I need a little help with this problem:
8/(y^3-4y).
The book says the answer is -2/y+1/(y-2)+1/y+2 but I'm not following how they got that answer.
I have a feeling my main problem is that my algebra is rusty .
$\displaystyle \frac{8}{y(y+2)(y-2)} = \frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-2}$
$\displaystyle 8 = A(y+2)(y-2) + By(y-2) + Cy(y+2)
$
let $\displaystyle y = 0$ ... $\displaystyle A = -2$
let $\displaystyle y = 2$ ... $\displaystyle C = 1$
let $\displaystyle y = -2$ ... $\displaystyle B = 1$
$\displaystyle \frac{8}{y(y+2)(y-2)} = -\frac{2}{y} + \frac{1}{y+2} + \frac{1}{y-2}$