Partial Fraction Decomposition by Repeated Synthetic Division

We present an efficient and elementary method to find the partial fraction decomposition of a rational function when the denominator is a product of two highly powered linear factors.


Introduction
Partial fraction decomposition is a classic topic with applications in calculus, differential equations, control theory, and other fields of mathematics.Theoretically, it is well-known that every rational function has a unique partial fraction decomposition as it is an easy exercise in abstract algebra.However, actually decomposing a rational function into partial fractions is computationally intensive.From the aspect of computation, there has been recent developments in this topic for general rational functions [1] as well as special cases [2]- [7].In this article, we present a method for the special case when the denominator is given as a product of two highly powered linear factors.

( ) (
) ( ) The case when the denominator is a power of a single linear factor has been treated in [2] and [5].
Our method is built on top of their methods with the observation that when the denominator is of the form n x the partial fraction decomposition is trivial.The method does not use any derivatives and the computation involves only simple algebraic operations associated with repeated synthetic division.So the method is applicable to both hand and machine calculuations.

Partial Fraction Decomposition
We separate the case when 0 α = from the case when 0 α ≠ .We will assume the factors in the denominator are monic since we can always factor out leading coefficients if necessary.We also assume that the degree of the numerator is less than the degree of the denominator for simplicity of presentation.However the method works in such a case with a little modification (by adding an extra step of back substitution in the end).
Case 1: The denominator is a product of m x and ( ) .
In this case, we find the constants backward from n B down to 1 B recursively using Heaviside cover-up method and synthetic division.By the Heaviside cover-up method, we get ( ) . ) , which is obtained by synthetic division with zero remainder.We repeat the process recursively to get all B i 's.
The numerator is divisible by 1 x − , so we apply synthetic division to simplify the function and get ( ) , and the answer is ( ) ( ) ( ) The whole process can be done as shown in Table 1.Remark 1.The remainder theorem says that the evaluation of ( ) i f β can be done by synthetic division as it is equal to the remainder when The method is also known as Horner's rule.For example, 4 B in Example 1 can be evaluated as follows.
( ) The denominator is a product of ( ) In this case, we take two steps.The first step is to make a substitution u x α = − and expand ( ) Then the problem is reduced to Case 1.The second step is to solve the reduced problem.
We substitute the linear factor with a higher degree because it would reduce the amount of work in the second step.
We can get the coefficients of ( ) through repeated synthetic division [2] as and 1 c is the remainder when the quotient is divided by x α − , and so on.The algorithm for this case is presented below for implementation in a computer.

Algorithm
Input: numerator ( ) We show how the method works for the following function.
( ) ( ) ( ) ( ) ( )  The whole process is described in Table 2.  Reductions Modulo u 2 + u + 1 Partial Fraction Decomposition multiplications and additions.In the second step of partial fraction decomposition, we use less number of synthetic divisions.For the evaluation of functions through synthetic division, the cost is the same.Therefore, the total computational cost is ( ) O n .This method is not the best algorithm in terms of asymptotic speed as the algorithm in [8] is performed in ( ) n steps.However, this method is still intersting because it uses only one technique (synthetic division) in the whole process and hand calculation is straightforward.

=
Then we subtract the last term from Equation (1) to get ( ) (

1 .
where f i 's are successive quotients from synthetic division.In the end, a function of the form We demonstrate how to decompose the following function.
By the Heaviside cover-up method,

−
Repeat the process to get 3 2 B = − and 23  B = , 1 2 B = and we are left with

Table 3 .
Synthetic Division for Example 3.

Table 1 .
Synthetic Division for Example 1.

Table 2 .
Synthetic Division for Example 2.