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

Nov 25, 2017

(4x^3)/(x^3+2x^2-x-2) = 4+2/(3(x-1))+2/(x+1)-32/(3(x+2)

#### Explanation:

Note that:

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

$\textcolor{w h i t e}{{x}^{3} + 2 {x}^{2} - x - 2} = {x}^{2} \left(x + 2\right) - 1 \left(x + 2\right)$

$\textcolor{w h i t e}{{x}^{3} + 2 {x}^{2} - x - 2} = \left({x}^{2} - 1\right) \left(x + 2\right)$

$\textcolor{w h i t e}{{x}^{3} + 2 {x}^{2} - x - 2} = \left(x - 1\right) \left(x + 1\right) \left(x + 2\right)$

So:

$\frac{4 {x}^{3}}{{x}^{3} + 2 {x}^{2} - x - 2} = 4 + \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 2}$

So:

$4 {x}^{3} = 4 \left({x}^{3} + 2 {x}^{2} - x - 2\right) + A \left(x + 1\right) \left(x + 2\right) + B \left(x - 1\right) \left(x + 2\right) + C \left(x - 1\right) \left(x + 1\right)$

Putting $x = 1$ we get:

$4 = 6 A \text{ }$ so $\text{ } A = \frac{2}{3}$

Putting $x = - 1$ we get:

$- 4 = - 2 B \text{ }$ so $\text{ } B = 2$

Putting $x = - 2$ we get:

$- 32 = 3 C \text{ }$ so $\text{ } C = - \frac{32}{3}$

So:

(4x^3)/(x^3+2x^2-x-2) = 4+2/(3(x-1))+2/(x+1)-32/(3(x+2)