### 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.
So, the equation from Step 2 can be rewritten as:

$\begin{gathered}\frac{A}{x}+\frac{Bx+C}{x^2+9}\\ \frac{A(x^2+9)}{x(x^2+9)}+\frac{(Bx+C)x}{(x^2+9)x}\\ \frac{A(x^2+9)+(Bx+C)x}{x^3+9x}\\ \frac{Ax^2+9A+Bx^2+Cx}{x^3+9x}\\ \frac{(A+B)x^2+Cx+9A}{x^3+9x}\end{gathered}$

$\frac{-2x^2+4x-18}{x^3+9x} = \frac{(A+B)x^2+Cx+9A}{x^3+9x}$

Step 4

Multiply both sides of the equation by the denominator, $x^3+9x$, to set the numerators equal to each other.
The like terms on either side of the equation must be equal. Use this fact to solve for $A$, $B$, and $C$.

Set the linear terms equal to each other to solve for $C$.

Set the constant terms equal to each other to solve for $A$.

Set the quadratic terms equal to each other to solve for $B$.

Notice that both sides of the equation are in the same form. So, set the coefficients equal to each other:
Substitute –2 for $A$:

$\begin{aligned}\cancel{x^3+9x}\frac{-2x^2+4x-18}{\cancel{x^3+9x}} &= \cancel{x^3+9x}\frac{(A+B)x^2+Cx+9A}{\cancel{x^3+9x}}\\-2x^2+4x-18&=(A+B)x^2+Cx+9A\end{aligned}$

$\overbrace{-2x^2}^{\text{Quadratic}}+\overbrace{4x}^{\text{Linear}}-\overbrace{18}^{\text{Constant}}=\overbrace{(A+B)x^2}^{\text{Quadratic}}+\overbrace{Cx}^{\text{Linear}}+\overbrace{9A}^{\text{Constant}}$

$\begin{aligned}Cx&=4x\\C&=4\end{aligned}$

$\begin{aligned}9A &= -18\\A &=-2\end{aligned}$

$(A + B)x^2=-2x^2$

$A+B=-2$

$\begin{aligned}-2 + B &= -2\\B &=0\end{aligned}$

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.
Add the sum using the original denominator.

$\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$

$\begin{aligned}\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}&=\frac{(Ax+B)(x^2+1)+Cx+D}{x^4+2x^2+1}\\&=\frac{Ax^3+Bx^2+(A+C)x+(B+D)}{x^4+2x^2+1}\end{aligned}$

Step 3

Solve for the unknown numerators by setting the numerators equal to each other.
The like terms on either side of the equation must be equal. Use this fact to solve for $A$, $B$, $C$, and $D$.

Set the cubic terms equal to each other to solve for $A$.

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 $B$.
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 $C$.
Substitute 1 for $A$ and solve for $C$.

Set the constant terms equal to each other to solve for $D$.
Substitute 3 for $B$ and solve for $D$.

$Ax^3+Bx^2+(A+C)x+(B+D)=x^3+3x^2+3x+2$

$\overbrace{Ax^3}^{\text{Cubic}}+\overbrace{Bx^2}^{\text{Quadratic}}+\overbrace{(A+C)x}^{\text{Linear}}+\overbrace{(B+D)}^{\text{Constant}}=\overbrace{x^3}^{\text{Cubic}}+\overbrace{3x^2}^{\text{Quadratic}}+\overbrace{3x}^{\text{Linear}}+\overbrace{2}^{\text{Constant}}$

$Ax^3=x^3$

$Bx^2=3x^2$

$A + C = 3$

$\begin{aligned} A+C&=3\\1+C&=3\\C&=2\end{aligned}$

$B + D = 2$

$\begin{aligned}B + D &= 2\\3+D&=2\\D&=-1\end{aligned}$

Solution

Substitute values for $A$, $B$, $C$, and $D$ into the partial fraction sum.

$\begin{aligned}\frac{x^3+3x^2+3x+2}{x^4+2x^2+1}&=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}\\&=\frac{1x+3}{x^2+1}+\frac{2x+(-1)}{(x^2+1)^2}\\&=\frac{x+3}{x^2+1}+\frac{2x-1}{(x^2+1)^2}\end{aligned}$