Answer
The partial fraction is, $\frac{3}{x}+\frac{-5}{\left( x-2 \right)}+\frac{4}{\left( x+2 \right)}$
Work Step by Step
$\begin{align}
& \frac{2{{x}^{2}}-18x-12}{{{x}^{3}}-4x}=\frac{2{{x}^{2}}-18x-12}{x\left( {{x}^{2}}-{{2}^{2}} \right)} \\
& =\frac{2{{x}^{2}}-18x-12}{x\left( x-2 \right)\left( x+2 \right)} \\
& =\frac{A}{x}+\frac{B}{\left( x-2 \right)}+\frac{C}{\left( x+2 \right)}
\end{align}$
Now, multiply $x\left( x-2 \right)\left( x+2 \right)$ on both sides:
$\begin{align}
& x\left( x-2 \right)\left( x+2 \right)\frac{2{{x}^{2}}-18x-12}{x\left( x-2 \right)\left( x+2 \right)}=x\left( x-2 \right)\left( x+2 \right)\frac{A}{x}+x\left( x-2 \right)\left( x+2 \right)\frac{B}{\left( x-2 \right)}+x\left( x-2 \right)\left( x+2 \right)\frac{C}{\left( x+2 \right)} \\
& 2{{x}^{2}}-18x-12=\left( x-2 \right)\left( x+2 \right)A+x\left( x+2 \right)B+x\left( x-2 \right)C \\
& =\left( {{x}^{2}}+2x-2x-4 \right)A+B{{x}^{2}}+2Bx+C{{x}^{2}}-2Cx \\
& =A{{x}^{2}}-4A+B{{x}^{2}}+2Bx+C{{x}^{2}}-2Cx
\end{align}$ $2{{x}^{2}}-18x-12={{x}^{2}}\left( A+B+C \right)+x\left( 2B-2C \right)-4A$ …… (1)
Then, compare the coefficient of equation (1):
$A+B+C=2$ …… (2)
$2B-2C=-18$
$B-C=-9$ …… (3)
$-4A=-12$ …… (4)
$A=3$ …… (5)
And put the value of A in equation (2):
$3+B+C=2$
$B+C=-1$ …… (6)
And add equation (6) with equation (3):
$B=-5$
Then, put the value of B in equation (6):
$\begin{align}
& -5+C=-1 \\
& C=-1+5 \\
& C=4
\end{align}$
Therefore,
$\frac{2{{x}^{2}}-18x-12}{{{x}^{3}}-4x}=\frac{3}{x}+\frac{-5}{\left( x-2 \right)}+\frac{4}{\left( x+2 \right)}$
Thus, the partial fraction of the given expression is $\frac{3}{x}+\frac{-5}{\left( x-2 \right)}+\frac{4}{\left( x+2 \right)}$.
