Answer
The partial decomposition is,
$\frac{8}{x-3}+\frac{-2}{x-2}+\frac{-3}{x+2}$
Work Step by Step
Consider the given partial decomposition
$\frac{3{{x}^{2}}+17x-38}{\left( x-3 \right)\left( x-2 \right)\left( x+2 \right)}=\frac{A}{x-3}+\frac{B}{x-2}+\frac{C}{x+2}$
Multiplying both sides by $\left( x-3 \right)\left( x-2 \right)\left( x+2 \right)$
we get,
$\begin{align}
& 3{{x}^{2}}+17x-38=A\left( x-2 \right)\left( x+2 \right)+B\left( x-3 \right)\left( x+2 \right)+C\left( x-3 \right)\left( x-2 \right) \\
& 3{{x}^{2}}+17x-38=A\left( {{x}^{2}}-4 \right)+B\left( {{x}^{2}}-x-6 \right)+C\left( {{x}^{2}}-5x+6 \right) \\
& 3{{x}^{2}}+17x-38=\left( A+B+C \right){{x}^{2}}+\left( -B-5C \right)x+\left( -4A-6B+6C \right) \\
\end{align}$
Therefore, equating the equal powers of $x$, we get,
$\begin{align}
& A+B+C=3 \\
& -B-5C=17 \\
& -4A-6B+6C=-38
\end{align}$
Adding $A+B+C=3$ and $-B-5C=17$
$\begin{align}
& A-4C=20 \\
& A=4C+20
\end{align}$
From $-B-5C=17$
$B=-17-5C$
Therefore, form $A+B+C=3$ to get,
$\begin{align}
& -4\left( 4C+20 \right)-6\left( -17-5C \right)+6C=-38 \\
& -16C-80+102+30C+6C=-38 \\
& 20C=-60 \\
& C=-3
\end{align}$
Therefore
$\begin{align}
& A=-4\times 3+20=8 \\
& B=-17+5\times 3=-2
\end{align}$
Therefore, the required partial fraction is
$\frac{3{{x}^{2}}+17x-38}{\left( x-3 \right)\left( x-2 \right)\left( x+2 \right)}=\frac{8}{x-3}+\frac{-2}{x-2}+\frac{-3}{x+2}$
