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

1 Answer
Jan 25, 2016

Imagine the partial fraction decomposition problem:

(2x-1)/(x^2-x-6)2x1x2x6

Here, the denominator would simplify into (x-3)(x+2)(x3)(x+2) so the decomposition would be set up as

(2x-1)/((x-3)(x+2))=A/(x-3)+B/(x-2)2x1(x3)(x+2)=Ax3+Bx2

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

Consider

(x+3)/(x^2+4x+4)x+3x2+4x+4

Since x^2+4x+4=(x+2)^2x2+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)x+3(x+2)(x+2)=Ax+2+Bx+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)^2x+3(x+2)2=Ax+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)^33x2(2x1)3=A2x1+B(2x1)2+C(2x1)3

This is very similar for irreducible quadratic factors, except for that they take the form Ax+BAx+B instead of just AA. 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)5x(x2+4)2(x3)=Ax+Bx2+4+Cx+D(x2+4)2+Ex3