Answer
$2y^{2}+y-2-\displaystyle \frac{ 2}{2y-1}$
Work Step by Step
$\begin{array}{ccccccccccc}
& & 2y^{2} & +y& -2 & \\
& &--&-- &--& \\
2y-1&) & 4y^{3} &+0 & -5y& +0 & & \\
& & 4y^{3} & -2y^{2} & & & \color{red}{\leftarrow \small{2y^{2}(2y-1) } } \\
& &--&-- & & & \color{red}{ \small{subtract}} \\
& & & 2y^{2} & -5y& +0 & \\
& & & 2y^{2} & -y & & \color{red}{\leftarrow \small{y(2y-1) } }\\
& & &--&-- & &\color{red}{ \small{subtract}} \\
& & & & -4y & +0 & \\
& & & & -4y & +2 & \color{red}{\leftarrow \small{-2(2y-1) } }\\
& & & &-- & -- &\color{red}{ \small{subtract}} \\
& & & & & -2 &
\end{array}$
Quotient = $2y^{2}+y-2$
Remainder = $ -2$.
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{ 4y^{3}-5y }{2y-1}$ = $2y^{2}+y-2-\displaystyle \frac{ 2}{2y-1}$
