Write the partial fraction decomposition of each rational expression.
(1) 3x/(x + 2) (x - 4)
(2) (2x + 4)/(x^3 - 1)
It's just writting the numerator by using factors of the denominator.
Or make it as $\displaystyle \frac{3x}{(x+2)(x-4)}=\frac{a}{x+2}+\frac{b}{x-4}$ and find $\displaystyle a,b.$ (I can't provide more help, probably someone else's gonna help a bit more.)
This is exactly how to solve the problem. Then multiply both sides by $\displaystyle (x+2)(x-4)$, yielding:
$\displaystyle 3x = a(x - 4) + b(x + 2)$
$\displaystyle 3x = ax - 4a + bx + 2b$
$\displaystyle 3x = (a+b)x + (2b - 4a)$
from which we deduce
$\displaystyle a + b = 3$
$\displaystyle 2b - 4a = 0$
and of course the solution is a = 1, b = 2, as Krizalid indicated.