Answer
$\dfrac{2x^2-18x-12}{x(x-2)(x+2)}=\dfrac{3}{x}-\dfrac{5}{x-2}+\dfrac{4}{x+2}$
Work Step by Step
We are given the fraction:
$\dfrac{2x^2-18x-12}{x(x-2)(x+2)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{2x^2-18x-12}{x(x-2)(x+2)}=\dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x+2}$
Multiply all terms by the least common denominator $x(x-1)(x+1)$:
$x(x-2)(x+2)\cdot\dfrac{2x^2-18x-12}{x(x-2)(x+2)}=x(x-2)(x+2)\cdot\dfrac{A}{x}+x(x-2)(x+2)\cdot\dfrac{B}{x-2}+x(x-2)(x+2)\cdot\dfrac{C}{x+2}$
$2x^2-18x-12=A(x-2)(x+2)+Bx(x+2)+Cx(x-2)$
$2x^2-18x-12=Ax^2-4A+Bx^2+2Bx+Cx^2-2Cx$
$2x^2-18x-12=(A+B+C)x^2+(2B-2C)x-4A$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B+C=2\\
2B-2C=-18\\
-4A=-12
\end{cases}$
Solve the system:
$A=3$
$\begin{cases}
3+B+C=2\\
B-C=-9
\end{cases}$
$\begin{cases}
B+C=-1\\
B-C=-9
\end{cases}$
$B+C+B-C=-1-9$
$2B=-10$
$B=-5$
$B+C=-1$
$-5+C=-1$
$C=4$
The partial fraction decomposition is:
$\dfrac{2x^2-18x-12}{x(x-2)(x+2)}=\dfrac{3}{x}-\dfrac{5}{x-2}+\dfrac{4}{x+2}$
