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