Answer
$-(x+3), \quad x\ne \frac{1}{2}$
Work Step by Step
Factor the numerator by grouping to obtain:
\begin{align*}2x^2+5x-3&=2x^2-x+6x-3\\
&=(2x^2-x)+(6x-3)\\
&=x(2x-1)+3(2x-1)\\
&=(2x-1)(x+3)
\end{align*}
By factoring both denominator and numerator and canceling the common factors, we obtain:
\begin{align*}
\require{cancel}
\dfrac{2x^2+5x-3}{1-2x}&=\dfrac{(2x-1)(x+3)}{-(2x-1)}\\
\\&=\dfrac{\cancel{(2x-1)}(x+3)}{-\cancel{(2x-1)}}\\
\\&=-(x+3), \quad x\ne\frac{1}{2}
\end{align*}
