{"id":2336,"date":"2021-09-29T16:29:05","date_gmt":"2021-09-29T16:29:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-properties-of-power-series\/"},"modified":"2021-11-19T03:20:28","modified_gmt":"2021-11-19T03:20:28","slug":"skills-review-for-properties-of-power-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-properties-of-power-series\/","title":{"raw":"Skills Review for Properties of Power Series","rendered":"Skills Review for Properties of Power Series"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use partial fraction decomposition for linear factors<\/li>\r\n \t<li>Simplify expressions using the power property of pxponents<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Properties of Power Series section, we will look at how to combine, differentiate, and integrate power series. Here we will review how to use partial fraction decomposition for linear factors and how to use the power property of exponents.\r\n<h2>Use Partial Fraction Decomposition for Linear Factors<\/h2>\r\n<strong><em>(also in Module 5, Skills Review for Infinite Series)<\/em><\/strong>\r\n\r\nPartial fraction <strong>decomposition<\/strong> is used to break up one fraction into two.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\underset{\\text{ }\\\\ \\text{Simplified sum}}{\\frac{x+7}{{x}^{2}-x - 6}}=\\underset{\\text{ }\\\\ \\text{Partial fraction decomposition}}{\\frac{2}{x - 3}+\\frac{-1}{x+2}}\\\\ \\text{ }\\end{align}[\/latex]<\/div>\r\nWe will investigate rational expressions with linear 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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a rational expression with distinct linear factors in the denominator, decompose it.<\/h3>\r\n<ol>\r\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\r\n<div style=\"text-align: center;\">[latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{A}_{1}}{\\left({a}_{1}x+{b}_{1}\\right)}+\\frac{{A}_{2}}{\\left({a}_{2}x+{b}_{2}\\right)}+\\cdots \\text{+}\\frac{{A}_{n}}{\\left({a}_{n}x+{b}_{n}\\right)}[\/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<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Decomposing a Rational Function\u00a0 with Distinct Linear Factors<\/h3>\r\nDecompose the given <strong>rational expression<\/strong> with distinct linear factors.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"828392\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"828392\"]\r\n\r\nWe will separate the denominator factors and give each numerator a symbolic label, like [latex]A,B[\/latex], or [latex]C[\/latex].\r\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>\r\nMultiply both sides of the equation by the common denominator to eliminate the fractions:\r\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>\r\nThe resulting equation is\r\n<p style=\"text-align: center;\">[latex]3x=A\\left(x - 1\\right)+B\\left(x+2\\right)[\/latex]<\/p>\r\nExpand the right side of the equation and collect like terms.\r\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>\r\nSet up a system of equations associating corresponding coefficients.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}3=A+B\\\\ 0=-A+2B\\end{gathered}[\/latex]<\/p>\r\nAdd the two equations and solve for [latex]B[\/latex].\r\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>\r\nSubstitute [latex]B=1[\/latex] into one of the original equations in the system.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3&amp;=A+1\\\\ 2&amp;=A\\end{align}[\/latex]<\/p>\r\nThus, the partial fraction decomposition is\r\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>\r\n&nbsp;\r\n\r\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]\u00a0or [latex]B-[\/latex]term equal 0. If we let [latex]x=1[\/latex], the\u00a0[latex]A-[\/latex] term becomes 0 and we can simply solve for [latex]B[\/latex].\r\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>\r\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.\r\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>\r\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.\r\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>\r\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.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the partial fraction decomposition of the following expression.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{\\left(x - 3\\right)\\left(x - 2\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"800291\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"800291\"]\r\n\r\n[latex]\\dfrac{3}{x - 3}-\\dfrac{2}{x - 2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174463[\/ohm_question]\r\n\r\n<\/div>\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<div class=\"textbox\">\r\n<h3>How To: Given a rational expression with repeated linear factors, decompose it.<\/h3>\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<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Decomposing with Repeated Linear Factors<\/h3>\r\nDecompose 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\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,\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<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the partial fraction decomposition of the expression with repeated linear factors.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{6x - 11}{{\\left(x - 1\\right)}^{2}}[\/latex]<\/p>\r\n[reveal-answer q=\"623632\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"623632\"]\r\n\r\n[latex]\\dfrac{6}{x - 1}-\\dfrac{5}{{\\left(x - 1\\right)}^{2}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]174464[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Simplify Expressions Using the Power Property for Exponents<\/h2>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Power Rule of Exponents<\/h3>\r\nFor any real number [latex]a[\/latex] and positive integers [latex]m[\/latex] and [latex]n[\/latex], the power rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: USING the Power Rule of Exponents<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]{\\left({x}^{2}\\right)}^{7}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"992335\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"992335\"]\r\n\r\nUse the power rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{\\left({x}^{2}\\right)}^{7}={x}^{2\\cdot 7}={x}^{14}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}={\\left(2t\\right)}^{5\\cdot 3}={\\left(2t\\right)}^{15}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}={\\left(-3\\right)}^{5\\cdot 11}={\\left(-3\\right)}^{55}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"ohm93399\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93399&amp;theme=oea&amp;iframe_resize_id=ohm93399&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93370&amp;theme=oea&amp;iframe_resize_id=mom80[\/embed]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use partial fraction decomposition for linear factors<\/li>\n<li>Simplify expressions using the power property of pxponents<\/li>\n<\/ul>\n<\/div>\n<p>In the Properties of Power Series section, we will look at how to combine, differentiate, and integrate power series. Here we will review how to use partial fraction decomposition for linear factors and how to use the power property of exponents.<\/p>\n<h2>Use Partial Fraction Decomposition for Linear Factors<\/h2>\n<p><strong><em>(also in Module 5, Skills Review for Infinite Series)<\/em><\/strong><\/p>\n<p>Partial fraction <strong>decomposition<\/strong> is used to break up one fraction into two.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\underset{\\text{ }\\\\ \\text{Simplified sum}}{\\frac{x+7}{{x}^{2}-x - 6}}=\\underset{\\text{ }\\\\ \\text{Partial fraction decomposition}}{\\frac{2}{x - 3}+\\frac{-1}{x+2}}\\\\ \\text{ }\\end{align}[\/latex]<\/div>\n<p>We will investigate rational expressions with linear 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<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]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{A}_{1}}{\\left({a}_{1}x+{b}_{1}\\right)}+\\frac{{A}_{2}}{\\left({a}_{2}x+{b}_{2}\\right)}+\\cdots \\text{+}\\frac{{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 exercises\">\n<h3>Example: Decomposing a Rational Function\u00a0 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\">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<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]\u00a0or [latex]B-[\/latex]term equal 0. If we let [latex]x=1[\/latex], the\u00a0[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=\"bcc-box bcc-success\">\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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174463\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174463&theme=oea&iframe_resize_id=ohm174463\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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>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=\"bcc-box bcc-success\">\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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm174464\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174464&theme=oea&iframe_resize_id=ohm174464\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Simplify Expressions Using the Power Property for Exponents<\/h2>\n<div class=\"textbox\">\n<h3>A General Note: The Power Rule of Exponents<\/h3>\n<p>For any real number [latex]a[\/latex] and positive integers [latex]m[\/latex] and [latex]n[\/latex], the power rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/div>\n<\/div>\n<div>\n<div class=\"textbox exercises\">\n<h3>Example: USING the Power Rule of Exponents<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]{\\left({x}^{2}\\right)}^{7}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q992335\">Show Solution<\/span><\/p>\n<div id=\"q992335\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the power rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]{\\left({x}^{2}\\right)}^{7}={x}^{2\\cdot 7}={x}^{14}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(2t\\right)}^{5}\\right)}^{3}={\\left(2t\\right)}^{5\\cdot 3}={\\left(2t\\right)}^{15}[\/latex]<\/li>\n<li>[latex]{\\left({\\left(-3\\right)}^{5}\\right)}^{11}={\\left(-3\\right)}^{5\\cdot 11}={\\left(-3\\right)}^{55}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm93399\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93399&amp;theme=oea&amp;iframe_resize_id=ohm93399&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm93370\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93370&#38;theme=oea&#38;iframe_resize_id=ohm93370&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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-2336\">\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>Modification and Revision. <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 Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/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>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/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":349141,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen 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