Answer
$\frac{-3x^2-3x+27}{(x+2)(x+3)(2x-3)}=\frac{-3}{x+2}+\frac{1}{x+3}+\frac{1}{2x-3}$,
Work Step by Step
$\frac{-3x^2-3x+27}{(x+2)(2x^2+3x-9)}$, factorising $2x^2+3x-9=(x+3)(2x-3)$:
$\frac{-3x^2-3x+27}{(x+2)(x+3)(2x-3)}=\frac{A}{x+2}+\frac{B}{x+3}+\frac{C}{2x-3}$,
$-3x^2-3x+27=A(x+3)(2x-3)+B(x+2)(2x-3)+C(x+2)(x+3)$,
$-3x^2-3x+27=A(2x^2+3x-9)+B(x^2+5x+6)+C(2x^2+x-6)$,
$-3x^2-3x+27=(2A+B+2C)x^2+(3A+5B+C)x+(-9A+6B-6C)=-3x^2-3x+27$
$\begin{cases}
& 2A & +B & +2C= -3&\\
& 3A & +5B & +C= -3&\\
& -9A & +6B & -6C =27&\\
\end{cases}$
Multiplying Equation $1$ by $-\frac{3}{2}$ and adding it to Equation $2$,
$\begin{cases}
& -3A & -\frac{3}{2}B & -3C =\frac{9}{2}\\
& 3A & +5B & +C = -3\\
&-- -- & -- -- & -- -- &\\
& +0A & \frac{7}{2}B & -2C = \frac{3}{2}
\end{cases}$
$7B -4C=3$, multiplying both sided by $2$.
Multiplying Equation $2$ by $3$ and Adding it to Equation $3$,
$\begin{cases}
& 9A & +15B & +3C =- 9\\
& -9A & +6B & -6C = 27\\
& -- -- & -- -- & -- -- &\\
& 0A & 21B & -3C = 18
\end{cases}$
$7B-C=6$, dividing both sides by $3$,
Multiplying the result of the first Addition by $-1$ and Adding it to the result of the second addition:
$\begin{cases}
& -7B & +4C = -3\\
& 7B & -C =6\\
& -- -- & -- --& \\
& 0B & 3C = 3
\end{cases}$
thus, $C=1$
Substituting back $C=1$ into the equation, $7B-1=6, B=1$. Substituting $B$ and $C$ into the first Equation, $2A+1+2=-3, A=-3$.
Therefore,
$\frac{-3x^2-3x+27}{(x+2)(x+3)(2x-3)}=\frac{-3}{x+2}+\frac{1}{x+3}+\frac{1}{2x-3}$,
