Assume we are working with complex numbers . To reduce a polynomial fraction to partial fractions is to rearrange it so it is the sum of terms of the form a/(x-b)n Suppose we have two polynomials P(x) and Q(x), and we consider the fraction F(x)=P(x)/Q(x). The Fundamental theorem of algebra tells us we can factorise Q(x) completly, so
p0,... pk are the poles of the fraction : they are where Q(x) is zero and thus where the fraction blows up. The aim of the game is to find numbers ai,j and a polynomial R(x) of degree less than that of Q such that :
F(x)=a0,1/(x-p0) + ... a0,n0/(x-p0)n0+...+ak,1/(x-pk) + ... ak,nk/(x-pk)nk + R(x)
R(x) is just the remainder from the division of P(x) by Q(x). Obtaining the ai,ni is fairly easy. Just multiply the whole fraction by (x-pi)ni. All the terms in the expression will now have a power of (x-pi) in front of them, except the term the term with ai,ni. Evaluating (x-pi)niF(x) at pi allows you to get ai,ni.

The trouble starts when you want to get the other coefficients. You can't just multiply by the appropriate polynomial and evaluate at the appropriate pole because some (x-p) terms will still be left in the denominator. One solution is to subtract ai,ni/(x-pi)ni from F(x). You can then multiply the resulting fraction G(x) by (x-pi)ni-1 and evaluate it at pi to obtain ai,ni-1. Repeat for all the multiplicities of all the poles and you will have F expressed in terms of partial fractions.

A simple example is F(x)=1/(x2-1). Its poles are 1 and -1.

Evaluating these 2 at 1 and -1 respectively gives 1/2 and -1/2.
Thus F(x) = 1/(2(x-1)) - 1/(2(x+1))

Although finding the coefficients is not always easy for large fractions, it makes many operations easier. In particular, integration is much easier, since we know the antiderivative of a/(x-b)^n.