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

Q(x)=(x-p_{0})^{n0}(x-p_{1})^{n1}...(x-p_{k})^{nk}

p

_{0},... p

_{k} 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 a

_{i,j} and a

polynomial R(x) of degree less than that of Q such that :

F(x)=a

_{0,1}/(x-p

_{0}) + ... a

_{0,n0}/(x-p

_{0})

^{n0}+...+a

_{k,1}/(x-p

_{k}) + ... a

_{k,nk}/(x-p

_{k})

^{nk} + R(x)

R(x) is just the

remainder from the

division of P(x) by Q(x). Obtaining the a

_{i,ni} is fairly easy. Just multiply the whole fraction by (x-p

_{i})

^{ni}. All the terms in the

expression will now have a power of (x-p

_{i}) in front of them, except the term the term with a

_{i,ni}. Evaluating (x-p

_{i})

^{ni}F(x) at p

_{i} allows you to get a

_{i,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 a_{i,ni}/(x-p_{i})^{ni} from F(x). You can then multiply the resulting fraction G(x) by (x-p_{i})^{ni-1} and evaluate it at p_{i} to obtain a_{i,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/(x^{2}-1). Its poles are 1 and -1.

F(x)(x-1)=1/(x+1)
F(x)(x+1)=1/(x-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.