Answer
$y^{3}+8y +4$
Work Step by Step
$\begin{array}{ccccccccccc}
& & y^{3} & & +8y & +4 \\
& &--&-- &--&-- \\
2y-1&) & 2y^{4}& -y^{3} &+16y^{2}& +0y& -4 & & \\
& & 2y^{4}& -y^{3}& & & & \color{red}{\leftarrow \small{y^{3}(2y-1) } } \\
& &--&-- & & & & \color{red}{ \small{subtract}} \\
& & & 0 & +16y^{2}& +0y& -4 \\
& & & & 16y^{2} & -8y & & \color{red}{\leftarrow \small{8y(2y-1) } }\\
& & & &-- & -- & & \color{red}{ \small{subtract}} \\
& & & & & 8y & -4 \\
& & & & &8y & -4 & \color{red}{\leftarrow \small{4(2y-1) } }\\
& & & & & -- & -- & \color{red}{ \small{subtract}} \\
& & & & & & 0
\end{array}$
Quotient = $ y^{3}+8y +4$
Remainder = $0$.
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{ 2y^{4}-y^{3} +16y^{2} -4 }{2y-1}$ = $y^{3}+8y +4$
