Answer
$2x^{2}-8x+38-\displaystyle \frac{156}{x+4}$
Work Step by Step
$\begin{array}{ccccccccccc}
& &2x^{2} & -8x& +38 & \\
& &--&-- &--& \\
x+4&) & 2x^{3} & & +6x& -4 & \color{blue}{\leftarrow \small{\text{no }x^{2}... } } \\
& & 2x^{3} & +8x^{2}& & & \color{red}{\leftarrow \small{2x^{2}(x+4) } } \\
& &--&-- & & & \color{red}{ \small{subtract}} \\
& & & -8x^{2}& +6x& -4& & \\
& & & -8x^{2}& -32x& & \color{red}{\leftarrow \small{-8x(x+4) } }\\
& & &--&-- & &\color{red}{ \small{subtract}} \\
& & & & 38x & -4 & \\
& & & & 38x & +152 & \color{red}{\leftarrow \small{38(x+4) } }\\
& & & &-- & -- &\color{red}{ \small{subtract}} \\
& & & & & -156&
\end{array}$
Quotient = $2x^{2}-8x+38$
Remainder = $ -156$.
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{ 2x^{3}+6x-4}{x+4}$ = $2x^{2}-8x+38-\displaystyle \frac{156}{x+4}$
