# How do you express (x^2 - 8x + 44) / ((x + 2) (x - 2)^2) in partial fractions?

Jan 2, 2018

$\frac{{x}^{2} - 8 x + 44}{\left(x + 2\right) \cdot {\left(x - 2\right)}^{2}} = \frac{4}{x + 2} - \frac{3}{x - 2} + \frac{8}{x - 2} ^ 2$

#### Explanation:

$\frac{{x}^{2} - 8 x + 44}{\left(x + 2\right) \cdot {\left(x - 2\right)}^{2}}$

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

After expanding denominator,

$A \cdot {\left(x - 2\right)}^{2} + B \cdot \left({x}^{2} - 4\right) + C \cdot \left(x + 2\right) = {x}^{2} - 8 x + 44$

Set $x = - 2$, $16 A = 64$, so $A = 4$

Set $x = 2$, $4 C = 32$, so $C = 8$

Set $x = 0$, $4 A - 4 B + 2 C = 44$, so $B = - 3$

Thus,

$\frac{{x}^{2} - 8 x + 44}{\left(x + 2\right) \cdot {\left(x - 2\right)}^{2}} = \frac{4}{x + 2} - \frac{3}{x - 2} + \frac{8}{x - 2} ^ 2$