# Partial Fractions - Repeated Factors

**Partial fraction decomposition** is a technique used to write a rational function as the sum of simpler rational expressions. A partial fraction has repeated factors when one of the denominator factors has multiplicity greater than 1:

$\frac{1}{x^3-x^2} \implies \frac{1}{x^2(x-1)} \implies \frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}.$

The process for repeated factors is slightly different than the process for linear, non-repeated factors.

## Repeated Factor Partial Decomposition Forms

The partial fraction decomposition form is slightly different when there are repeated factors.

Partial Fraction Decomposition Form for Repeated Factors:

A factor is repeated if it has multiplicity greater than 1.

For each non-repeated factor in the denominator, follow the process for linear factors.

If $k$ is the multiplicity of the repeated factor, write $k$ rational expressions, each of which has that factor raised to a different power in the denominator.

If the repeated factor is linear, then each of these rational expressions will have a constant numerator coefficient.

Once the form is down, you can follow the same process as with linear factors to solve for the coefficients.

Find the partial fraction decomposition of the following rational expression:

$\frac{x^2+x+1}{x^3+3x^2+3x+1}.$

The denominator can be factored as a perfect cube:

$x^3+3x^2+3x+1=(x+1)^3.$

The multiplicity of this factor is 3. Therefore, 3 rational expressions are needed in the partial fraction decomposition, each of which has $(x+1)$ raised to a different positive integer power up to 3. Since $(x+1)$ is a linear factor, each of these rational expressions will have a constant numerator coefficient:

$\frac{x^2+x+1}{x^3+3x^2+3x+1}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}.$

To solve for the coefficients, combine the fractions on the right hand side of the equation:

$\frac{x^2+x+1}{x^3+3x^2+3x+1}=\frac{A(x+1)^2+B(x+1)+C}{(x+1)^3}.$

Since the denominators are equal, the numerators must also be equal:

$\begin{aligned} x^2+x+1 &= A(x+1)^2+B(x+1)+C. \end{aligned}$

Setting $x=-1$ in this equation gives $C=1.$ Group the remaining terms by degree:

$\begin{aligned} x^2+x+1 &= A(x+1)^2+B(x+1)+1 \\ x^2+x+1 &=Ax^2+2Ax+A+Bx+B+1 \\ x^2+x+1 &= Ax^2+(2A+B)x+ (A+B+1). \end{aligned}$

This gives the following system of equations:

$\begin{aligned} A &= 1 \\ 2A+B &= 1. \end{aligned}$

Solving this system of equations gives $A=1$ and $B=-1.$ Then the partial fraction decomposition is

$\frac{x^2+x+1}{x^3+3x^2+3x+1}=\frac{1}{x+1}-\frac{1}{(x+1)^2}+\frac{1}{(x+1)^3}.\ _\square$

Find the partial fraction decomposition of the following rational expression:

$\frac{16}{x^4-8x^2+16}.$

The denominator can be factored as a perfect square:

$x^4-8x^2+16=(x^2-4)^2,$

and then factored as difference of squares:

$(x^2-4)^2=(x-2)^2(x+2)^2.$

Each of these factors has multiplicity 2. Therefore, there should be 4 rational expressions in the partial fraction decomposition. The factors are linear, so each rational expression has a constant numerator coefficient:

$\frac{16}{x^4-8x^2+16}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x+2}+\frac{D}{(x+2)^2}.$

Solve for the coefficients by combining the fractions on the right hand side of the equation:

$\frac{16}{x^4-8x^2+16}=\frac{A(x-2)(x+2)^2+B(x+2)^2+C(x-2)^2(x+2)+D(x-2)^2}{(x-2)^2(x+2)^2}.$

Since the denominators are equal, the numerators are also equal:

$16=A(x-2)(x+2)^2+B(x+2)^2+C(x-2)^2(x+2)+D(x-2)^2.$

Setting $x=2$ in this equation gives $B=1.$ Setting $x=-2$ in this equation gives $D=1.$ The other coefficients must be solved for by grouping the terms by degree:

$\begin{aligned} 16 &= A(x-2)(x+2)^2+(x+2)^2+C(x-2)^2(x+2)+(x-2)^2 \\ 16 &= A(x^3+2x^2-4x-8)+(x^2+4x+4)+C(x^3-2x^2-4x+8)+(x^2-4x+4) \\ 16 &= (A+C)x^3+(2A-2C+2)x^2+(-4A-4C)x+(-8A+8C+8). \end{aligned}$

This gives the system of equations

$\begin{aligned} A+C &= 0 \\ 2A-2C+2 &= 0. \end{aligned}$

Solving this system of equations gives $A=-\frac{1}{2}$ and $C=\frac{1}{2}.$ Then the partial fraction decomposition is

$\frac{16}{x^4-8x^2+16}=\frac{-1}{2(x-2)}+\frac{1}{(x-2)^2}+\frac{1}{2(x+2)}+\frac{1}{(x+2)^2}.\ _\square$

When the following expression is expressed as a sum of partial fractions, what form will it take? $\frac{x^3-x+1}{x^4-x^3}$

**Cite as:**Partial Fractions - Repeated Factors.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/partial-fractions-repeated-factors/