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

##### 1 Answer

Imagine the partial fraction decomposition problem:

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

Here, the denominator would simplify into

#(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+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

For example:

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