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College Algebra/Partial Fraction Decomposition/Quadratic Factors in the Denominator

Partial Fraction Decomposition

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Quadratic Factors in the Denominator

Nonrepeated Irreducible Factors

A rational function with a nonrepeated, irreducible quadratic factor in the denominator can be decomposed into partial fractions.
If a denominator factors into an irreducible quadratic, that factor cannot be written in a simpler form as a product of factors.
Step-By-Step Example
Decompose a Rational Expression with an Irreducible Quadratic Factor
Write the rational expression as a sum of partial fractions.
−2x2+4x−18x3+9x\frac{-2x^2+4x-18}{x^3+9x}
Step 1
Factor the denominator.
−2x2+4x−18x3+9x=−2x2+4x−18x(x2+9)\frac{-2x^2+4x-18}{x^3+9x} = \frac{-2x^2+4x-18}{x(x^2+9)}
Step 2
Write the fraction as a sum with unknown numerators, using each linear factor and each irreducible quadratic factor exactly once. In the partial fraction with a quadratic denominator, the numerator will be linear. Use Bx+CBx+C as the unknown linear expression.
−2x2+4x−18x3+9x=Ax+Bx+Cx2+9\frac{-2x^2+4x-18}{x^3+9x} = \frac{A}{x} + \frac{Bx+C}{x^2+9}
Step 3
Using the process for partial fraction decomposition, find the unknown numerators. Add the sum using the original denominator.
Ax+Bx+Cx2+9A(x2+9)x(x2+9)+(Bx+C)x(x2+9)xA(x2+9)+(Bx+C)xx3+9xAx2+9A+Bx2+Cxx3+9x(A+B)x2+Cx+9Ax3+9x\begin{gathered}\frac{A}{x}+\frac{Bx+C}{x^2+9}\\ \frac{A(x^2+9)}{x(x^2+9)}+\frac{(Bx+C)x}{(x^2+9)x}\\ \frac{A(x^2+9)+(Bx+C)x}{x^3+9x}\\ \frac{Ax^2+9A+Bx^2+Cx}{x^3+9x}\\ \frac{(A+B)x^2+Cx+9A}{x^3+9x}\end{gathered}
So, the equation from Step 2 can be rewritten as:
−2x2+4x−18x3+9x=(A+B)x2+Cx+9Ax3+9x\frac{-2x^2+4x-18}{x^3+9x} = \frac{(A+B)x^2+Cx+9A}{x^3+9x}
Step 4
Multiply both sides of the equation by the denominator, x3+9xx^3+9x, to set the numerators equal to each other.
x3+9x−2x2+4x−18x3+9x=x3+9x(A+B)x2+Cx+9Ax3+9x−2x2+4x−18=(A+B)x2+Cx+9A\begin{aligned}\cancel{x^3+9x}\frac{-2x^2+4x-18}{\cancel{x^3+9x}} &= \cancel{x^3+9x}\frac{(A+B)x^2+Cx+9A}{\cancel{x^3+9x}}\\-2x^2+4x-18&=(A+B)x^2+Cx+9A\end{aligned}
The like terms on either side of the equation must be equal. Use this fact to solve for AA, BB, and CC.
−2x2⏞Quadratic+4x⏞Linear−18⏞Constant=(A+B)x2⏞Quadratic+Cx⏞Linear+9A⏞Constant\overbrace{-2x^2}^{\text{Quadratic}}+\overbrace{4x}^{\text{Linear}}-\overbrace{18}^{\text{Constant}}=\overbrace{(A+B)x^2}^{\text{Quadratic}}+\overbrace{Cx}^{\text{Linear}}+\overbrace{9A}^{\text{Constant}}
Set the linear terms equal to each other to solve for CC.
Cx=4xC=4\begin{aligned}Cx&=4x\\C&=4\end{aligned}
Set the constant terms equal to each other to solve for AA.
9A=−18A=−2\begin{aligned}9A &= -18\\A &=-2\end{aligned}
Set the quadratic terms equal to each other to solve for BB.
(A+B)x2=−2x2(A + B)x^2=-2x^2
Notice that both sides of the equation are in the same form. So, set the coefficients equal to each other:
A+B=−2A+B=-2
Substitute –2 for AA:
−2+B=−2B=0\begin{aligned}-2 + B &= -2\\B &=0\end{aligned}
Solution
Substitute values for AA, BB, and CC into the partial fraction sum, and simplify.
−2x2+4x−18x3+9x=Ax+Bx+Cx2+9=−2x+0x+4x2+9=4x2+9−2x\begin{aligned}\frac{-2x^2+4x-18}{x^3+9x}&=\frac{A}{x}+\frac{Bx+C}{x^2+9}\\&=\frac{-2}{x}+\frac{0x+4}{x^2+9}\\&=\frac{4}{x^2+9}-\frac{2}{x}\end{aligned}

Repeated Irreducible Factors

A rational expression with a repeated, irreducible quadratic factor in the denominator can be decomposed into partial fractions.
There are also times when factoring the denominator leads to a repeated irreducible quadratic factor. The process is altered to include additional addends with the repeated factors in the denominator.
Step-By-Step Example
Decompose a Rational Expression with a Repeated Irreducible Quadratic Factor
Write the rational expression as a sum of partial fractions.
x3+3x2+3x+2x4+2x2+1\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}
Step 1
Factor the denominator.
x3+3x2+3x+2x4+2x2+1=x3+3x2+3x+2(x2+1)2\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}=\frac{x^3+3x^2+3x+2}{(x^2+1)^2}
Step 2
Write the fraction as a sum with unknown numerators. The irreducible quadratic factor is repeated two times, so the sum will contain two partial fractions with linear numerators. Use Ax+BAx+B and Cx+DCx + D as the unknown numerators.
x3+3x2+3x+2x4+2x2+1=Ax+Bx2+1+Cx+D(x2+1)2\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}
Add the sum using the original denominator.
Ax+Bx2+1+Cx+D(x2+1)2=(Ax+B)(x2+1)+Cx+Dx4+2x2+1=Ax3+Bx2+(A+C)x+(B+D)x4+2x2+1\begin{aligned}\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}&=\frac{(Ax+B)(x^2+1)+Cx+D}{x^4+2x^2+1}\\&=\frac{Ax^3+Bx^2+(A+C)x+(B+D)}{x^4+2x^2+1}\end{aligned}
Step 3
Solve for the unknown numerators by setting the numerators equal to each other.
Ax3+Bx2+(A+C)x+(B+D)=x3+3x2+3x+2Ax^3+Bx^2+(A+C)x+(B+D)=x^3+3x^2+3x+2
The like terms on either side of the equation must be equal. Use this fact to solve for AA, BB, CC, and DD.
Ax3⏞Cubic+Bx2⏞Quadratic+(A+C)x⏞Linear+(B+D)⏞Constant=x3⏞Cubic+3x2⏞Quadratic+3x⏞Linear+2⏞Constant\overbrace{Ax^3}^{\text{Cubic}}+\overbrace{Bx^2}^{\text{Quadratic}}+\overbrace{(A+C)x}^{\text{Linear}}+\overbrace{(B+D)}^{\text{Constant}}=\overbrace{x^3}^{\text{Cubic}}+\overbrace{3x^2}^{\text{Quadratic}}+\overbrace{3x}^{\text{Linear}}+\overbrace{2}^{\text{Constant}}
Set the cubic terms equal to each other to solve for AA.
Ax3=x3Ax^3=x^3
Notice that both sides of the equation have the same form. Set the coefficients equal to each other to get A=1A = 1. Next, set the quadratic terms equal to each other to solve for BB.
Bx2=3x2Bx^2=3x^2
Notice that both sides of the equation have the same form. Set the coefficients equal to each other to get B=3B = 3. Now, set the linear terms equal to each other to solve for CC.
A+C=3A + C = 3
Substitute 1 for AA and solve for CC.
A+C=31+C=3C=2\begin{aligned} A+C&=3\\1+C&=3\\C&=2\end{aligned}
Set the constant terms equal to each other to solve for DD.
B+D=2B + D = 2
Substitute 3 for BB and solve for DD.
B+D=23+D=2D=−1\begin{aligned}B + D &= 2\\3+D&=2\\D&=-1\end{aligned}
Solution
Substitute values for AA, BB, CC, and DD into the partial fraction sum.
x3+3x2+3x+2x4+2x2+1=Ax+Bx2+1+Cx+D(x2+1)2=1x+3x2+1+2x+(−1)(x2+1)2=x+3x2+1+2x−1(x2+1)2\begin{aligned}\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}&=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}\\&=\frac{1x+3}{x^2+1}+\frac{2x+(-1)}{(x^2+1)^2}\\&=\frac{x+3}{x^2+1}+\frac{2x-1}{(x^2+1)^2}\end{aligned}
<Linear Factors in the Denominator