{"id":391,"date":"2019-07-15T22:44:42","date_gmt":"2019-07-15T22:44:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/chapter\/distinct-and-repeated-linear-factors\/"},"modified":"2019-07-15T22:44:42","modified_gmt":"2019-07-15T22:44:42","slug":"distinct-and-repeated-linear-factors","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/chapter\/distinct-and-repeated-linear-factors\/","title":{"raw":"Linear Factors","rendered":"Linear Factors"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Decompose a rational expression with distinct linear factors.<\/li>\n \t<li>Decompose a rational expression with repeated linear factors.<\/li>\n<\/ul>\n<\/div>\nRecall the algebra regarding adding and subtracting rational expressions.\n<div class=\"textbox examples\">\n<h3>recall how to add rational expressions<\/h3>\n<ol>\n \t<li>Factor the numerator and denominator.<\/li>\n \t<li>Find the LCD of the expressions.<\/li>\n \t<li>Multiply the expressions by a form of 1 that changes the denominators to the LCD.<\/li>\n \t<li>Add or subtract the numerators over the common denominator.<\/li>\n \t<li>Simplify.<\/li>\n<\/ol>\n<\/div>\nThese operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at <strong>partial fraction decomposition<\/strong>, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified <strong>rational expression<\/strong> to the original expressions, called the <strong>partial fractions<\/strong>. Some types of rational expressions require solving a system of equations in order to decompose them.\n\nFor example, suppose we add the following fractions:\n<p style=\"text-align: center\">[latex]\\dfrac{2}{x - 3}+\\dfrac{-1}{x+2}[\/latex]<\/p>\nWe would first need to find a common denominator,\n<p style=\"text-align: center\">[latex]\\left(x+2\\right)\\left(x - 3\\right)[\/latex].<\/p>\nNext, we would write each expression with this common denominator and find the sum of the terms.\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;\\frac{2}{x - 3}\\left(\\frac{x+2}{x+2}\\right)+\\frac{-1}{x+2}\\left(\\frac{x - 3}{x - 3}\\right) \\\\[2mm] &amp;=\\frac{2x+4-x+3}{\\left(x+2\\right)\\left(x - 3\\right)} \\\\[2mm] &amp;=\\frac{x+7}{{x}^{2}-x - 6} \\end{align}[\/latex]<\/p>\nPartial fraction <strong>decomposition<\/strong> is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.\n<p style=\"text-align: center\">[latex]\\begin{align} \\dfrac{x+7}{{x}^{2}-x - 6}&amp;=\\dfrac{2}{x - 3}+\\dfrac{-1}{x+2} \\\\[2mm]\\text{Simplified sum}&amp;\\hspace{6mm}\\text{Partial fraction decomposition} \\end{align}[\/latex]<\/p>\nWe will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.\n\nWhen the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of [latex]{x}^{2}-x - 6[\/latex] are [latex]\\left(x - 3\\right)\\left(x+2\\right)[\/latex], the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nThe techniques to decompose fractions are challenging and will require some practice to become familiar to you. The explanations of how to perform the decomposition may be confusing the first time (or more) that you read through them. If so, work through the given examples on paper first, slowly, step by step. Then return to the General Note and How To boxes to gain an understanding. It will take time and effort, so don't be discouraged if it takes multiple attempts for each example.\n\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Partial Fraction Decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}:Q\\left(x\\right)[\/latex] Has Nonrepeated Linear Factors<\/h3>\nThe <strong>partial fraction decomposition<\/strong> of [latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] when [latex]Q\\left(x\\right)[\/latex] has nonrepeated linear factors and the degree of [latex]P\\left(x\\right)[\/latex] is less than the degree of [latex]Q\\left(x\\right)[\/latex] is\n<p style=\"text-align: center\">[latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{A}_{1}}{\\left({a}_{1}x+{b}_{1}\\right)}+\\dfrac{{A}_{2}}{\\left({a}_{2}x+{b}_{2}\\right)}+\\dfrac{{A}_{3}}{\\left({a}_{3}x+{b}_{3}\\right)}+\\cdot \\cdot \\cdot +\\dfrac{{A}_{n}}{\\left({a}_{n}x+{b}_{n}\\right)}[\/latex].<\/p>\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a rational expression with distinct linear factors in the denominator, decompose it.<\/h3>\n<ol>\n \t<li>Use a variable for the original numerators, usually [latex]A,B,[\/latex] or [latex]C[\/latex], depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use [latex]{A}_{n}[\/latex] for each numerator\n<div style=\"text-align: center\">[latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{A}_{1}}{\\left({a}_{1}x+{b}_{1}\\right)}+\\dfrac{{A}_{2}}{\\left({a}_{2}x+{b}_{2}\\right)}+\\cdots \\text{+}\\dfrac{{A}_{n}}{\\left({a}_{n}x+{b}_{n}\\right)}[\/latex]<\/div><\/li>\n \t<li>Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\n \t<li>Expand the right side of the equation and collect like terms.<\/li>\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>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<h3>recall how to eliminate denominators from an equation using the lcd<\/h3>\nThe following example relies upon the technique of multiplying both sides of a rational equation by the lowest common denominator (LCD) to eliminate the denominators from the equation.\n<ol>\n \t<li>Factor all denominators in the equation<\/li>\n \t<li>Find the LCD<\/li>\n \t<li>Multiply the whole equation by the LCD (each term). As you do, cancel out the denominator in each term. If the LCD has been chose correctly, there will be no denominators remaining at the end of this step.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Decomposing a Rational Expression with Distinct Linear Factors<\/h3>\nDecompose the given <strong>rational expression<\/strong> with distinct linear factors.\n<p style=\"text-align: center\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/p>\n[reveal-answer q=\"828392\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"828392\"]\n\nWe will separate the denominator factors and give each numerator a symbolic label, like [latex]A,B[\/latex], or [latex]C[\/latex].\n<p style=\"text-align: center\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{A}{\\left(x+2\\right)}+\\dfrac{B}{\\left(x - 1\\right)}[\/latex]<\/p>\nMultiply both sides of the equation by the common denominator to eliminate the fractions:\n<p style=\"text-align: center\">[latex]\\cancel{\\left(x+2\\right)}\\cancel{\\left(x - 1\\right)}\\left[\\dfrac{3x}{\\cancel{\\left(x+2\\right)}\\cancel{\\left(x - 1\\right)}}\\right]=\\cancel{\\left(x+2\\right)}\\left(x - 1\\right)\\left[\\dfrac{A}{\\cancel{\\left(x+2\\right)}}\\right]+\\left(x+2\\right)\\cancel{\\left(x - 1\\right)}\\left[\\dfrac{B}{\\cancel{\\left(x - 1\\right)}}\\right][\/latex]<\/p>\nThe resulting equation is\n<p style=\"text-align: center\">[latex]3x=A\\left(x - 1\\right)+B\\left(x+2\\right)[\/latex]<\/p>\nExpand the right side of the equation and collect like terms.\n<p style=\"text-align: center\">[latex]\\begin{gathered}3x=Ax-A+Bx+2B\\\\ 3x=\\left(A+B\\right)x-A+2B\\end{gathered}[\/latex]<\/p>\nSet up a system of equations associating corresponding coefficients.\n<p style=\"text-align: center\">[latex]\\begin{gathered}3=A+B\\\\ 0=-A+2B\\end{gathered}[\/latex]<\/p>\n\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nAssociating corresponding coefficients is a well-used mathematical technique. In this example,&nbsp;if\n<p style=\"text-align: center\">[latex](3)x + 0 = (A+B)x + (-A+2B)[\/latex]<\/p>\nthen it must be true that\n<p style=\"text-align: center\">[latex]3=A+B[\/latex] and [latex]0=-A+2B[\/latex].<\/p>\n\n<\/div>\nAdd the two equations and solve for [latex]B[\/latex].\n<p style=\"text-align: center\">[latex]\\begin{align}3&amp;=A+B \\\\ 0&amp;=-A+2B \\\\ \\hline 3&amp;=0+3B \\\\[4mm] B&amp;=1 \\end{align}[\/latex]<\/p>\nSubstitute [latex]B=1[\/latex] into one of the original equations in the system.\n<p style=\"text-align: center\">[latex]\\begin{align}3&amp;=A+1\\\\ 2&amp;=A\\end{align}[\/latex]<\/p>\nThus, the partial fraction decomposition is\n<p style=\"text-align: center\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{2}{\\left(x+2\\right)}+\\dfrac{1}{\\left(x - 1\\right)}[\/latex]<\/p>\n&nbsp;\n\nAnother method to use to solve for [latex]A[\/latex] or [latex]B[\/latex] is by considering the equation that resulted from eliminating the fractions and substituting a value for [latex]x[\/latex] that will make either the [latex]A-[\/latex]&nbsp;or [latex]B-[\/latex]term equal 0. If we let [latex]x=1[\/latex], the&nbsp;[latex]A-[\/latex] term becomes 0 and we can simply solve for [latex]B[\/latex].\n<p style=\"text-align: center\">[latex]\\begin{align}3x&amp;=A\\left(x - 1\\right)+B\\left(x+2\\right) \\\\ 3\\left(1\\right)&amp;=A\\left[\\left(1\\right)-1\\right]+B\\left[\\left(1\\right)+2\\right] \\\\ 3&amp;=0+3B\\hfill \\\\ B&amp;=1 \\end{align}[\/latex]<\/p>\nNext, either substitute [latex]B=1[\/latex] into the equation and solve for [latex]A[\/latex], or make the [latex]B-[\/latex]term 0 by substituting [latex]x=-2[\/latex] into the equation.\n<p style=\"text-align: center\">[latex]\\begin{align}3x&amp;=A\\left(x - 1\\right)+B\\left(x+2\\right) \\\\ 3\\left(-2\\right)&amp;=A\\left[\\left(-2\\right)-1\\right]+B\\left[\\left(-2\\right)+2\\right] \\\\ -6&amp;=-3A+0 \\\\ \\frac{-6}{-3}&amp;=A \\\\ A&amp;=2 \\end{align}[\/latex]<\/p>\nWe obtain the same values for [latex]A[\/latex] and [latex]B[\/latex] using either method, so the decompositions are the same using either method.\n<p style=\"text-align: center\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{2}{\\left(x+2\\right)}+\\dfrac{1}{\\left(x - 1\\right)}[\/latex]<\/p>\nAlthough this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the <strong>Heaviside method<\/strong>, named after Charles Heaviside, a pioneer in the study of electronics.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFind the partial fraction decomposition of the following expression.\n<p style=\"text-align: center\">[latex]\\dfrac{x}{\\left(x - 3\\right)\\left(x - 2\\right)}[\/latex]<\/p>\n[reveal-answer q=\"800291\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"800291\"]\n\n[latex]\\dfrac{3}{x - 3}-\\dfrac{2}{x - 2}[\/latex]\n\n[\/hidden-answer]\n\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=18135&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\n\n<\/div>\nIn this video, you will see another example of how to find a partial fraction decomposition when you have distinct linear factors.\n\nhttps:\/\/youtu.be\/WoVdOcuSI0I\n<h2>Decomposing P(x)\/ Q(x), Where Q(x) Has Repeated Linear Factors<\/h2>\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.\n<div class=\"textbox\">\n<h3>A General Note: Partial Fraction Decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}:Q\\left(x\\right)[\/latex] Has Repeated Linear Factors<\/h3>\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\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>\nWrite the denominator powers in increasing order.\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a rational expression with repeated linear factors, decompose it.<\/h3>\n<ol>\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.\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>\n \t<li>Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\n \t<li>Expand the right side of the equation and collect like terms.<\/li>\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>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Decomposing with Repeated Linear Factors<\/h3>\nDecompose the given rational expression with repeated linear factors.\n<p style=\"text-align: center\">[latex]\\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}[\/latex]<\/p>\n[reveal-answer q=\"969334\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"969334\"]\n\nThe denominator factors are [latex]x{\\left(x - 2\\right)}^{2}[\/latex]. 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]. Thus,\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>\nNext, we multiply both sides by the common denominator.\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>\nOn the right side of the equation, we expand and collect like terms.\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]\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>\nNext, we compare the coefficients of both sides. This will give the system of equations in three variables:\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&amp; \\hfill \\text{(1)}\\\\ \\hfill -4A - 2B+C=2&amp; \\hfill \\text{(2)}\\\\ \\hfill 4A=4&amp; \\hfill \\text{(3)}\\end{array}[\/latex]<\/p>\nSolving for [latex]A[\/latex] , we have\n<p style=\"text-align: center\">[latex]\\begin{align}4A&amp;=4 \\\\ A&amp;=1 \\end{align}[\/latex]<\/p>\nSubstitute [latex]A=1[\/latex] into equation (1).\n<p style=\"text-align: center\">[latex]\\begin{align}A+B=-1 \\\\ \\left(1\\right)+B=-1 \\\\ B=-2 \\end{align}[\/latex]<\/p>\nThen, to solve for [latex]C[\/latex], substitute the values for [latex]A[\/latex] and [latex]B[\/latex] into equation (2).\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>\nThus,\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[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFind the partial fraction decomposition of the expression with repeated linear factors.\n<p style=\"text-align: center\">[latex]\\dfrac{6x - 11}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\n[reveal-answer q=\"623632\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"623632\"]\n\n[latex]\\dfrac{6}{x - 1}-\\dfrac{5}{{\\left(x - 1\\right)}^{2}}[\/latex]\n\n[\/hidden-answer]\n\n<\/div>\nIn this video, you will see an example of how to find the partial fraction decomposition of a rational expression with repeated linear factors.\n\nhttps:\/\/youtu.be\/6DdwGw_5dvk\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Decompose a rational expression with distinct linear factors.<\/li>\n<li>Decompose a rational expression with repeated linear factors.<\/li>\n<\/ul>\n<\/div>\n<p>Recall the algebra regarding adding and subtracting rational expressions.<\/p>\n<div class=\"textbox examples\">\n<h3>recall how to add rational expressions<\/h3>\n<ol>\n<li>Factor the numerator and denominator.<\/li>\n<li>Find the LCD of the expressions.<\/li>\n<li>Multiply the expressions by a form of 1 that changes the denominators to the LCD.<\/li>\n<li>Add or subtract the numerators over the common denominator.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<p>These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at <strong>partial fraction decomposition<\/strong>, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified <strong>rational expression<\/strong> to the original expressions, called the <strong>partial fractions<\/strong>. Some types of rational expressions require solving a system of equations in order to decompose them.<\/p>\n<p>For example, suppose we add the following fractions:<\/p>\n<p style=\"text-align: center\">[latex]\\dfrac{2}{x - 3}+\\dfrac{-1}{x+2}[\/latex]<\/p>\n<p>We would first need to find a common denominator,<\/p>\n<p style=\"text-align: center\">[latex]\\left(x+2\\right)\\left(x - 3\\right)[\/latex].<\/p>\n<p>Next, we would write each expression with this common denominator and find the sum of the terms.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&\\frac{2}{x - 3}\\left(\\frac{x+2}{x+2}\\right)+\\frac{-1}{x+2}\\left(\\frac{x - 3}{x - 3}\\right) \\\\[2mm] &=\\frac{2x+4-x+3}{\\left(x+2\\right)\\left(x - 3\\right)} \\\\[2mm] &=\\frac{x+7}{{x}^{2}-x - 6} \\end{align}[\/latex]<\/p>\n<p>Partial fraction <strong>decomposition<\/strong> is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\dfrac{x+7}{{x}^{2}-x - 6}&=\\dfrac{2}{x - 3}+\\dfrac{-1}{x+2} \\\\[2mm]\\text{Simplified sum}&\\hspace{6mm}\\text{Partial fraction decomposition} \\end{align}[\/latex]<\/p>\n<p>We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.<\/p>\n<p>When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of [latex]{x}^{2}-x - 6[\/latex] are [latex]\\left(x - 3\\right)\\left(x+2\\right)[\/latex], the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>The techniques to decompose fractions are challenging and will require some practice to become familiar to you. The explanations of how to perform the decomposition may be confusing the first time (or more) that you read through them. If so, work through the given examples on paper first, slowly, step by step. Then return to the General Note and How To boxes to gain an understanding. It will take time and effort, so don&#8217;t be discouraged if it takes multiple attempts for each example.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Partial Fraction Decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}:Q\\left(x\\right)[\/latex] Has Nonrepeated Linear Factors<\/h3>\n<p>The <strong>partial fraction decomposition<\/strong> of [latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] when [latex]Q\\left(x\\right)[\/latex] has nonrepeated linear factors 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 style=\"text-align: center\">[latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{A}_{1}}{\\left({a}_{1}x+{b}_{1}\\right)}+\\dfrac{{A}_{2}}{\\left({a}_{2}x+{b}_{2}\\right)}+\\dfrac{{A}_{3}}{\\left({a}_{3}x+{b}_{3}\\right)}+\\cdot \\cdot \\cdot +\\dfrac{{A}_{n}}{\\left({a}_{n}x+{b}_{n}\\right)}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a rational expression with distinct linear factors in the denominator, decompose it.<\/h3>\n<ol>\n<li>Use a variable for the original numerators, usually [latex]A,B,[\/latex] or [latex]C[\/latex], depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use [latex]{A}_{n}[\/latex] for each numerator\n<div style=\"text-align: center\">[latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{{A}_{1}}{\\left({a}_{1}x+{b}_{1}\\right)}+\\dfrac{{A}_{2}}{\\left({a}_{2}x+{b}_{2}\\right)}+\\cdots \\text{+}\\dfrac{{A}_{n}}{\\left({a}_{n}x+{b}_{n}\\right)}[\/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<\/div>\n<div class=\"textbox examples\">\n<h3>recall how to eliminate denominators from an equation using the lcd<\/h3>\n<p>The following example relies upon the technique of multiplying both sides of a rational equation by the lowest common denominator (LCD) to eliminate the denominators from the equation.<\/p>\n<ol>\n<li>Factor all denominators in the equation<\/li>\n<li>Find the LCD<\/li>\n<li>Multiply the whole equation by the LCD (each term). As you do, cancel out the denominator in each term. If the LCD has been chose correctly, there will be no denominators remaining at the end of this step.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Decomposing a Rational Expression with Distinct Linear Factors<\/h3>\n<p>Decompose the given <strong>rational expression<\/strong> with distinct linear factors.<\/p>\n<p style=\"text-align: center\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q828392\">Show Solution<\/span><\/p>\n<div id=\"q828392\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will separate the denominator factors and give each numerator a symbolic label, like [latex]A,B[\/latex], or [latex]C[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{A}{\\left(x+2\\right)}+\\dfrac{B}{\\left(x - 1\\right)}[\/latex]<\/p>\n<p>Multiply both sides of the equation by the common denominator to eliminate the fractions:<\/p>\n<p style=\"text-align: center\">[latex]\\cancel{\\left(x+2\\right)}\\cancel{\\left(x - 1\\right)}\\left[\\dfrac{3x}{\\cancel{\\left(x+2\\right)}\\cancel{\\left(x - 1\\right)}}\\right]=\\cancel{\\left(x+2\\right)}\\left(x - 1\\right)\\left[\\dfrac{A}{\\cancel{\\left(x+2\\right)}}\\right]+\\left(x+2\\right)\\cancel{\\left(x - 1\\right)}\\left[\\dfrac{B}{\\cancel{\\left(x - 1\\right)}}\\right][\/latex]<\/p>\n<p>The resulting equation is<\/p>\n<p style=\"text-align: center\">[latex]3x=A\\left(x - 1\\right)+B\\left(x+2\\right)[\/latex]<\/p>\n<p>Expand the right side of the equation and collect like terms.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}3x=Ax-A+Bx+2B\\\\ 3x=\\left(A+B\\right)x-A+2B\\end{gathered}[\/latex]<\/p>\n<p>Set up a system of equations associating corresponding coefficients.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}3=A+B\\\\ 0=-A+2B\\end{gathered}[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Associating corresponding coefficients is a well-used mathematical technique. In this example,&nbsp;if<\/p>\n<p style=\"text-align: center\">[latex](3)x + 0 = (A+B)x + (-A+2B)[\/latex]<\/p>\n<p>then it must be true that<\/p>\n<p style=\"text-align: center\">[latex]3=A+B[\/latex] and [latex]0=-A+2B[\/latex].<\/p>\n<\/div>\n<p>Add the two equations and solve for [latex]B[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}3&=A+B \\\\ 0&=-A+2B \\\\ \\hline 3&=0+3B \\\\[4mm] B&=1 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]B=1[\/latex] into one of the original equations in the system.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}3&=A+1\\\\ 2&=A\\end{align}[\/latex]<\/p>\n<p>Thus, the partial fraction decomposition is<\/p>\n<p style=\"text-align: center\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{2}{\\left(x+2\\right)}+\\dfrac{1}{\\left(x - 1\\right)}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Another method to use to solve for [latex]A[\/latex] or [latex]B[\/latex] is by considering the equation that resulted from eliminating the fractions and substituting a value for [latex]x[\/latex] that will make either the [latex]A-[\/latex]&nbsp;or [latex]B-[\/latex]term equal 0. If we let [latex]x=1[\/latex], the&nbsp;[latex]A-[\/latex] term becomes 0 and we can simply solve for [latex]B[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}3x&=A\\left(x - 1\\right)+B\\left(x+2\\right) \\\\ 3\\left(1\\right)&=A\\left[\\left(1\\right)-1\\right]+B\\left[\\left(1\\right)+2\\right] \\\\ 3&=0+3B\\hfill \\\\ B&=1 \\end{align}[\/latex]<\/p>\n<p>Next, either substitute [latex]B=1[\/latex] into the equation and solve for [latex]A[\/latex], or make the [latex]B-[\/latex]term 0 by substituting [latex]x=-2[\/latex] into the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}3x&=A\\left(x - 1\\right)+B\\left(x+2\\right) \\\\ 3\\left(-2\\right)&=A\\left[\\left(-2\\right)-1\\right]+B\\left[\\left(-2\\right)+2\\right] \\\\ -6&=-3A+0 \\\\ \\frac{-6}{-3}&=A \\\\ A&=2 \\end{align}[\/latex]<\/p>\n<p>We obtain the same values for [latex]A[\/latex] and [latex]B[\/latex] using either method, so the decompositions are the same using either method.<\/p>\n<p style=\"text-align: center\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}=\\dfrac{2}{\\left(x+2\\right)}+\\dfrac{1}{\\left(x - 1\\right)}[\/latex]<\/p>\n<p>Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the <strong>Heaviside method<\/strong>, named after Charles Heaviside, a pioneer in the study of electronics.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the partial fraction decomposition of the following expression.<\/p>\n<p style=\"text-align: center\">[latex]\\dfrac{x}{\\left(x - 3\\right)\\left(x - 2\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q800291\">Show Solution<\/span><\/p>\n<div id=\"q800291\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{3}{x - 3}-\\dfrac{2}{x - 2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm18135\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=18135&#38;theme=oea&#38;iframe_resize_id=ohm18135&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In this video, you will see another example of how to find a partial fraction decomposition when you have distinct linear factors.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1: Partial Fraction Decomposition (Linear Factors)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/WoVdOcuSI0I?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Decomposing P(x)\/ Q(x), Where Q(x) Has 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<div class=\"textbox\">\n<h3>A General Note: Partial Fraction Decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}:Q\\left(x\\right)[\/latex] Has 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 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>Write the denominator powers in increasing order.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a rational expression with repeated linear factors, decompose it.<\/h3>\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<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Decomposing with Repeated Linear Factors<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q969334\">Show Solution<\/span><\/p>\n<div id=\"q969334\" class=\"hidden-answer\" style=\"display: none\">\n<p>The denominator factors are [latex]x{\\left(x - 2\\right)}^{2}[\/latex]. 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]. 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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the partial fraction decomposition of the expression with repeated linear factors.<\/p>\n<p style=\"text-align: center\">[latex]\\dfrac{6x - 11}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q623632\">Show Solution<\/span><\/p>\n<div id=\"q623632\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\dfrac{6}{x - 1}-\\dfrac{5}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In this video, you will see an example of how to find the partial fraction decomposition of a rational expression with repeated linear factors.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 3: Partial Fraction Decomposition (Repeated Linear Factors)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/6DdwGw_5dvk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/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-391\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Ex 1: Partial Fraction Decomposition (Linear Factors) . <strong>Authored by<\/strong>: Sousa, James (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/WoVdOcuSI0I\">https:\/\/youtu.be\/WoVdOcuSI0I<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 3: Partial Fraction Decomposition (Repeated Linear Factors) . <strong>Authored by<\/strong>: Sousa, James (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/6DdwGw_5dvk\">https:\/\/youtu.be\/6DdwGw_5dvk<\/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 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":17533,"menu_order":23,"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\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et 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