# How do you express (4x^2+6x-2)/((x-1)(x+1)^2) in partial fractions?

Jul 6, 2018

$\frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) {\left(x + 1\right)}^{2}} = \frac{2}{x + 1} + \frac{2}{x + 1} ^ 2 + \frac{2}{x - 1}$

#### Explanation:

We consider the following ansatz

$\frac{4 {x}^{2} + 6 x - 2}{\left(x - 1\right) \cdot {\left(x + 1\right)}^{2}} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 1} ^ 2$
multiplying by

$\left(x - 1\right) {\left(x + 1\right)}^{2}$

we get

$4 {x}^{2} + 6 x - 2 = A {\left(x + 1\right)}^{2} + B \left(x + 1\right) \left(x - 1\right) + C \left(x - 1\right)$

ultiplying out and rearranging

we get

$4 {x}^{2} + 6 x - 2 = {x}^{2} \left(A + B\right) + x \left(2 A + C\right) + C \left(x - 1\right)$
this leads us to the following System

$A + B = 4$

$2 A + C = 6$

$A - B - C = - 2$

solving this System we get $A = B = C = 2$