Answer
$\displaystyle \frac{6x^{2}-14x-27}{(x+2)(x-3)^{2}}=\frac{A}{x+2}+\frac{B}{x-3} +\displaystyle \frac{C}{(x-3)^{2}}$
Work Step by Step
For the factor $(x+2),$
see p.541. Decomposition with Distinct Linear Factors In the Denominator:
Include one partial fraction with a constant numerator for each distinct linear factor in the denominator.
On the RHS, the corresponding term will be $\displaystyle \frac{A}{x+2}$
For the factor $(x-3)^{2}$
see p.544. Decomposition with a Repeated Linear Factor In the Denominator
Include one partial fraction with a constant numerator for each power of a repeated linear factor in the denominator.
On the RHS, the corresponding terms will be
$\displaystyle \frac{B}{x-3}$ +$\displaystyle \frac{C}{(x-3)^{2}}$
Setup only:
$\displaystyle \frac{6x^{2}-14x-27}{(x+2)(x-3)^{2}}=\frac{A}{x+2}+\frac{B}{x-3} +\displaystyle \frac{C}{(x-3)^{2}}$
