# How do you express  (x^3 + 2) / (x^4 -16) in partial fractions?

##### 1 Answer
Nov 20, 2016

The answer is $= \frac{\frac{3}{16}}{x + 2} + \frac{\frac{5}{16}}{x - 2} + \frac{\frac{x}{2} - \frac{1}{4}}{{x}^{2} + 4}$

#### Explanation:

Let's factorise the denominator

${x}^{4} - 16 = \left({x}^{2} - 4\right) \left({x}^{2} + 4\right) = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 4\right)$

Therefore,

$\frac{{x}^{3} + 2}{{x}^{4} - 16} = \frac{{x}^{3} + 2}{\left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 4\right)}$

The decomposition in partial fractions is

$\frac{{x}^{3} + 2}{{x}^{4} - 16} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{{x}^{2} + 4}$

$\frac{A \left(x - 2\right) \left({x}^{2} + 4\right) + B \left(x + 2\right) \left({x}^{2} + 4\right) + \left(C x + D\right) \left(x - 2\right) \left(x + 2\right)}{\left(x - 2\right) \left(x + 2\right) \left({x}^{2} + 4\right)}$

Therefore,

$\left({x}^{3} + 2\right) = \left(A \left(x - 2\right) \left({x}^{2} + 4\right) + B \left(x + 2\right) \left({x}^{2} + 4\right) + \left(C x + D\right) \left(x - 2\right) \left(x + 2\right)\right)$

let $x = 2$
$10 = 32 B$ ; $\implies$ $B = \frac{5}{16}$

Let $x = - 2$
$- 6 = - 32 A$ ; $\implies$$A = \frac{3}{16}$

Coefficients of ${x}^{3}$
$1 = A + B + C$ , $\implies$ $C = 1 - \frac{5}{16} - \frac{3}{16} = \frac{8}{16} = \frac{1}{2}$

Let $x = 0$
$2 = - 8 A + 8 B - 4 D$ : $\implies$ $4 D = - 2 - \frac{3}{2} + \frac{5}{2} = - 1$
$D = - \frac{1}{4}$

$\frac{{x}^{3} + 2}{{x}^{4} - 16} = \frac{\frac{3}{16}}{x + 2} + \frac{\frac{5}{16}}{x - 2} + \frac{\frac{x}{2} - \frac{1}{4}}{{x}^{2} + 4}$