Answer
$\frac{2}{x-2} + \frac{3}{x+2} - \frac{1}{2x-1}$
Work Step by Step
$\frac{9x^2- 9x + 6}{(2x^{3} - x^{2} - 8x + 4)}$ = $\frac{A}{x-2} + \frac{B}{x+2} + \frac{C}{2x-1}$
Factor by grouping : $(x^{2} -4) (2x -1)$ = $(x-2)(x+2)(2x-1)$
Multiply both sides by (x-2) (x+2) (2x-1)
$9x^2- 9x + 6 = A(x+2) (2x-1) + B (x-2) (2x-1) + C (x-2) (x+2)$
Substitute x=2
12A = 24
A = 2
Substitute x=-2
20B = 69
B=3
Substitute x= 1/2
-15C/4 = 15/4
C = -1
$\frac{2}{x-2} + \frac{3}{x+2} - \frac{1}{2x-1}$
