Answer
The partial fraction is $\frac{4}{7\left( x-3 \right)}-\frac{8}{7\left( 2x+1 \right)}$.
Work Step by Step
We find the partial fraction of the stated expression as given below,
$\begin{align}
& \frac{4}{2{{x}^{2}}-5x-3}=\frac{4}{\left( x-3 \right)\left( 2x+1 \right)} \\
& =\frac{A}{\left( x-3 \right)}+\frac{B}{\left( 2x+1 \right)}
\end{align}$
Now, multiply $\left( x-3 \right)\left( 2x+1 \right)$ on both sides:
$\begin{align}
& \left( x-3 \right)\left( 2x+1 \right)\frac{4}{\left( x-3 \right)\left( 2x+1 \right)}=\left( x-3 \right)\left( 2x+1 \right)\frac{A}{\left( x-3 \right)}+\left( x-3 \right)\left( 2x+1 \right)\frac{B}{\left( 2x+1 \right)} \\
& 4=\left( 2x+1 \right)A+\left( x-3 \right)B \\
& =2Ax+A+Bx-3B \\
& 4=x\left( 2A+B \right)+A-3B
\end{align}$ …… (I)
Then, compare the coefficient of equation (I):
$2A+B=0$ …… (II)
$A-3B=4$ …… (III)
Then, multiply equation (II) by 3 and add with equation (III),
Then,
$\begin{align}
& 7A=4 \\
& A=\frac{4}{7}
\end{align}$
Now, put the value of A in equation (II):
$\begin{align}
& 2\left( \frac{4}{7} \right)+B=0 \\
& B=-\frac{8}{7}
\end{align}$
Therefore,
$\frac{4}{2{{x}^{2}}-5x-3}=\frac{4}{7\left( x-3 \right)}-\frac{8}{7\left( 2x+1 \right)}$
Thus, the partial fraction of the provided expression is $\frac{4}{7\left( x-3 \right)}-\frac{8}{7\left( 2x+1 \right)}$.
