1}-\\frac{7}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\n<p>55.\u00a0[latex]\\frac{1}{x+1}-\\frac{2}{{\\left(x+1\\right)}^{2}}+\\frac{5}{{\\left(x+1\\right)}^{3}}[\/latex]<\/p>\n<p>57.\u00a0[latex]\\frac{5}{x - 2}-\\frac{3}{10\\left(x+2\\right)}+\\frac{7}{x+8}-\\frac{7}{10\\left(x - 8\\right)}[\/latex]<\/p>\n<p>59.\u00a0[latex]-\\frac{5}{4x}-\\frac{5}{2\\left(x+2\\right)}+\\frac{11}{2\\left(x+4\\right)}+\\frac{5}{4\\left(x+4\\right)}[\/latex]<\/p>\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-1783\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1783","chapter","type-chapter","status-publish","hentry"],"part":1775,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1783","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1783\/revisions"}],"predecessor-version":[{"id":2255,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1783\/revisions\/2255"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1775"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1783\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/media?parent=1783"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1783"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1783"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/license?post=1783"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}