I know what the answer is but just can't seem to do the actual decomposition.
Can anyone help?
4x - 4 / (x^{2}-4x+4)
$\displaystyle \frac{4x-4}{x^2-4x+4} = \frac{4(x-1)}{(x-2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2}$
$\displaystyle 4(x-1) = A(x-2) + B$
let $\displaystyle x = 2$ ...
$\displaystyle B = 4$
let $\displaystyle x = 1$
$\displaystyle 0 = -A + 4$
$\displaystyle A = 4$
$\displaystyle \frac{4x-4}{x^2-4x+4} = \frac{4}{x-2} + \frac{4}{(x-2)^2}$