{"id":1779,"date":"2015-11-12T18:30:45","date_gmt":"2015-11-12T18:30:45","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1779"},"modified":"2015-11-12T18:30:45","modified_gmt":"2015-11-12T18:30:45","slug":"decomposing-px-qx-where-qx-has-a-nonrepeated-irreducible-quadratic-factor","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/decomposing-px-qx-where-qx-has-a-nonrepeated-irreducible-quadratic-factor\/","title":{"raw":"Decomposing P(x) \/ Q(x), Where Q(x) Has a Nonrepeated Irreducible Quadratic Factor","rendered":"Decomposing P(x) \/ Q(x), Where Q(x) Has a Nonrepeated Irreducible Quadratic Factor"},"content":{"raw":"

So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators [latex]A,B[\/latex], or [latex]C[\/latex] representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic<\/span><\/strong> expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as [latex]Ax+B,Bx+C[\/latex], etc.\n<\/p>

\n

A General Note: Decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}:Q\\left(x\\right)[\/latex] Has a Nonrepeated Irreducible Quadratic Factor<\/h3>\nThe partial fraction decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] such that [latex]Q\\left(x\\right)[\/latex] has a nonrepeated 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 written as\n
[latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{A}_{1}x+{B}_{1}}{\\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\\right)}+\\frac{{A}_{2}x+{B}_{2}}{\\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\\right)}+\\cdot \\cdot \\cdot +\\frac{{A}_{n}x+{B}_{n}}{\\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{n}\\right)}[\/latex]<\/div>\nThe decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: [latex]A,B,C[\/latex], and so on.\n\n<\/div>\n
\n

How To: Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.<\/h3>\n
  1. Use variables such as [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.\n
    [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{A}{ax+b}+\\frac{{A}_{1}x+{B}_{1}}{\\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\\right)}+\\frac{{A}_{2}x+{B}_{2}}{\\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\\right)}+\\cdot \\cdot \\cdot +\\frac{{A}_{n}x+{B}_{n}}{\\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{n}\\right)}[\/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 3: Decomposing [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] When Q(x)<\/em> Contains a Nonrepeated Irreducible Quadratic Factor<\/h3>\nFind a partial fraction decomposition of the given expression.\n
    [latex]\\frac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}[\/latex]<\/div>\n<\/div>\n
    \n

    Solution<\/h3>\nWe have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,\n
    [latex]\\frac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}=\\frac{A}{\\left(x+3\\right)}+\\frac{Bx+C}{\\left({x}^{2}+x+2\\right)}[\/latex]<\/div>\nWe follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.\n
    [latex]\\begin{array}{l}\\left(x+3\\right)\\left({x}^{2}+x+2\\right)\\left[\\frac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}\\right]=\\left[\\frac{A}{\\left(x+3\\right)}+\\frac{Bx+C}{\\left({x}^{2}+x+2\\right)}\\right]\\left(x+3\\right)\\left({x}^{2}+x+2\\right)\\hfill \\\\ \\text{ }8{x}^{2}+12x - 20=A\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right)\\hfill \\end{array}[\/latex]<\/div>\nNotice we could easily solve for [latex]A[\/latex] by choosing a value for [latex]x[\/latex] that will make the [latex]Bx+C[\/latex] term equal 0. Let [latex]x=-3[\/latex] and substitute it into the equation.\n
    [latex]\\begin{array}{l}\\text{ }8{x}^{2}+12x - 20=A\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right)\\hfill \\\\ \\text{ }8{\\left(-3\\right)}^{2}+12\\left(-3\\right)-20=A\\left({\\left(-3\\right)}^{2}+\\left(-3\\right)+2\\right)+\\left(B\\left(-3\\right)+C\\right)\\left(\\left(-3\\right)+3\\right)\\hfill \\\\ \\text{ }16=8A\\hfill \\\\ \\text{ }A=2\\hfill \\end{array}[\/latex]<\/div>\nNow that we know the value of [latex]A[\/latex], substitute it back into the equation. Then expand the right side and collect like terms.\n
    [latex]\\begin{array}{l}\\hfill \\\\ 8{x}^{2}+12x - 20=2\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right)\\hfill \\\\ 8{x}^{2}+12x - 20=2{x}^{2}+2x+4+B{x}^{2}+3B+Cx+3C\\hfill \\\\ 8{x}^{2}+12x - 20=\\left(2+B\\right){x}^{2}+\\left(2+3B+C\\right)x+\\left(4+3C\\right)\\hfill \\end{array}[\/latex]<\/div>\nSetting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.\n
    [latex]\\begin{array}{ll}\\text{ }2+B=8\\hfill & \\text{(1)}\\hfill \\\\ 2+3B+C=12\\hfill & \\text{(2)}\\hfill \\\\ \\text{ }4+3C=-20\\hfill & \\text{(3)}\\hfill \\end{array}[\/latex]<\/div>\nSolve for [latex]B[\/latex] using equation (1) and solve for [latex]C[\/latex] using equation (3).\n
    [latex]\\begin{array}{ll}\\text{ }2+B=8\\hfill & \\text{(1)}\\hfill \\\\ \\text{ }B=6\\hfill & \\hfill \\\\ \\hfill & \\hfill \\\\ 4+3C=-20\\hfill & \\text{(3)}\\hfill \\\\ \\text{ }3C=-24\\hfill & \\hfill \\\\ \\text{ }C=-8\\hfill & \\hfill \\end{array}[\/latex]<\/div>\nThus, the partial fraction decomposition of the expression is\n
    [latex]\\frac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}=\\frac{2}{\\left(x+3\\right)}+\\frac{6x - 8}{\\left({x}^{2}+x+2\\right)}[\/latex]<\/div>\n<\/div>\n
    \n

    Q & A<\/h3>\n

    Could we have just set up a system of equations to solve Example 3?<\/strong><\/em><\/h3>\nYes, we could have solved it by setting up a system of equations without solving for [latex]A[\/latex] first. The expansion on the right would be:<\/em>\n
    [latex]\\begin{array}{l}\\begin{array}{l}\\\\ 8{x}^{2}+12x - 20=A{x}^{2}+Ax+2A+B{x}^{2}+3B+Cx+3C\\end{array}\\hfill \\\\ 8{x}^{2}+12x - 20=\\left(A+B\\right){x}^{2}+\\left(A+3B+C\\right)x+\\left(2A+3C\\right)\\hfill \\end{array}[\/latex]<\/div>\nSo the system of equations would be:<\/em>\n
    [latex]\\begin{array}{l}\\text{ }A+B=8\\hfill \\\\ A+3B+C=12\\hfill \\\\ \\text{ }2A+3C=-20\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
    \n

    Try It 3<\/h3>\nFind the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.\n
    [latex]\\frac{5{x}^{2}-6x+7}{\\left(x - 1\\right)\\left({x}^{2}+1\\right)}[\/latex]<\/div>\n
    Solution<\/a><\/div>\n<\/div>","rendered":"

    So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators [latex]A,B[\/latex], or [latex]C[\/latex] representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic<\/span><\/strong> expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as [latex]Ax+B,Bx+C[\/latex], etc.\n<\/p>\n

    \n

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

    The partial fraction decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] such that [latex]Q\\left(x\\right)[\/latex] has a nonrepeated 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 written as<\/p>\n

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

    The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: [latex]A,B,C[\/latex], and so on.<\/p>\n<\/div>\n

    \n

    How To: Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.<\/h3>\n
      \n
    1. Use variables such as [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.\n
      [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{A}{ax+b}+\\frac{{A}_{1}x+{B}_{1}}{\\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\\right)}+\\frac{{A}_{2}x+{B}_{2}}{\\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\\right)}+\\cdot \\cdot \\cdot +\\frac{{A}_{n}x+{B}_{n}}{\\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{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<\/div>\n
      \n

      Example 3: Decomposing [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] When Q(x)<\/em> Contains a Nonrepeated Irreducible Quadratic Factor<\/h3>\n

      Find a partial fraction decomposition of the given expression.<\/p>\n

      [latex]\\frac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}[\/latex]<\/div>\n<\/div>\n
      \n

      Solution<\/h3>\n

      We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,<\/p>\n

      [latex]\\frac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}=\\frac{A}{\\left(x+3\\right)}+\\frac{Bx+C}{\\left({x}^{2}+x+2\\right)}[\/latex]<\/div>\n

      We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.<\/p>\n

      [latex]\\begin{array}{l}\\left(x+3\\right)\\left({x}^{2}+x+2\\right)\\left[\\frac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}\\right]=\\left[\\frac{A}{\\left(x+3\\right)}+\\frac{Bx+C}{\\left({x}^{2}+x+2\\right)}\\right]\\left(x+3\\right)\\left({x}^{2}+x+2\\right)\\hfill \\\\ \\text{ }8{x}^{2}+12x - 20=A\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right)\\hfill \\end{array}[\/latex]<\/div>\n

      Notice we could easily solve for [latex]A[\/latex] by choosing a value for [latex]x[\/latex] that will make the [latex]Bx+C[\/latex] term equal 0. Let [latex]x=-3[\/latex] and substitute it into the equation.<\/p>\n

      [latex]\\begin{array}{l}\\text{ }8{x}^{2}+12x - 20=A\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right)\\hfill \\\\ \\text{ }8{\\left(-3\\right)}^{2}+12\\left(-3\\right)-20=A\\left({\\left(-3\\right)}^{2}+\\left(-3\\right)+2\\right)+\\left(B\\left(-3\\right)+C\\right)\\left(\\left(-3\\right)+3\\right)\\hfill \\\\ \\text{ }16=8A\\hfill \\\\ \\text{ }A=2\\hfill \\end{array}[\/latex]<\/div>\n

      Now that we know the value of [latex]A[\/latex], substitute it back into the equation. Then expand the right side and collect like terms.<\/p>\n

      [latex]\\begin{array}{l}\\hfill \\\\ 8{x}^{2}+12x - 20=2\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right)\\hfill \\\\ 8{x}^{2}+12x - 20=2{x}^{2}+2x+4+B{x}^{2}+3B+Cx+3C\\hfill \\\\ 8{x}^{2}+12x - 20=\\left(2+B\\right){x}^{2}+\\left(2+3B+C\\right)x+\\left(4+3C\\right)\\hfill \\end{array}[\/latex]<\/div>\n

      Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.<\/p>\n

      [latex]\\begin{array}{ll}\\text{ }2+B=8\\hfill & \\text{(1)}\\hfill \\\\ 2+3B+C=12\\hfill & \\text{(2)}\\hfill \\\\ \\text{ }4+3C=-20\\hfill & \\text{(3)}\\hfill \\end{array}[\/latex]<\/div>\n

      Solve for [latex]B[\/latex] using equation (1) and solve for [latex]C[\/latex] using equation (3).<\/p>\n

      [latex]\\begin{array}{ll}\\text{ }2+B=8\\hfill & \\text{(1)}\\hfill \\\\ \\text{ }B=6\\hfill & \\hfill \\\\ \\hfill & \\hfill \\\\ 4+3C=-20\\hfill & \\text{(3)}\\hfill \\\\ \\text{ }3C=-24\\hfill & \\hfill \\\\ \\text{ }C=-8\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n

      Thus, the partial fraction decomposition of the expression is<\/p>\n

      [latex]\\frac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}=\\frac{2}{\\left(x+3\\right)}+\\frac{6x - 8}{\\left({x}^{2}+x+2\\right)}[\/latex]<\/div>\n<\/div>\n
      \n

      Q & A<\/h3>\n

      Could we have just set up a system of equations to solve Example 3?<\/strong><\/em><\/h3>\n

      Yes, we could have solved it by setting up a system of equations without solving for [latex]A[\/latex] first. The expansion on the right would be:<\/em><\/p>\n

      [latex]\\begin{array}{l}\\begin{array}{l}\\\\ 8{x}^{2}+12x - 20=A{x}^{2}+Ax+2A+B{x}^{2}+3B+Cx+3C\\end{array}\\hfill \\\\ 8{x}^{2}+12x - 20=\\left(A+B\\right){x}^{2}+\\left(A+3B+C\\right)x+\\left(2A+3C\\right)\\hfill \\end{array}[\/latex]<\/div>\n

      So the system of equations would be:<\/em><\/p>\n

      [latex]\\begin{array}{l}\\text{ }A+B=8\\hfill \\\\ A+3B+C=12\\hfill \\\\ \\text{ }2A+3C=-20\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Try It 3<\/h3>\n

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

      [latex]\\frac{5{x}^{2}-6x+7}{\\left(x - 1\\right)\\left({x}^{2}+1\\right)}[\/latex]<\/div>\n
      Solution<\/a><\/div>\n<\/div>\n\n\t\t\t
      \n\t\t\t

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