# How do you write the partial fraction decomposition of the rational expression  (8x^2 - 4x - 8)/(x^4 + 2x^3)?

Feb 2, 2018

$\frac{8 {x}^{2} - 4 x - 8}{{x}^{3} \left(x + 2\right)} = - \frac{4}{x + 2} + \frac{4}{x} - \frac{4}{x} ^ 3$

#### Explanation:

We have: $\frac{8 {x}^{2} - 4 x - 8}{{x}^{4} + 2 {x}^{3}} = \frac{8 {x}^{2} - 4 x - 8}{{x}^{3} \left(x + 2\right)}$

$\frac{8 {x}^{2} - 4 x - 8}{{x}^{3} \left(x + 2\right)} = \frac{A}{x + 2} + \frac{B}{x} + \frac{C}{x} ^ 2 + \frac{D}{x} ^ 3$

$\textcolor{w h i t e}{\frac{8 {x}^{2} - 4 x - 8}{{x}^{3} \left(x + 2\right)}} = \frac{A x + B \left(x + 2\right)}{x \left(x + 2\right)} + \frac{C}{x} ^ 2 + \frac{D}{x} ^ 3$

$\textcolor{w h i t e}{\frac{8 {x}^{2} - 4 x - 8}{{x}^{3} \left(x + 2\right)}} = \frac{A {x}^{2} + B x \left(x + 2\right) + C \left(x + 2\right)}{{x}^{2} \left(x + 2\right)} + \frac{D}{x} ^ 3$

$\textcolor{w h i t e}{\frac{8 {x}^{2} - 4 x - 8}{{x}^{3} \left(x + 2\right)}} = \frac{A {x}^{3} + B {x}^{2} \left(x + 2\right) + C x \left(x + 2\right) + D \left(x + 2\right)}{{x}^{3} \left(x + 2\right)}$

$8 {x}^{2} - 4 x - 8 = A {x}^{3} + B {x}^{2} \left(x + 2\right) + C x \left(x + 2\right) + D \left(x + 2\right)$

Putting in $x = 0$ gives us:
$- 8 = D \left(2\right)$
$D = - 4$

$8 {x}^{2} - 4 x - 8 = A {x}^{3} + B {x}^{2} \left(x + 2\right) + C x \left(x + 2\right) - 4 \left(x + 2\right)$

Now putting in $x = - 2$ gives us:
$8 {\left(- 2\right)}^{2} - 4 \left(- 2\right) - 8 = A {\left(- 2\right)}^{3} + B {\left(- 2\right)}^{2} \left(- 2 + 2\right) + C \left(- 2\right) \left(- 2 + 2\right) - 4 \left(- 2 + 2\right)$
$32 = - 8 A$
$A = - 4$

$8 {x}^{2} - 4 x - 8 = - 4 {x}^{3} + B {x}^{2} \left(x + 2\right) + C x \left(x + 2\right) - 4 \left(x + 2\right)$

There are no numbers that can give us just $B$ or $C$, so we must find one in terms of the other, and so we shall put in $x = 1$ to get:
$- 4 = - 4 + 3 B + 3 C - 12$
$12 = 3 \left(B + C\right)$
$B = 4 - C$ (we'll shall put this back in the original equation to help find $C$.

$8 {x}^{2} - 4 x - 8 = - 4 {x}^{3} + \left(4 - C\right) {x}^{2} \left(x + 2\right) + C x \left(x + 2\right) - 4 \left(x + 2\right)$

We need to put in a random value, i.e. $x = 2$

$8 {\left(2\right)}^{2} - 4 \left(2\right) - 8 = - 4 {\left(2\right)}^{3} + \left(4 - C\right) {\left(2\right)}^{2} \left(\left(2\right) + 2\right) + C \left(2\right) \left(\left(2\right) + 2\right) - 4 \left(\left(2\right) + 2\right)$

$16 = 8 c + 16 \left(4 - C\right) - 48$

$64 = 16 \left(4 - C\right)$

$4 = 4 - C$

$C = 0$

Remember that $B = 4 - C$

$B = 4 - 0 = 4$

$\frac{8 {x}^{2} - 4 x - 8}{{x}^{3} \left(x + 2\right)} = - \frac{4}{x + 2} + \frac{4}{x} + \frac{0}{x} ^ 2 - \frac{4}{x} ^ 3$

$\textcolor{w h i t e}{\frac{8 {x}^{2} - 4 x - 8}{{x}^{3} \left(x + 2\right)}} = - \frac{4}{x + 2} + \frac{4}{x} - \frac{4}{x} ^ 3$