{"id":1213,"date":"2025-07-23T21:00:13","date_gmt":"2025-07-23T21:00:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1213"},"modified":"2026-03-20T19:39:46","modified_gmt":"2026-03-20T19:39:46","slug":"partial-fractions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/partial-fractions-learn-it-3\/","title":{"raw":"Partial Fractions: Learn It 3","rendered":"Partial Fractions: Learn It 3"},"content":{"raw":"

Nonrepeated Irreducible Quadratic Factor<\/h2>\r\nSo 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.\r\n\r\nNow 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,Cx+D[\/latex], etc.\r\n\r\n
\r\n

partial fraction decomposition: nonrepeated irreducible quadratic factor<\/h3>\r\nThe partial fraction decomposition of [latex]\\dfrac{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\r\n\r\n \r\n

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

How To: Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.<\/strong>\r\n
    \r\n \t
  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.\r\n
    [latex]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{A}{ax+b}+\\dfrac{{A}_{1}x+{B}_{1}}{\\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\\right)}+\\dfrac{{A}_{2}x+{B}_{2}}{\\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\\right)}+\\cdot \\cdot \\cdot +\\dfrac{{A}_{n}x+{B}_{n}}{\\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{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>
    Find a partial fraction decomposition of the given expression.\r\n

    [latex]\\dfrac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}[\/latex]<\/p>\r\n[reveal-answer q=\"785023\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"785023\"]\r\n\r\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,\r\n

    [latex]\\dfrac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}=\\dfrac{A}{\\left(x+3\\right)}+\\dfrac{Bx+C}{\\left({x}^{2}+x+2\\right)}[\/latex]<\/p>\r\nWe follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.\r\n

    [latex]\\begin{align}\\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) \\\\[2mm] 8{x}^{2}+12x - 20&=A\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right) \\end{align}[\/latex]<\/p>\r\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.\r\n

    [latex]\\begin{align}8{x}^{2}+12x - 20&=A\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right) \\\\ 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) \\\\ 16&=8A \\\\ A&=2 \\end{align}[\/latex]<\/p>\r\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.\r\n

    [latex]\\begin{align} &8{x}^{2}+12x - 20=2\\left({x}^{2}+x+2\\right)+\\left(Bx+C\\right)\\left(x+3\\right) \\\\ &8{x}^{2}+12x - 20=2{x}^{2}+2x+4+B{x}^{2}+3B+Cx+3C \\\\ &8{x}^{2}+12x - 20=\\left(2+B\\right){x}^{2}+\\left(2+3B+C\\right)x+\\left(4+3C\\right) \\end{align}[\/latex]<\/p>\r\nSetting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.\r\n

    [latex]\\begin{align}2+B=8 && \\text{(1)} \\\\ 2+3B+C=12 && \\text{(2)} \\\\ 4+3C=-20 && \\text{(3)} \\end{align}[\/latex]<\/p>\r\nSolve for [latex]B[\/latex] using equation (1) and solve for [latex]C[\/latex] using equation (3).\r\n

    [latex]\\begin{align}2+B=8 && \\text{(1)} \\\\ B=6 \\\\ \\\\ 4+3C=-20 && \\text{(3)} \\\\ 3C=-24 \\\\ C=-8 \\end{align}[\/latex]<\/p>\r\nThus, the partial fraction decomposition of the expression is\r\n

    [latex]\\dfrac{8{x}^{2}+12x - 20}{\\left(x+3\\right)\\left({x}^{2}+x+2\\right)}=\\dfrac{2}{\\left(x+3\\right)}+\\dfrac{6x - 8}{\\left({x}^{2}+x+2\\right)}[\/latex]<\/p>\r\n \r\n\r\nCould we have just set up a system of equations to solve for [latex]A, B, \\text{ and } C[\/latex]?<\/strong>\r\n\r\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>\r\n

    [latex]\\begin{align} 8{x}^{2}+12x - 20&=A{x}^{2}+Ax+2A+B{x}^{2}+3B+Cx+3C \\\\ 8{x}^{2}+12x - 20&=\\left(A+B\\right){x}^{2}+\\left(A+3B+C\\right)x+\\left(2A+3C\\right) \\end{align}[\/latex]<\/p>\r\nSo the system of equations would be:<\/em>\r\n

    [latex]\\begin{align}A+B=8 \\\\ A+3B+C=12 \\\\ 2A+3C=-20 \\end{align}[\/latex]<\/p>\r\nIf you solve this system of equation, you will also find that [latex]A = 2, B = 6, \\text{ and } C = -8[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

    Sometimes, a combination of methods is helpful to obtain the decomposition. In the example above, you could have set the system up first, then used [latex]x=-3[\/latex] to obtain the value for [latex]A[\/latex], and it would be quick work to obtain [latex]B[\/latex] and [latex]C[\/latex] from the system.Remember that creativity is a key component of doing mathematics and that there is often more than one good way to reach a conclusion.<\/section>
    [ohm_question hide_question_numbers=1]321640[\/ohm_question]<\/section>\r\n
    [ohm_question hide_question_numbers=1]321641[\/ohm_question]<\/section>
    [ohm_question hide_question_numbers=1]321642[\/ohm_question]<\/section><\/div>","rendered":"

    Nonrepeated Irreducible Quadratic Factor<\/h2>\n

    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.<\/p>\n

    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,Cx+D[\/latex], etc.<\/p>\n

    \n

    partial fraction decomposition: nonrepeated irreducible quadratic factor<\/h3>\n

    The partial fraction decomposition of [latex]\\dfrac{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

     <\/p>\n

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

     <\/p>\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<\/section>\n

    How To: Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.<\/strong><\/p>\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]\\dfrac{P\\left(x\\right)}{Q\\left(x\\right)}=\\dfrac{A}{ax+b}+\\dfrac{{A}_{1}x+{B}_{1}}{\\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\\right)}+\\dfrac{{A}_{2}x+{B}_{2}}{\\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\\right)}+\\cdot \\cdot \\cdot +\\dfrac{{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<\/section>\n
      Find a partial fraction decomposition of the given expression.<\/p>\n

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