# Decompose \frac {48-8x}{x^4-16} into partial fraction.

## Question:

Decompose {eq}\frac {48-8x}{x^4-16} {/eq} into partial fraction.

## Partial Fraction:

There are different types of partial fractions, i.e. linear partial fraction, repeated linear partial fraction and a quadratic factor. The partial factor which is in the form of {eq}\displaystyle \frac{P}{pu+q} {/eq} is called a linear factor of a partial fraction.

Solving for $$\displaystyle P = \frac {48-8x}{x^4-16}$$

Factor:

$$\displaystyle = \frac{-8x+48}{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}$$

First we are going to find its partial fractions.

{eq}\displaystyle \frac{48-8x}{\left(x^2+4\right)\left(x+2\right)\left(x-2\right)}=\frac{a_1x+a_0}{x^2+4}+\frac{a_2}{x+2}+\frac{a_3}{x-2} {/eq}

Simplify:

{eq}\displaystyle 48-8x=\left(a_1x+a_0\right)\left(x+2\right)\left(x-2\right)+a_2\left(x^2+4\right)\left(x-2\right)+a_3\left(x^2+4\right)\left(x+2\right) {/eq}

By using its root -2, we get the value of {eq}a_2=-2 {/eq} and another root 2, we get the value of {eq}a_3=1. {/eq}

Put in the value of {eq}a_2 {/eq} and {eq}a_3. {/eq}

{eq}\displaystyle 48-8x=\left(a_1x+a_0\right)\left(x+2\right)\left(x-2\right)+\left(-2\right)\left(x^2+4\right)\left(x-2\right)+1\cdot \left(x^2+4\right)\left(x+2\right) {/eq}

Expand:

{eq}\displaystyle 48-8x=a_1x^3-x^3+a_0x^2+6x^2-4a_1x-4x+24-4a_0 {/eq}

Group the elements accourding to powers of x.

{eq}\displaystyle -8x+48=x^3\left(a_1-1\right)+x^2\left(a_0+6\right)+x\left(-4a_1-4\right)+\left(-4a_0+24\right) {/eq}

Equate the coefficients of similar terms on both sides to create a list of equations:

{eq}\displaystyle 24-4a_0=48\\ \displaystyle -4a_1-4=-8\\ \displaystyle a_0+6=0\\ \displaystyle a_1-1=0. {/eq}

Now, solving above equations we get the value of {eq}a_1=1 {/eq} and {eq}a_0=-6. {/eq}

So, partial fractions are given by

$$\displaystyle = \frac{x-6}{x^2+4}+\frac{1}{x-2}-\frac{2}{x+2}.$$