{"id":135,"date":"2025-02-13T22:44:11","date_gmt":"2025-02-13T22:44:11","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/partial-fractions\/"},"modified":"2026-03-20T19:25:36","modified_gmt":"2026-03-20T19:25:36","slug":"partial-fractions","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/partial-fractions\/","title":{"raw":"Partial Fractions: Learn It 1","rendered":"Partial Fractions: Learn It 1"},"content":{"raw":"
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  • Decompose [latex]\\frac{{P( x )}}{{ Q( x )}}[\/latex] , where Q( x ) has only linear factors.<\/li>\r\n \t
  • Decompose [latex]\\frac{{P( x )}}{{ Q( x )}}[\/latex] , where Q( x ) has an irreducible quadratic factor.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section>\r\n

    Linear Factors<\/h2>\r\nRecall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression.\r\n\r\n
    How to: Add Rational Expressions<\/strong>\r\n
      \r\n \t
    1. Factor the numerator and denominator.<\/li>\r\n \t
    2. Find the LCD of the expressions.<\/li>\r\n \t
    3. Multiply the expressions by a form of 1 that changes the denominators to the LCD.<\/li>\r\n \t
    4. Add or subtract the numerators over the common denominator.<\/li>\r\n \t
    5. Simplify.<\/li>\r\n<\/ol>\r\n<\/section>In this section, we will look at 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 rational expression<\/strong> to the original expressions, called the partial fractions<\/strong>. Some types of rational expressions require solving a system of equations in order to decompose them.\r\n\r\n
      For example, suppose we add the following fractions:\r\n

      [latex]\\dfrac{2}{x - 3}+\\dfrac{-1}{x+2}[\/latex]<\/p>\r\nWe would first need to find a common denominator,\r\n

      [latex]\\left(x+2\\right)\\left(x - 3\\right)[\/latex].<\/p>\r\nNext, we would write each expression with this common denominator and find the sum of the terms.\r\n

      [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>\r\nPartial fraction decomposition<\/strong> is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.\r\n

      [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>\r\n\r\n<\/section>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.\r\n\r\n

      \r\n

      partial fraction decomposition: nonrepeated linear factors<\/h3>\r\nThe 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\r\n\r\n \r\n

      [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>\r\n\r\n<\/section>

      How To: Given a rational expression with distinct linear factors in the denominator, decompose it.<\/strong>\r\n
        \r\n \t
      1. 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
        [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>\r\n \t
      2. Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\r\n \t
      3. Expand the right side of the equation and collect like terms.<\/li>\r\n \t
      4. 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>
        Decompose the given rational expression with nonrepeated linear factors.\r\n

        [latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"199986\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"199986\"]\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

        [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

        [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

        [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

        [latex]\\begin{gather*} 3x = Ax - A + Bx + 2B \\\\ 3x = (A + B)x - A + 2B \\\\ 3x + 0 = (A + B)x + (-A + 2B) \\end{gather*}[\/latex]<\/p>\r\nSet up a system of equations associating corresponding coefficients.\r\n

        [latex]\\begin{align*} \\text{Coefficient of } x: & \\quad 3 = A + B \\\\ \\text{Constant term: } & \\quad 0 = -A + 2B \\end{align*}[\/latex]<\/p>\r\nSolving the system of equation.\r\n

        [latex]\\begin{align*} 3 &= A + B \\\\ 0 &= -A + 2B \\\\ \\hline\u00a0 3 &= 3B \\\\ B &= 1 \\\\ \\text{Substitute } B = 1 \\text{ into } 3 = A + B: \\\\ 3 &= A + 1 \\\\ A &= 3 - 1 \\\\ A &= 2 \\end{align*}[\/latex]<\/p>\r\nThus, the partial fraction decomposition is\r\n

        [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[\/hidden-answer]\r\n\r\n[reveal-answer q=\"208444\"]Heaviside Method[\/reveal-answer]\r\n[hidden-answer a=\"208444\"]\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 [latex]0[\/latex]. If we let [latex]x=1[\/latex], the\u00a0[latex]A-[\/latex] term becomes [latex]0[\/latex] and we can simply solve for [latex]B[\/latex].\r\n

        [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>\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

        [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>\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

        [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 Heaviside method<\/strong>, named after Charles Heaviside, a pioneer in the study of electronics.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

        [ohm_question hide_question_numbers=1]321634[\/ohm_question]<\/section><\/div>\r\n
        [ohm_question hide_question_numbers=1]321635[\/ohm_question]<\/section>
        [ohm_question hide_question_numbers=1]321636[\/ohm_question]<\/section>\r\n
        \r\n \t
        <\/dd>\r\n<\/dl>","rendered":"
        \n
        \n
        \n
          \n
        • Decompose [latex]\\frac{{P( x )}}{{ Q( x )}}[\/latex] , where Q( x ) has only linear factors.<\/li>\n
        • Decompose [latex]\\frac{{P( x )}}{{ Q( x )}}[\/latex] , where Q( x ) has an irreducible quadratic factor.<\/li>\n<\/ul>\n<\/div>\n<\/section>\n

          Linear Factors<\/h2>\n

          Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression.<\/p>\n

          How to: Add Rational Expressions<\/strong><\/p>\n
            \n
          1. Factor the numerator and denominator.<\/li>\n
          2. Find the LCD of the expressions.<\/li>\n
          3. Multiply the expressions by a form of 1 that changes the denominators to the LCD.<\/li>\n
          4. Add or subtract the numerators over the common denominator.<\/li>\n
          5. Simplify.<\/li>\n<\/ol>\n<\/section>\n

            In this section, we will look at 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 rational expression<\/strong> to the original expressions, called the partial fractions<\/strong>. Some types of rational expressions require solving a system of equations in order to decompose them.<\/p>\n

            For example, suppose we add the following fractions:<\/p>\n

            [latex]\\dfrac{2}{x - 3}+\\dfrac{-1}{x+2}[\/latex]<\/p>\n

            We would first need to find a common denominator,<\/p>\n

            [latex]\\left(x+2\\right)\\left(x - 3\\right)[\/latex].<\/p>\n

            Next, we would write each expression with this common denominator and find the sum of the terms.<\/p>\n

            [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

            Partial fraction 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

            [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<\/section>\n

            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

            \n

            partial fraction decomposition: nonrepeated linear factors<\/h3>\n

            The 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>\n

            [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<\/section>\n

            How To: Given a rational expression with distinct linear factors in the denominator, decompose it.<\/strong><\/p>\n
              \n
            1. 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
              [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
            2. Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\n
            3. Expand the right side of the equation and collect like terms.<\/li>\n
            4. 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
              Decompose the given rational expression with nonrepeated linear factors.<\/p>\n

              [latex]\\dfrac{3x}{\\left(x+2\\right)\\left(x - 1\\right)}[\/latex]<\/p>\n