Answer
$\frac{3x^2+12x-20}{(x+2)^2(x-2)^2}=\frac{-1}{x+2}+\frac{-2}{(x+2)^2}+\frac{1}{x-2}+\frac{1}{(x-2)^2}$
Work Step by Step
$\frac{3x^2+12x-20}{x^4-8x^2+16}$=factorising $x^4-8x^2+16=(x^2-4)(x^2-4)=(x-2)(x+2)(x-2)(x+2)=(x-2)^2(x+2)^2$,
$\frac{3x^2+12x-20}{(x+2)^2(x-2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}$,
$A(x+2)(x-2)^2+B(x-2)^2+C(x-2)(x+2)^2+D(x+2)^2$,
$A(x^3-2x^2-4x+8)+B(x^2-4x+4)+C(x^3-2x^2-4x-8)+D(x^2+4x+4)$,
$(A+C)x^3+(-2A+B+2C+D)x^2+(-4A-4B-4C+4D)x+(8A+4B-8C+4D)=3x^2+12x-20$,
$\begin{cases}
A+C=0\\
-2A +B +2C+D = 3\\
-4A -4B -4C +4D=12\\
8A +4B -8C +4D= -20
\end{cases}$
Multiplying Equation 1 by 2 and adding it to Equation 2.
$\begin{cases}
2A +2C=0\\
-2A +B +2C+D=3\\
-- -- -- -- -- -- -- --\\
0A +B +4C +D=3
\end{cases}$
Multiplying Equation 3 by 2 and adding it to Equation 4.
$\begin{cases}
& -8A -8B-8C+8D=24\\
& 8A +4B -8C +4D=-20\\
& -- -- -- -- -- -- -- --\\
& 0A -4B -16C +12D=4
\end{cases}$
Multiplying Equation 1 by -8 and adding it to Equation 4.
$\begin{cases}
-8A-8C=0\\
8A+4B-8C+4D=-20\\
-- -- -- -- -- -- --\\
0A -4B -16C +4D=-20
\end{cases}$
Adding the Addition result of Equation 3, Equation 4 and the Addition result of Equation 1, Equation 4
$\begin{cases}
& -4B -16C +12D=4\\
& 4B -16C +4D=-20\\
& -- -- -- -- -- --\\
& -32C +16D=-16
\end{cases}$
Multiplying the Addition result of Equation 1 and Equation 2 by 4 and adding it to the Addition result of Equation 1 and Equation 4.
$\begin{cases}
4B+16C+4D=12\\
-4B-16C+12D=4\\
-- -- -- -- -- -- --\\
0B+0C+16D=16
\end{cases}$
thus, $D=1$, Substituting back into the Equation, $-2C+1=-1, C=1$.
Substituting back into the Equation, $A+1=0, A=-1$.
Substituting back into the Equation, $2+B+2+1=3, B=-2$
Therefore,
$\frac{3x^2+12x-20}{(x+2)^2(x-2)^2}=\frac{-1}{x+2}+\frac{-2}{(x+2)^2}+\frac{1}{x-2}+\frac{1}{(x-2)^2}$
