Answer
$ \displaystyle \frac{6}{x-3}+\frac{3}{x+5}$
Work Step by Step
First, factor the trinomial in the denominator
(Two factors of $-15$ that add to $+2$ are $5$ and $-3$)
$\displaystyle \frac{9x+21}{(x-4)(x+5)}=...$
(see p.541:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.)
$\displaystyle \frac{9x+21}{(x-4)(x+5)}=\frac{A}{x-3}+\frac{B}{x+5}$
multiply both sides with $(x-3)(x+5)\quad $(LCD)
$9x+21=A(x+5)+B(x-3)$
$ 9x+21=Ax+5A+Bx-34B\qquad$ /group like terms...
$9x+21=x(A+B)+(5A-3B)$
... we now equate the coefficients of like powers...
$\left\{\begin{array}{ll}
I. & 9=A+B\\
II. & 21=5A-3B
\end{array}\right.\qquad $solve the system
$(II+3\times I)$ eliminates $B...$
$21+27=3A-5A+3B-3B$
$48=8A$
$A=6$
back substitute: $9=A+B$
$9=6+B$
$B=3$
Rewrite the setup decomposition with $A=6, B=3$
$\displaystyle \frac{9x+21}{(x-4)(x+5)}= \frac{6}{x-3}+\frac{3}{x+5}$
