Question

partial fraction decomposition
7x^2+12x+108/ x(x+6)(x-6)

Answer 1

The partial fraction decomposition that you're after is of the form
\[\large \frac{7x^2+12x+108}{x(x+6)(x-6)} = \frac{A}{x} + \frac{B}{x+6} + \frac{C}{x-6}.\tag{1}\]
Multiply both sides of the equation by \(\large x(x+6)(x-6)\) to find that
\[\large \begin{aligned}7x^2+12x+108 &= A(x+6)(x-6) + Bx(x-6) + Cx(x+6) \\ &= A(x^2-36) +B(x^2-6x) + C(x^2+6)\\ &= Ax^2-36A + Bx^2-6Bx + Cx^2+6C\\ &= (A+B+C)x^2 +(C-6B)x+6C-36A\end{aligned}\] Comparing coefficients of both sides leaves us with the system of equations
\[\large \left\{\begin{aligned} A+B+C &= 7\\ C-6B &= 12\\ 6C-36A &= 108\end{aligned}\right.\] Once you solve for A, B, and C, plug them into equation (1) and you'll have your partial fraction decomposition.

Does this make sense? :-)