Answer
$2y^{3}+3y^{2}-4y +1 $
Work Step by Step
$\begin{array}{ccccccccccc}
& & 2y^{3} & +3y^{2}& -4y & +1 \\
& &--&-- &--&-- \\
2y-3&) & 4y^{4}& +0 &-17y^{2}& +14y& -3 & & \\
& & 4y^{4}& -6y^{3}& & & & \color{red}{\leftarrow \small{2y^{3}(2y-3) } } \\
& &--&-- & & & & \color{red}{ \small{subtract}} \\
& & & 6y^{3} & -17y^{2}& +14y& -3 \\
& & & 6y^{3} & -9y^{2} & & & \color{red}{\leftarrow \small{3y^{2}( 2y-3) } }\\
& & &--&-- & & & \color{red}{ \small{subtract}} \\
& & & & -8y^{2} & +14y& -3 \\
& & & & -8y^{2} & +12y & & \color{red}{\leftarrow \small{-4y( 2y-3) } }\\
& & & &-- & -- & & \color{red}{ \small{subtract}} \\
& & & & & 2y & -3 & \\
& & & & & 2y & -3 & \color{red}{\leftarrow \small{1( 2y-3) } }\\
& & & &-- & -- & & \color{red}{ \small{subtract}} \\
& & & & & & 0
\end{array}$
Quotient = $2y^{3}+3y^{2}-4y +1$
Remainder = $0$.
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{ 4y^{4} -17y^{2} -14y -3 }{2y-3}$ = $2y^{3}+3y^{2}-4y +1 $
