Question

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

Answer 1

Imagine the partial fraction decomposition problem:

`(2x-1)/(x^2-x-6)`

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

`(2x-1)/((x-3)(x+2))=A/(x-3)+B/(x-2)`

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

Consider

`(x+3)/(x^2+4x+4)`

Since `x^2+4x+4=(x+2)^2`, you'd be tempted to set up the decomposition as

`(x+3)/((x+2)(x+2))=A/(x+2)+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

`(x+3)/(x+2)^2=A/(x+2)+B/(x+2)^2`

Which can indeed be solved.

An example with a third power:

`(3x-2)/(2x-1)^3=A/(2x-1)+B/(2x-1)^2+C/(2x-1)^3`

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

For example:

`(5x)/((x^2+4)^2(x-3))=(Ax+B)/(x^2+4)+(Cx+D)/(x^2+4)^2+E/(x-3)`