# Partial Fraction Method

First Method

The method for computing partial fraction decomposition applies to all rational functions with one qualification:

‘The degree of the numerator must be less than the degree of the denominator’

Works for Proper Rational Expressions, where the degree of the numerator is less than the bottom.

Note: When order/degree of numerator is not less than the degree of the denominator or also equal to. Then there is problem and students get confused. So to solve that problem we divide the numerator with denominator (use long division) and the result that we obtain from  division we apply partial fraction on it. As show in example below.

[latex s=3]F(x) = \frac{6s^2+3s+2}{2s^2+14s+20}[/latex]

The above fraction has numerator and denominator that are both 2nd order. Before performing a partial fraction expansion, the fraction must be manipulated. So that the order of the numerator is less than that of the denominator

[latex s=2]s^2+14s+20\div{6s^2+3s+2}[/latex]

by dividing we get the remainder $\textbf{39s+58}$ and quotient is $\textbf{3}$ so,

[latex s=3]F(x) = 3+\frac{-39s-58}{2s^2+14s+20}[/latex]

Partial fraction expansion can now be applied to the remaining fractional on the term

[latex s=3]F(x) = \frac{-39s-58}{2s^2+14s+20}[/latex]

There are four case of partial fraction

Case 1. When there is distinct linear factor in denominator. It is in the form of  $(ax+b)$

Case 1 Example

Case 2. When there is repeated linear factor in denominator. It is in the form of  $(ax+b)^2$

Case 2 Example

Case 3. When there is quadratic term in denominator. It is in the form of  $(ax^2+bx+c)$

Case 3 Example

Case 4. When there is repeated quadratic term in denominator. It is in the form of  $(ax^2+bx+c)^2$

Case 4 Example

Note: These rules can be mixed together in any way. Means that the question given to you either repeated linear and with quadratic term or combination of some other cases.

## Second Method

This method is called residue or cover-up method. This method is easy and there is no need for calculation as in above example. Lets see the example to understand.

Example for Cover-up methods if repeated linear terms are present:

For that we only take derivative of that term and put the value again as shown below.

Example for Cover-up methods if complex root are present:

Like us !