# How do you express (2x^2-18x-12)/(x^3-4x) in partial fractions?

Oct 30, 2016

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

#### Explanation:

Expand:
$\frac{2 {x}^{2} - 18 x - 12}{{x}^{3} - 4 x} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 2}$

Multiply both sides by $x \left(x + 2\right) \left(x - 2\right)$

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

Let x = 0 to make B and C disappear:

$- 12 = A \left(2\right) \left(- 2\right)$

$A = 3$

Let x = -2 to make A and C disappear:

$2 {\left(- 2\right)}^{2} - 18 \left(- 2\right) - 12 = B \left(- 2\right) \left(- 2 - 2\right)$

$32 = B \left(8\right)$

$B = 4$

Let x = 2 to make A and B disappear:

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

$- 40 = C \left(8\right)$

$C = - 5$

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