This document describes how to use synthetic division and
partial fraction

expansion to reduce a rational function to its canonical form. Synthetic
division

and partial fraction expansion are implemented in Matlab's residue function,

which is a good way to experiment with them.

**1 Partial fractions**

Suppose we have a rational function

We would like to represent it in a simpler form. It turns
out that any rational

function can be decomposed in the partial fraction expansion

Here K(s) is a polynomial, and C_{i}(s) and P_{i}(s) are
\simple" polynomials. K(s)

is called the direct term; it is necessary only if n≥ m, and if it exists it has

degree (n - m).

The denominator polynomials P _{i} depend on the roots of A ,
which are also

called poles of the rational function. (Roots of B are called zeros of the
rational

function.) Each isolated root x_{i} of A results in a denominator polynomial of

the form P_{i}(s) = s - x_{i}; each complex conjugate pair of roots x_{i}±
y_{i}i gives a

denominator of the form
Multiple roots result in

terms of higher degree; for example a real root xi with multiplicity k gives a

denominator

To determine the direct term we can use synthetic division
(see below). So

for now let us as sume n < m. In this case we can use the Heaviside method

(also called the cover-up method) to determine the coefficient polynomials C_{i}.

The simplest case is an isolated root x_{i} of A(s). In this case, C_{i} is a
constant,

and we have

where the second equation holds because every term on the
right-hand side

contains a factor (s - x_{i}) except for the term (s - x_{i})C_{i}/(s - x_{i}). So, we
can

determine C_{i} by deleting one of the factors (s - x_{i}) of A_{i} from our rational

function, and evaluating the result at x_{i}.

For example, suppose we have

We will then have a term in our expansion

To determine a, we evaluate 1/(s^{2} + 1) at s = 2. This
tells us that a = 1/5, so

our term is

This way of determining coe cients gives the method its
name: we \covered

up" the factor 1/(s - 2) of B(s)/A(s) and evaluated the remaining expression

at s = 2.

If our denominator has a repeated root or a complex
conjugate pair of roots

(or even a repeated conjugate pair), then we will have a factor P _{i}(s) in the
de-

nominator which has degree d > 1. This factor will result in a term C_{i}(s)/P_{i}(s)

in our expansion, where degree(C_{i}) < d. In this case we can determine the

coefficients of C_{i} by evaluating P_{i}(s)B(s)/A(s) at the d points where P_{i} is
zero;

this will result in d equations in the d unknown coefficients .

For example, consider again the rational function

The factor (s^{2} + 1) leads to a term in our expansion

To determine a and b, we evaluate 1/(s-2) at the two
points at which (s^{2} +1)

is 0, namely ± i. This gets us two equations,

Solving these equations gives a = -1/5 and b = -2/5; combining the new term

with our previous result tells us that our final expansion is

**2 Synthetic division**

We are given a rational function B(s)/A(s) with numerator degree n and de-

nominator degree m. If n≥ m, we can pull out a quotient term K(s), leaving

a remainder term R(s) with degree(R) < m, so that

The process is ana logous to long division, and is called
synthetic division. We

will illustrate it by example: suppose we start with

We are looking for K(s) and R(s), with degree(R) < 2, so
that

B(s) = K(s)A(s) + R(s)

To get the highest- order term of B(s) (namely s^{3}) right,
we can see that we

have to multiply A(s) by s. If we set K_{1}(s) = s, we have

R_{1}(s) = B(s) - K_{1}(s)A(s) = (s^{3} - s^{2} + s + 1) - (s^{3} -
4s^{2} + 3s) = 3s^{2} - 2s + 1

This gets us partway to our goal: R_{1}(s) has a smaller
degree than B(s) did,

but not small enough. But, we can repeat the process: to get rid of the leading

term of R_{1}(s) (namely 3s^{2}), we can multiply A(s) by 3. Setting K_{2}(s) = s + 3,

we have

R_{2}(s) = B(s)-K_{2}(s)A(s) = (s^{3} -s^{2} +s+1)-(s^{3} -4s^{2}
+3s)-(3s^{2} -12s+9)

Cancelling terms gives R_{2}(s) = 10s-8, which has sufficiently low degree, so we

can take R = R_{2} and K = K_{2}.