{"id":1214,"date":"2025-07-23T21:00:10","date_gmt":"2025-07-23T21:00:10","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1214"},"modified":"2026-03-20T19:32:03","modified_gmt":"2026-03-20T19:32:03","slug":"partial-fractions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/partial-fractions-learn-it-2\/","title":{"raw":"Partial Fractions: Learn It 2","rendered":"Partial Fractions: Learn It 2"},"content":{"raw":"<h2>Repeated Linear Factors<\/h2>\r\nSome fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>partial fraction decomposition: repeated linear factors<\/h3>\r\nThe partial fraction decomposition of [latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex], when [latex]Q\\left(x\\right)[\/latex] has a repeated linear factor occurring [latex]n[\/latex] times and the degree of [latex]P\\left(x\\right)[\/latex] is less than the degree of [latex]Q\\left(x\\right)[\/latex], is\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{A}_{1}}{\\left(ax+b\\right)}+\\dfrac{{A}_{2}}{{\\left(ax+b\\right)}^{2}}+\\dfrac{{A}_{3}}{{\\left(ax+b\\right)}^{3}}+\\cdot \\cdot \\cdot +\\dfrac{{A}_{n}}{{\\left(ax+b\\right)}^{n}}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nWrite the denominator powers in increasing order.\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a rational expression with repeated linear factors, decompose it.<\/strong>\r\n<ol>\r\n \t<li>Use a variable like [latex]A,B[\/latex], or [latex]C[\/latex] for the numerators and account for increasing powers of the denominators.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{A}_{1}}{\\left(ax+b\\right)}+\\dfrac{{A}_{2}}{{\\left(ax+b\\right)}^{2}}+ \\text{. }\\text{. }\\text{. + }\\dfrac{{A}_{n}}{{\\left(ax+b\\right)}^{n}}[\/latex]<\/div><\/li>\r\n \t<li>Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\r\n \t<li>Expand the right side of the equation and collect like terms.<\/li>\r\n \t<li>Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Decompose the given rational expression with repeated linear factors.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}[\/latex]<\/p>\r\n[reveal-answer q=\"969334\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"969334\"]\r\n\r\nFactoring the denominator: [latex]x(x^2-4x+4) = x{\\left(x - 2\\right)}^{2}[\/latex].\r\n\r\nTo allow for the repeated factor of [latex]\\left(x - 2\\right)[\/latex], the decomposition will include three denominators: [latex]x,\\left(x - 2\\right)[\/latex], and [latex]{\\left(x - 2\\right)}^{2}[\/latex].\r\n\r\nThus,\r\n<p style=\"text-align: center;\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\\dfrac{A}{x}+\\dfrac{B}{\\left(x - 2\\right)}+\\dfrac{C}{{\\left(x - 2\\right)}^{2}}[\/latex]<\/p>\r\nNext, we multiply both sides by the common denominator.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x{\\left(x - 2\\right)}^{2}\\left[\\dfrac{-{x}^{2}+2x+4}{x{\\left(x - 2\\right)}^{2}}\\right]=\\left[\\dfrac{A}{x}+\\dfrac{B}{\\left(x - 2\\right)}+\\dfrac{C}{{\\left(x - 2\\right)}^{2}}\\right]x{\\left(x - 2\\right)}^{2} \\\\[2mm] -{x}^{2}+2x+4=A{\\left(x - 2\\right)}^{2}+Bx\\left(x - 2\\right)+Cx \\end{gathered}[\/latex]<\/p>\r\nOn the right side of the equation, we expand and collect like terms.\r\n<p style=\"text-align: center;\">[latex]-{x}^{2}+2x+4=A\\left({x}^{2}-4x+4\\right)+B\\left({x}^{2}-2x\\right)+Cx[\/latex]\r\n[latex]\\begin{align}&amp;=A{x}^{2}-4Ax+4A+B{x}^{2}-2Bx+Cx \\\\ &amp;=\\left(A+B\\right){x}^{2}+\\left(-4A - 2B+C\\right)x+4A \\end{align}[\/latex]<\/p>\r\nNext, we compare the coefficients of both sides. This will give the system of equations in three variables:\r\n<p style=\"text-align: center;\">[latex]-{x}^{2}+2x+4=\\left(A+B\\right){x}^{2}+\\left(-4A - 2B+C\\right)x+4A[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rr}\\hfill A+B=-1&amp; \\hfill \\text{(1)}\\\\ \\hfill -4A - 2B+C=2&amp; \\hfill \\text{(2)}\\\\ \\hfill 4A=4&amp; \\hfill \\text{(3)}\\end{array}[\/latex]<\/p>\r\nSolving for [latex]A[\/latex] , we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}4A&amp;=4 \\\\ A&amp;=1 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]A=1[\/latex] into equation (1).\r\n<p style=\"text-align: center;\">[latex]\\begin{align}A+B=-1 \\\\ \\left(1\\right)+B=-1 \\\\ B=-2 \\end{align}[\/latex]<\/p>\r\nThen, to solve for [latex]C[\/latex], substitute the values for [latex]A[\/latex] and [latex]B[\/latex] into equation (2).\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-4A - 2B+C=2\\\\ -4\\left(1\\right)-2\\left(-2\\right)+C=2\\\\ -4+4+C=2\\\\ C=2\\end{align}[\/latex]<\/p>\r\nThus,\r\n<p style=\"text-align: center;\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\\dfrac{1}{x}-\\dfrac{2}{\\left(x - 2\\right)}+\\dfrac{2}{{\\left(x - 2\\right)}^{2}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321637[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321638[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\n[ohm_question hide_question_numbers=1]321639[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section>","rendered":"<h2>Repeated Linear Factors<\/h2>\n<p>Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>partial fraction decomposition: repeated linear factors<\/h3>\n<p>The partial fraction decomposition of [latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex], when [latex]Q\\left(x\\right)[\/latex] has a repeated linear factor occurring [latex]n[\/latex] times and the degree of [latex]P\\left(x\\right)[\/latex] is less than the degree of [latex]Q\\left(x\\right)[\/latex], is<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{A}_{1}}{\\left(ax+b\\right)}+\\dfrac{{A}_{2}}{{\\left(ax+b\\right)}^{2}}+\\dfrac{{A}_{3}}{{\\left(ax+b\\right)}^{3}}+\\cdot \\cdot \\cdot +\\dfrac{{A}_{n}}{{\\left(ax+b\\right)}^{n}}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Write the denominator powers in increasing order.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a rational expression with repeated linear factors, decompose it.<\/strong><\/p>\n<ol>\n<li>Use a variable like [latex]A,B[\/latex], or [latex]C[\/latex] for the numerators and account for increasing powers of the denominators.\n<div style=\"text-align: center;\">[latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{A}_{1}}{\\left(ax+b\\right)}+\\dfrac{{A}_{2}}{{\\left(ax+b\\right)}^{2}}+ \\text{. }\\text{. }\\text{. + }\\dfrac{{A}_{n}}{{\\left(ax+b\\right)}^{n}}[\/latex]<\/div>\n<\/li>\n<li>Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\n<li>Expand the right side of the equation and collect like terms.<\/li>\n<li>Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Decompose the given rational expression with repeated linear factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q969334\">Show Solution<\/button><\/p>\n<div id=\"q969334\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factoring the denominator: [latex]x(x^2-4x+4) = x{\\left(x - 2\\right)}^{2}[\/latex].<\/p>\n<p>To allow for the repeated factor of [latex]\\left(x - 2\\right)[\/latex], the decomposition will include three denominators: [latex]x,\\left(x - 2\\right)[\/latex], and [latex]{\\left(x - 2\\right)}^{2}[\/latex].<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\\dfrac{A}{x}+\\dfrac{B}{\\left(x - 2\\right)}+\\dfrac{C}{{\\left(x - 2\\right)}^{2}}[\/latex]<\/p>\n<p>Next, we multiply both sides by the common denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x{\\left(x - 2\\right)}^{2}\\left[\\dfrac{-{x}^{2}+2x+4}{x{\\left(x - 2\\right)}^{2}}\\right]=\\left[\\dfrac{A}{x}+\\dfrac{B}{\\left(x - 2\\right)}+\\dfrac{C}{{\\left(x - 2\\right)}^{2}}\\right]x{\\left(x - 2\\right)}^{2} \\\\[2mm] -{x}^{2}+2x+4=A{\\left(x - 2\\right)}^{2}+Bx\\left(x - 2\\right)+Cx \\end{gathered}[\/latex]<\/p>\n<p>On the right side of the equation, we expand and collect like terms.<\/p>\n<p style=\"text-align: center;\">[latex]-{x}^{2}+2x+4=A\\left({x}^{2}-4x+4\\right)+B\\left({x}^{2}-2x\\right)+Cx[\/latex]<br \/>\n[latex]\\begin{align}&=A{x}^{2}-4Ax+4A+B{x}^{2}-2Bx+Cx \\\\ &=\\left(A+B\\right){x}^{2}+\\left(-4A - 2B+C\\right)x+4A \\end{align}[\/latex]<\/p>\n<p>Next, we compare the coefficients of both sides. This will give the system of equations in three variables:<\/p>\n<p style=\"text-align: center;\">[latex]-{x}^{2}+2x+4=\\left(A+B\\right){x}^{2}+\\left(-4A - 2B+C\\right)x+4A[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rr}\\hfill A+B=-1& \\hfill \\text{(1)}\\\\ \\hfill -4A - 2B+C=2& \\hfill \\text{(2)}\\\\ \\hfill 4A=4& \\hfill \\text{(3)}\\end{array}[\/latex]<\/p>\n<p>Solving for [latex]A[\/latex] , we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}4A&=4 \\\\ A&=1 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]A=1[\/latex] into equation (1).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}A+B=-1 \\\\ \\left(1\\right)+B=-1 \\\\ B=-2 \\end{align}[\/latex]<\/p>\n<p>Then, to solve for [latex]C[\/latex], substitute the values for [latex]A[\/latex] and [latex]B[\/latex] into equation (2).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-4A - 2B+C=2\\\\ -4\\left(1\\right)-2\\left(-2\\right)+C=2\\\\ -4+4+C=2\\\\ C=2\\end{align}[\/latex]<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\\dfrac{1}{x}-\\dfrac{2}{\\left(x - 2\\right)}+\\dfrac{2}{{\\left(x - 2\\right)}^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321637\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321637&theme=lumen&iframe_resize_id=ohm321637&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321638\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321638&theme=lumen&iframe_resize_id=ohm321638&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p><iframe loading=\"lazy\" id=\"ohm321639\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321639&theme=lumen&iframe_resize_id=ohm321639&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n","protected":false},"author":13,"menu_order":28,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":131,"module-header":"learn_it","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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1214"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1214\/revisions"}],"predecessor-version":[{"id":5951,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1214\/revisions\/5951"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/131"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1214\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1214"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1214"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1214"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1214"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}