Find the ordered triplet (A,B,C) if:
(6x^2 - 8x -20) / [(x+2)(x-3)(x-2)] = (A/(x+2)) + (B/(x-3)) + (C/(x-2))
**hint: mulitply both sides by (x+2)(x-3)(x-2)
Multiply through and you get,
$\displaystyle 6x^2-8x-20=A(x-3)(x-2)+B(x+2)(x-2)+C(x+2)(x-3)$
Now this is a partial fraction decomposition which can always be achived. Thus this is true for all $\displaystyle x$.
If $\displaystyle x=2$
$\displaystyle 6(2)^2-8(2)-20=A(2-3)(0)+B(2+2)(0)+C(2)(-1)$ you can solve for $\displaystyle C$
If $\displaystyle x=3$
$\displaystyle 6(3)^3-8(3)-20=A(0)(3-2)+B(3+2)(3-2)+C(5)(0)$ you can solve for $\displaystyle B$
If $\displaystyle x=-2$
$\displaystyle 6(-2)^2-8(-2)-20=A(-5)(-4)+B(0)(-2-2)+C(0)(-5)$ you can solve for $\displaystyle A$