{"id":1780,"date":"2015-11-12T18:30:45","date_gmt":"2015-11-12T18:30:45","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1780"},"modified":"2015-11-12T18:30:45","modified_gmt":"2015-11-12T18:30:45","slug":"decomposing-px-qx-when-qx-has-a-repeated-irreducible-quadratic-factor","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/decomposing-px-qx-when-qx-has-a-repeated-irreducible-quadratic-factor\/","title":{"raw":"Decomposing P(x) \/ Q(x), When Q(x) Has a Repeated Irreducible Quadratic Factor","rendered":"Decomposing P(x) \/ Q(x), When Q(x) Has a Repeated Irreducible Quadratic Factor"},"content":{"raw":"

Now that we can decompose a simplified rational expression<\/strong> with an irreducible quadratic<\/strong> factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.\n<\/p>

\n

A General Note: Decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] When Q(x)<\/em> Has a Repeated Irreducible Quadratic Factor<\/h3>\nThe partial fraction decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex], when [latex]Q\\left(x\\right)[\/latex] has a repeated irreducible quadratic factor and the degree of [latex]P\\left(x\\right)[\/latex] is less than the degree of [latex]Q\\left(x\\right)[\/latex], is\n
[latex]\\frac{P\\left(x\\right)}{{\\left(a{x}^{2}+bx+c\\right)}^{n}}=\\frac{{A}_{1}x+{B}_{1}}{\\left(a{x}^{2}+bx+c\\right)}+\\frac{{A}_{2}x+{B}_{2}}{{\\left(a{x}^{2}+bx+c\\right)}^{2}}+\\frac{{A}_{3}x+{B}_{3}}{{\\left(a{x}^{2}+bx+c\\right)}^{3}}+\\cdot \\cdot \\cdot +\\frac{{A}_{n}x+{B}_{n}}{{\\left(a{x}^{2}+bx+c\\right)}^{n}}[\/latex]<\/div>\nWrite the denominators in increasing powers.\n\n<\/div>\n
\n

How To: Given a rational expression that has a repeated irreducible factor, decompose it.<\/h3>\n
  1. Use variables like [latex]A,B[\/latex], or [latex]C[\/latex] for the constant numerators over linear factors, and linear expressions such as [latex]{A}_{1}x+{B}_{1},{A}_{2}x+{B}_{2}[\/latex], etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as\n
    [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{A}{ax+b}+\\frac{{A}_{1}x+{B}_{1}}{\\left(a{x}^{2}+bx+c\\right)}+\\frac{{A}_{2}x+{B}_{2}}{{\\left(a{x}^{2}+bx+c\\right)}^{2}}+\\cdots +\\text{ }\\frac{{A}_{n}+{B}_{n}}{{\\left(a{x}^{2}+bx+c\\right)}^{n}}[\/latex]<\/div><\/li>\n\t
  2. Multiply both sides of the equation by the common denominator to eliminate fractions.<\/li>\n\t
  3. Expand the right side of the equation and collect like terms.<\/li>\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>\n<\/ol><\/div>\n
    \n

    Example 4: Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator<\/h3>\nDecompose the given expression that has a repeated irreducible factor in the denominator.\n
    [latex]\\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\\left({x}^{2}+1\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n
    \n

    Solution<\/h3>\nThe factors of the denominator are [latex]x,\\left({x}^{2}+1\\right)[\/latex], and [latex]{\\left({x}^{2}+1\\right)}^{2}[\/latex]. Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form [latex]Ax+B[\/latex]. So, let\u2019s begin the decomposition.\n
    [latex]\\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\\left({x}^{2}+1\\right)}^{2}}=\\frac{A}{x}+\\frac{Bx+C}{\\left({x}^{2}+1\\right)}+\\frac{Dx+E}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex]<\/div>\nWe eliminate the denominators by multiplying each term by [latex]x{\\left({x}^{2}+1\\right)}^{2}[\/latex]. Thus,\n
    [latex]{x}^{4}+{x}^{3}+{x}^{2}-x+1=A{\\left({x}^{2}+1\\right)}^{2}+\\left(Bx+C\\right)\\left(x\\right)\\left({x}^{2}+1\\right)+\\left(Dx+E\\right)\\left(x\\right)[\/latex]<\/div>\nExpand the right side.\n
    [latex]\\begin{array}{l} \\text{ }{x}^{4}+{x}^{3}+{x}^{2}-x+1=A\\left({x}^{4}+2{x}^{2}+1\\right)+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\\hfill \\\\ \\text{ }=A{x}^{4}+2A{x}^{2}+A+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\\hfill \\end{array}[\/latex]<\/div>\nNow we will collect like terms.\n
    [latex]{x}^{4}+{x}^{3}+{x}^{2}-x+1=\\left(A+B\\right){x}^{4}+\\left(C\\right){x}^{3}+\\left(2A+B+D\\right){x}^{2}+\\left(C+E\\right)x+A[\/latex]<\/div>\nSet up the system of equations matching corresponding coefficients on each side of the equal sign.\n
    [latex]\\begin{array}{l}\\text{ }A+B=1\\hfill \\\\ \\text{ }C=1\\hfill \\\\ 2A+B+D=1\\hfill \\\\ \\text{ }C+E=-1\\hfill \\\\ \\text{ }A=1\\hfill \\end{array}[\/latex]<\/div>\nWe can use substitution from this point. Substitute [latex]A=1[\/latex] into the first equation.\n
    [latex]\\begin{array}{l}1+B=1\\hfill \\\\ \\text{ }B=0\\hfill \\end{array}[\/latex]<\/div>\nSubstitute [latex]A=1[\/latex] and [latex]B=0[\/latex] into the third equation.\n
    [latex]\\begin{array}{l}2\\left(1\\right)+0+D=1\\hfill \\\\ \\text{ }D=-1\\hfill \\end{array}[\/latex]<\/div>\nSubstitute [latex]C=1[\/latex] into the fourth equation.\n
    [latex]\\begin{array}{r}\\hfill 1+E=-1\\\\ \\hfill \\text{ }E=-2\\end{array}[\/latex]<\/div>\nNow we have solved for all of the unknowns on the right side of the equal sign. We have [latex]A=1[\/latex], [latex]B=0[\/latex], [latex]C=1[\/latex], [latex]D=-1[\/latex], and [latex]E=-2[\/latex]. We can write the decomposition as follows:\n
    [latex]\\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\\left({x}^{2}+1\\right)}^{2}}=\\frac{1}{x}+\\frac{1}{\\left({x}^{2}+1\\right)}-\\frac{x+2}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n
    \n

    Try It 4<\/h3>\nFind the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.\n
    [latex]\\frac{{x}^{3}-4{x}^{2}+9x - 5}{{\\left({x}^{2}-2x+3\\right)}^{2}}[\/latex]<\/div>\n
    Solution<\/a><\/div>\n<\/div>","rendered":"

    Now that we can decompose a simplified rational expression<\/strong> with an irreducible quadratic<\/strong> factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.\n<\/p>\n

    \n

    A General Note: Decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] When Q(x)<\/em> Has a Repeated Irreducible Quadratic Factor<\/h3>\n

    The partial fraction decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex], when [latex]Q\\left(x\\right)[\/latex] has a repeated irreducible quadratic factor 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

    [latex]\\frac{P\\left(x\\right)}{{\\left(a{x}^{2}+bx+c\\right)}^{n}}=\\frac{{A}_{1}x+{B}_{1}}{\\left(a{x}^{2}+bx+c\\right)}+\\frac{{A}_{2}x+{B}_{2}}{{\\left(a{x}^{2}+bx+c\\right)}^{2}}+\\frac{{A}_{3}x+{B}_{3}}{{\\left(a{x}^{2}+bx+c\\right)}^{3}}+\\cdot \\cdot \\cdot +\\frac{{A}_{n}x+{B}_{n}}{{\\left(a{x}^{2}+bx+c\\right)}^{n}}[\/latex]<\/div>\n

    Write the denominators in increasing powers.<\/p>\n<\/div>\n

    \n

    How To: Given a rational expression that has a repeated irreducible factor, decompose it.<\/h3>\n
      \n
    1. Use variables like [latex]A,B[\/latex], or [latex]C[\/latex] for the constant numerators over linear factors, and linear expressions such as [latex]{A}_{1}x+{B}_{1},{A}_{2}x+{B}_{2}[\/latex], etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as\n
      [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{A}{ax+b}+\\frac{{A}_{1}x+{B}_{1}}{\\left(a{x}^{2}+bx+c\\right)}+\\frac{{A}_{2}x+{B}_{2}}{{\\left(a{x}^{2}+bx+c\\right)}^{2}}+\\cdots +\\text{ }\\frac{{A}_{n}+{B}_{n}}{{\\left(a{x}^{2}+bx+c\\right)}^{n}}[\/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<\/div>\n
      \n

      Example 4: Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator<\/h3>\n

      Decompose the given expression that has a repeated irreducible factor in the denominator.<\/p>\n

      [latex]\\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\\left({x}^{2}+1\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n
      \n

      Solution<\/h3>\n

      The factors of the denominator are [latex]x,\\left({x}^{2}+1\\right)[\/latex], and [latex]{\\left({x}^{2}+1\\right)}^{2}[\/latex]. Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form [latex]Ax+B[\/latex]. So, let\u2019s begin the decomposition.<\/p>\n

      [latex]\\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\\left({x}^{2}+1\\right)}^{2}}=\\frac{A}{x}+\\frac{Bx+C}{\\left({x}^{2}+1\\right)}+\\frac{Dx+E}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex]<\/div>\n

      We eliminate the denominators by multiplying each term by [latex]x{\\left({x}^{2}+1\\right)}^{2}[\/latex]. Thus,<\/p>\n

      [latex]{x}^{4}+{x}^{3}+{x}^{2}-x+1=A{\\left({x}^{2}+1\\right)}^{2}+\\left(Bx+C\\right)\\left(x\\right)\\left({x}^{2}+1\\right)+\\left(Dx+E\\right)\\left(x\\right)[\/latex]<\/div>\n

      Expand the right side.<\/p>\n

      [latex]\\begin{array}{l} \\text{ }{x}^{4}+{x}^{3}+{x}^{2}-x+1=A\\left({x}^{4}+2{x}^{2}+1\\right)+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\\hfill \\\\ \\text{ }=A{x}^{4}+2A{x}^{2}+A+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\\hfill \\end{array}[\/latex]<\/div>\n

      Now we will collect like terms.<\/p>\n

      [latex]{x}^{4}+{x}^{3}+{x}^{2}-x+1=\\left(A+B\\right){x}^{4}+\\left(C\\right){x}^{3}+\\left(2A+B+D\\right){x}^{2}+\\left(C+E\\right)x+A[\/latex]<\/div>\n

      Set up the system of equations matching corresponding coefficients on each side of the equal sign.<\/p>\n

      [latex]\\begin{array}{l}\\text{ }A+B=1\\hfill \\\\ \\text{ }C=1\\hfill \\\\ 2A+B+D=1\\hfill \\\\ \\text{ }C+E=-1\\hfill \\\\ \\text{ }A=1\\hfill \\end{array}[\/latex]<\/div>\n

      We can use substitution from this point. Substitute [latex]A=1[\/latex] into the first equation.<\/p>\n

      [latex]\\begin{array}{l}1+B=1\\hfill \\\\ \\text{ }B=0\\hfill \\end{array}[\/latex]<\/div>\n

      Substitute [latex]A=1[\/latex] and [latex]B=0[\/latex] into the third equation.<\/p>\n

      [latex]\\begin{array}{l}2\\left(1\\right)+0+D=1\\hfill \\\\ \\text{ }D=-1\\hfill \\end{array}[\/latex]<\/div>\n

      Substitute [latex]C=1[\/latex] into the fourth equation.<\/p>\n

      [latex]\\begin{array}{r}\\hfill 1+E=-1\\\\ \\hfill \\text{ }E=-2\\end{array}[\/latex]<\/div>\n

      Now we have solved for all of the unknowns on the right side of the equal sign. We have [latex]A=1[\/latex], [latex]B=0[\/latex], [latex]C=1[\/latex], [latex]D=-1[\/latex], and [latex]E=-2[\/latex]. We can write the decomposition as follows:<\/p>\n

      [latex]\\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\\left({x}^{2}+1\\right)}^{2}}=\\frac{1}{x}+\\frac{1}{\\left({x}^{2}+1\\right)}-\\frac{x+2}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n
      \n

      Try It 4<\/h3>\n

      Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.<\/p>\n

      [latex]\\frac{{x}^{3}-4{x}^{2}+9x - 5}{{\\left({x}^{2}-2x+3\\right)}^{2}}[\/latex]<\/div>\n
      Solution<\/a><\/div>\n<\/div>\n\n\t\t\t
      \n\t\t\t

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