# Partial Fractions

Step 1:

Firstly, we have to check that the fraction $\displaystyle \frac{P(x)}{Q(x)}$ is a proper fraction. (Degree of P(x) strictly smaller than Degree of Q(x))

For example,

$\displaystyle \frac{3}{2x+9}$: Proper fraction

$\displaystyle \frac{8x^2}{x^2+3}$: Improper fraction

$\displaystyle \frac{2x^5+1}{3x^3+x}$: Improper fraction

If the partial fraction is an improper fraction, we need to use long division.

Step 2:

Secondly, we have to see which type of partial fractions it is. There are three main types of partial fractions.

## Distinct Linear Factors:

$\displaystyle\boxed{\frac{px+q}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d}}$

## Repeated Linear Factors:

$\displaystyle\boxed{\frac{px+q}{(ax+b)(cx+d)^2}=\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{(cx+d)^2}}$

(Quadratic Factor must be irreducible, i.e. cannot be factorized further)

E.g.

$\displaystyle\boxed{\frac{2x-1}{(x^2+3)(x+1)}=\frac{Ax+B}{x^2+3}+\frac{C}{x+1}}$

## Worked Example

Q: Express $\displaystyle\frac{8}{2x^2-5x-3}$ in partial fractions.

Step 1: Factorize denominator

$\displaystyle\frac{8}{2x^2-5x-3}=\frac{8}{(2x+1)(x-3)}$

Step 2: Use Partial Fractions formula for Distinct Linear Factors

$\displaystyle\frac{8}{(2x+1)(x-3)}=\frac{A}{2x+1}+\frac{B}{x-3}$

Step 3: Multiply throughout by $(2x+1)(x-3)$

$8=A(x-3)+B(2x+1)$

Step 4: Choose two suitable values of x.

Let x=3, then 8=B(7), thus B=8/7

Let x=-1/2, then 8=A(-3.5), thus A=-16/7

Step 5: Write down the partial fraction and check using substition (very important to check your answer)

$\displaystyle\frac{8}{2x^2-5x-3}=\frac{-16}{7(2x+1)}+\frac{8}{7(x-3)}$ (Ans)

Checking:

Let x=9,

$\displaystyle\frac{8}{2x^2-5x-3}=\frac{8}{114}=\frac{4}{57}$

$\displaystyle\frac{-16}{7(2x+1)}+\frac{8}{7(x-3)}=\frac{4}{57}$

Since both give the same value, the check is a success!