#### Answer

$-\displaystyle \frac{8}{7(2x+1)}+\frac{4}{7(x-3)}$

#### Work Step by Step

First, factor the trinomial in the denominator
(Two factors of $2\cdot(-3)=-6$ that add to $-5$ are $1$ and $-6$)
$2x^{2}-5x-3=2x^{2}+x-6x-3$
$=x(2x+1)-3(2x+1)$
$=(2x+1)(x-3)$
(see p.541:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.)
$\displaystyle \frac{4}{(2x+1)(x-3)}=\frac{A}{(2x+1)}+\frac{B}{(x-3)}$
multiply both sides with $(2x-1)(x-3)\quad $(LCD)
$4=A(x-3)+B(2x+1)$
$ 4=Ax-3A+2Bx+B\qquad$ /group like terms...
$0x+4=x(A+2B)+(-3A+B)$
... we now equate the coefficients of like powers...
$\left\{\begin{array}{ll}
I. & 0=A+2B\\
II. & 4=-3A+B
\end{array}\right.\qquad $solve the system
$(II+3\times I)$ eliminates $A...$
$4+0=-2A+2A+B+6B$
$4=7B$
$B=\displaystyle \frac{4}{7}$
back substitute: $0=A+2B$
$0=A+\displaystyle \frac{8}{7}$
$A=-\displaystyle \frac{8}{7}$
Rewrite the setup decomposition with $A=-\displaystyle \frac{8}{7}, B=\frac{4}{7}$
$\displaystyle \frac{4}{(2x+1)(x-3)}=\frac{-\frac{8}{7}}{(2x+1)}+\frac{\frac{4}{7}}{(x-3)}$
$\displaystyle \frac{4}{(2x+1)(x-3)}=-\frac{8}{7(2x+1)}+\frac{4}{7(x-3)}$