Answer
$\frac{3x^2+5x-13}{(3x+2)(x-2)^2}=\frac{-135}{192x+128}+\frac{109}{64x-128}+\frac{9}{8(x-2)^2}$,
Work Step by Step
$\frac{3x^2+5x-13}{(3x+2)(x^2-4x+4)}$, factorising $4x^2-4x+4=(x-2)(x-2)=(x-2)^2$:
$\frac{3x^2+5x-13}{(3x+2)(x+2)(x-2)}=\frac{A}{3x+2}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$,
$3x^2+5x-13=A(x-2)(x-2)+B(3x+2)(x-2)+C(3x+2)$,
$3x^2+5x-13=A(x^2-4x+4)+B(3x^2-4x-4)+C(3x+2)$,
$3x^2+5x-13=(A+3B)x^2+(-4A-4B+3C)x+(4A-4B+2C)=3x^2+5x-13$
$\begin{cases}
& A & +3B & = 3&\\
& -4A & -4B & +3C= 5&\\
& 4A & -4B & +2C =-13&\\
\end{cases}$
Adding Equation $2$ and Equation $3$,
$\begin{cases}
& -4A & -4B & +3C =5\\
& 4A & -4B & +2C = -13\\
&-- -- & -- -- & -- -- &\\
& +0A & -8B & +5C = -8
\end{cases}$
Multiplying Equation $1$ by $4$ and Adding it to Equation $2$,
$\begin{cases}
& 4A & +12B & +0C =12\\
& -4A & -4B & +3C = 5\\
& -- -- & -- -- & -- -- &\\
& 0A & 8B & +3C = 17
\end{cases}$
Adding the result of the first Addition and the result of the second addition:
$\begin{cases}
& 8B & +3C = 17\\
& -8B & +5C =-8\\
& -- -- & -- --& \\
& 0B & 8C = 9
\end{cases}$
thus, $C=\frac{9}{8}$
Substituting back $C=\frac{9}{8}$ into the equation, $8B+\frac{27}{8}=17, B=\frac{109}{64}$. Substituting $B$ into the first Equation, $A+\frac{327}{64}=, A=\frac{-135}{64}$.
Therefore,
$\frac{3x^2+5x-13}{(3x+2)(x-2)^2}=\frac{-135}{192x+128}+\frac{109}{64x-128}+\frac{9}{8(x-2)^2}$,
