# How do you find the partial fraction decomposition when you have repeated quadratic or linear factors?

Jan 25, 2016

Imagine the partial fraction decomposition problem:

$\frac{2 x - 1}{{x}^{2} - x - 6}$

Here, the denominator would simplify into $\left(x - 3\right) \left(x + 2\right)$ so the decomposition would be set up as

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

However, when the denominator has a repeated factor, something slightly different happens.

Consider

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

Since ${x}^{2} + 4 x + 4 = {\left(x + 2\right)}^{2}$, you'd be tempted to set up the decomposition as

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

However, this will not work. This is remedied by "counting up" to whatever the exponent is on the repeated factor. This would look like

$\frac{x + 3}{x + 2} ^ 2 = \frac{A}{x + 2} + \frac{B}{x + 2} ^ 2$

Which can indeed be solved.

An example with a third power:

$\frac{3 x - 2}{2 x - 1} ^ 3 = \frac{A}{2 x - 1} + \frac{B}{2 x - 1} ^ 2 + \frac{C}{2 x - 1} ^ 3$

This is very similar for irreducible quadratic factors, except for that they take the form $A x + B$ instead of just $A$. The same rules from earlier apply for "counting up" to the exponent.

For example:

$\frac{5 x}{{\left({x}^{2} + 4\right)}^{2} \left(x - 3\right)} = \frac{A x + B}{{x}^{2} + 4} + \frac{C x + D}{{x}^{2} + 4} ^ 2 + \frac{E}{x - 3}$