Answer
The partial fraction decomposition is ${{x}^{3}}+4{{x}^{2}}+12x+32+\frac{80}{x-2}+\frac{32}{{{\left( x-2 \right)}^{2}}}$.
Work Step by Step
Let us consider the expression,
$\frac{{{x}^{5}}}{{{x}^{2}}-4x+4}$
Firstly, solve by long division method,
${{x}^{2}}-4x+4\overset{{{x}^{3}}+4{{x}^{2}}+12x+32}{\overline{\left){\begin{align}
& {{x}^{5}} \\
& {{x}^{5}}\text{ }-\text{4}{{x}^{4}}+4{{x}^{3}} \\
& -\text{ }+\text{ }- \\
& \overline{\begin{align}
& \text{ 4}{{x}^{4}}-4{{x}^{3}} \\
& \text{ 4}{{x}^{4}}-\text{16}{{x}^{3}}+16{{x}^{2}} \\
& \text{ }-\text{ + }- \\
& \overline{\begin{align}
& \text{ 12}{{x}^{3}}-16{{x}^{2}} \\
& \text{ 12}{{x}^{3}}-48{{x}^{2}}+48x \\
& \text{ + }- \\
& \overline{\begin{align}
& \text{ 32}{{x}^{2}}-48x \\
& \text{ 32}{{x}^{2}}-128x+128 \\
& \text{ }-\text{ + }- \\
& \overline{\text{ 80}x-128} \\
\end{align}} \\
\end{align}} \\
\end{align}} \\
\end{align}}\right.}}$
So,
$\frac{{{x}^{5}}}{{{x}^{2}}-4x+4}={{x}^{3}}+4{{x}^{2}}+12x+32+\frac{80x-128}{{{\left( x-2 \right)}^{2}}}$
Then, partial fraction of $\frac{80x-128}{{{x}^{2}}-4x+4}=\frac{80x-128}{{{\left( x-2 \right)}^{2}}}$
And to find partial fraction decomposition,
$\frac{80x-128}{{{\left( x-2 \right)}^{2}}}=\frac{A}{x-2}+\frac{B}{{{\left( x-2 \right)}^{2}}}$ …… (1)
And multiply both sides by ${{\left( x-2 \right)}^{2}}$,
$\begin{align}
& 80x-128=A\left( x-2 \right)+B \\
& 80x-128=Ax+2A+B \\
\end{align}$
And equate the terms,
$\begin{align}
& A=80 \\
& -2A+B=-128
\end{align}$
Then, solving the above equation,
$A=80,B=32$
And put these values in equation (1),
$\frac{80x-128}{{{\left( x-2 \right)}^{2}}}=\frac{80}{x-2}+\frac{32}{{{\left( x-2 \right)}^{2}}}$
Thus, the partial fraction decomposition of the remainder term is,
${{x}^{3}}+4{{x}^{2}}+12x+32+\frac{80}{x-2}+\frac{32}{{{\left( x-2 \right)}^{2}}}$.
