# How do you write the partial fraction decomposition of the rational expression  (2x^2-18x-12)/(x^3-4x)?

Aug 7, 2016

$\frac{2 {x}^{2} - 18 x - 12}{{x}^{3} - 4 x} = \frac{3}{x} - \frac{5}{x - 2} + \frac{4}{x + 2}$

#### Explanation:

$\frac{2 {x}^{2} - 18 x - 12}{{x}^{3} - 4 x}$

$= \frac{2 {x}^{2} - 18 x - 12}{x \left(x - 2\right) \left(x + 2\right)}$

$= \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}$

Use Heaviside's cover-up method to find:

$A = \frac{2 {\left(0\right)}^{2} - 18 \left(0\right) - 12}{\left(\left(0\right) - 2\right) \left(\left(0\right) + 2\right)} = \frac{- 12}{- 4} = 3$

$B = \frac{2 {\left(2\right)}^{2} - 18 \left(2\right) - 12}{\left(2\right) \left(\left(2\right) + 2\right)} = \frac{8 - 36 - 12}{8} = \frac{- 40}{8} = - 5$

$C = \frac{2 {\left(- 2\right)}^{2} - 18 \left(- 2\right) - 12}{\left(- 2\right) \left(\left(- 2\right) - 2\right)} = \frac{8 + 36 - 12}{8} = \frac{32}{8} = 4$

So:

$\frac{2 {x}^{2} - 18 x - 12}{{x}^{3} - 4 x} = \frac{3}{x} - \frac{5}{x - 2} + \frac{4}{x + 2}$