Answer
$2x^{2}+4x+5+\displaystyle \frac{9}{2x+3}$
Work Step by Step
$\begin{array}{ccccccccccc}
& &2x^{2} & +4x& +5 & \\
& &--&-- &--& \\
2x-4&) & 4x^{3} & & -6x& -11 & \color{blue}{\leftarrow \small{\text{no }x^{2}... } } \\
& & 4x^{3} & -8x^{2}& & & \color{red}{\leftarrow \small{2x^{2}(2x-43) } } \\
& &--&-- & & & \color{red}{ \small{subtract}} \\
& & & 8x^{2}& -6x& -11 & & \\
& & & 8x^{2} & -16x& & \color{red}{\leftarrow \small{4x(2x-4) } }\\
& & &--&-- & &\color{red}{ \small{subtract}} \\
& & & & 10x & -11 & \\
& & & & 10x & -20 & \color{red}{\leftarrow \small{5(2x-4) } }\\
& & & &-- & -- &\color{red}{ \small{subtract}} \\
& & & & & 9 &
\end{array}$
Quotient = $2x^{2}+4x+5$
Remainder = $ 9$.
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{ 4x^{3}-6x-11}{2x-4}$ = $2x^{2}+4x+5+\displaystyle \frac{9}{2x+3}$
