## Decomposing P(x)/ Q(x), Where Q(x) Has Repeated Linear Factors

Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.

### A General Note: Partial Fraction Decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)$ Has Repeated Linear Factors

The partial fraction decomposition of $\frac{P\left(x\right)}{Q\left(x\right)}$, when $Q\left(x\right)$ has a repeated linear factor occurring $n$ times and the degree of $P\left(x\right)$ is less than the degree of $Q\left(x\right)$, is

$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left(ax+b\right)}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+\frac{{A}_{3}}{{\left(ax+b\right)}^{3}}+\cdot \cdot \cdot +\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}$

Write the denominator powers in increasing order.

### How To: Given a rational expression with repeated linear factors, decompose it.

1. Use a variable like $A,B$, or $C$ for the numerators and account for increasing powers of the denominators.
$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left(ax+b\right)}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+ \text{. }\text{. }\text{. + }\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
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.

### Example 2: Decomposing with Repeated Linear Factors

Decompose the given rational expression with repeated linear factors.

$\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}$

### Solution

The denominator factors are $x{\left(x - 2\right)}^{2}$. To allow for the repeated factor of $\left(x - 2\right)$, the decomposition will include three denominators: $x,\left(x - 2\right)$, and ${\left(x - 2\right)}^{2}$. Thus,

$\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\frac{A}{x}+\frac{B}{\left(x - 2\right)}+\frac{C}{{\left(x - 2\right)}^{2}}$

Next, we multiply both sides by the common denominator.

$\begin{array}{l}x{\left(x - 2\right)}^{2}\left[\frac{-{x}^{2}+2x+4}{x{\left(x - 2\right)}^{2}}\right]=\left[\frac{A}{x}+\frac{B}{\left(x - 2\right)}+\frac{C}{{\left(x - 2\right)}^{2}}\right]x{\left(x - 2\right)}^{2}\hfill \\ \text{ }-{x}^{2}+2x+4=A{\left(x - 2\right)}^{2}+Bx\left(x - 2\right)+Cx\hfill \end{array}$

On the right side of the equation, we expand and collect like terms.

$\begin{array}{l}-{x}^{2}+2x+4=A\left({x}^{2}-4x+4\right)+B\left({x}^{2}-2x\right)+Cx\hfill \\ \text{ }=A{x}^{2}-4Ax+4A+B{x}^{2}-2Bx+Cx\hfill \\ \text{ }=\left(A+B\right){x}^{2}+\left(-4A - 2B+C\right)x+4A\hfill \end{array}$

Next, we compare the coefficients of both sides. This will give the system of equations in three variables:

$-{x}^{2}+2x+4=\left(A+B\right){x}^{2}+\left(-4A - 2B+C\right)x+4A$
$\begin{array}{rr}\hfill A+B=-1& \hfill \text{(1)}\\ \hfill -4A - 2B+C=2& \hfill \text{(2)}\\ \hfill 4A=4& \hfill \text{(3)}\end{array}$

Solving for $A$ , we have

$\begin{array}{l}4A=4\hfill \\ \text{ }A=1\hfill \end{array}$

Substitute $A=1$ into equation (1).

$\begin{array}{l}\text{ }A+B=-1\hfill \\ \left(1\right)+B=-1\hfill \\ \text{ }B=-2\hfill \end{array}$

Then, to solve for $C$, substitute the values for $A$ and $B$ into equation (2).

$\begin{array}{r}\hfill \text{ }-4A - 2B+C=2\\ \hfill -4\left(1\right)-2\left(-2\right)+C=2\\ \hfill \text{ }-4+4+C=2\\ \hfill \text{ }C=2\end{array}$

Thus,

$\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\frac{1}{x}-\frac{2}{\left(x - 2\right)}+\frac{2}{{\left(x - 2\right)}^{2}}$

### Try It 2

Find the partial fraction decomposition of the expression with repeated linear factors.

$\frac{6x - 11}{{\left(x - 1\right)}^{2}}$