# 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.
$\frac{-2x^2+4x-18}{x^3+9x}$
Step 1
Factor the denominator.
$\frac{-2x^2+4x-18}{x^3+9x} = \frac{-2x^2+4x-18}{x(x^2+9)}$
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 $Bx+C$ as the unknown linear expression.
$\frac{-2x^2+4x-18}{x^3+9x} = \frac{A}{x} + \frac{Bx+C}{x^2+9}$
Step 3
Using the process for partial fraction decomposition, find the unknown numerators. Add the sum using the original denominator.
$\begin{gathered}\frac{A}{x}+\frac{Bx+C}{x^2+9}\\ \frac{A(x^2+9)}{x(x^2+9)}+\frac{(Bx+C)x}{(x^2+9)x}\\ \frac{A(x^2+9)+(Bx+C)x}{x^3+9x}\\ \frac{Ax^2+9A+Bx^2+Cx}{x^3+9x}\\ \frac{(A+B)x^2+Cx+9A}{x^3+9x}\end{gathered}$
So, the equation from Step 2 can be rewritten as:
$\frac{-2x^2+4x-18}{x^3+9x} = \frac{(A+B)x^2+Cx+9A}{x^3+9x}$
Step 4
Multiply both sides of the equation by the denominator, $x^3+9x$, to set the numerators equal to each other.
\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 $A$, $B$, and $C$.
$\overbrace{-2x^2}^{\text{Quadratic}}+\overbrace{4x}^{\text{Linear}}-\overbrace{18}^{\text{Constant}}=\overbrace{(A+B)x^2}^{\text{Quadratic}}+\overbrace{Cx}^{\text{Linear}}+\overbrace{9A}^{\text{Constant}}$
Set the linear terms equal to each other to solve for $C$.
\begin{aligned}Cx&=4x\\C&=4\end{aligned}
Set the constant terms equal to each other to solve for $A$.
\begin{aligned}9A &= -18\\A &=-2\end{aligned}
Set the quadratic terms equal to each other to solve for $B$.
$(A + B)x^2=-2x^2$
Notice that both sides of the equation are in the same form. So, set the coefficients equal to each other:
$A+B=-2$
Substitute –2 for $A$:
\begin{aligned}-2 + B &= -2\\B &=0\end{aligned}
Solution
Substitute values for $A$, $B$, and $C$ into the partial fraction sum, and simplify.
\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.
$\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}$
Step 1
Factor the denominator.
$\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}=\frac{x^3+3x^2+3x+2}{(x^2+1)^2}$
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 $Ax+B$ and $Cx + D$ as the unknown numerators.
$\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$
Add the sum using the original denominator.
\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.
$Ax^3+Bx^2+(A+C)x+(B+D)=x^3+3x^2+3x+2$
The like terms on either side of the equation must be equal. Use this fact to solve for $A$, $B$, $C$, and $D$.
$\overbrace{Ax^3}^{\text{Cubic}}+\overbrace{Bx^2}^{\text{Quadratic}}+\overbrace{(A+C)x}^{\text{Linear}}+\overbrace{(B+D)}^{\text{Constant}}=\overbrace{x^3}^{\text{Cubic}}+\overbrace{3x^2}^{\text{Quadratic}}+\overbrace{3x}^{\text{Linear}}+\overbrace{2}^{\text{Constant}}$
Set the cubic terms equal to each other to solve for $A$.
$Ax^3=x^3$
Notice that both sides of the equation have the same form. Set the coefficients equal to each other to get $A = 1$. Next, set the quadratic terms equal to each other to solve for $B$.
$Bx^2=3x^2$
Notice that both sides of the equation have the same form. Set the coefficients equal to each other to get $B = 3$. Now, set the linear terms equal to each other to solve for $C$.
$A + C = 3$
Substitute 1 for $A$ and solve for $C$.
\begin{aligned} A+C&=3\\1+C&=3\\C&=2\end{aligned}
Set the constant terms equal to each other to solve for $D$.
$B + D = 2$
Substitute 3 for $B$ and solve for $D$.
\begin{aligned}B + D &= 2\\3+D&=2\\D&=-1\end{aligned}
Solution
Substitute values for $A$, $B$, $C$, and $D$ into the partial fraction sum.
\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}