Answer
$4x^2+3x-8+\dfrac{18}{x^2+3}$
Work Step by Step
$ \begin{array}{lllllllllll}
& & 4x^{2} & +3x & -8 & & & & \color{red}{ \small{quotient}} & & \\
& & -- & -- & -- & -- & -- & & & & \\
x^{2}+3 & ) & 4x^{4} & +3x^{3} & +4x^{2} & +9x & -6 & & & & \\
& & 4x^{4} & & +12x^{2} & & & & \color{red}{\leftarrow \small{4x^{2}(x^{2}+3) } } & & \\
& & -- & -- & -- & & & & \color{red}{ \small{subtract}} & & \\
& & & 3x^{3} & -8x^{2} & +9x & -6 & & & & \\
& & & 3x^{3} & & +9x & & & \color{red}{\leftarrow \small{3x(x^{2}+3) } } & & \\
& & & -- & -- & -- & & & & & \\
& & & & -8x^{2} & & -6 & & & & \\
& & & & -8x^{2} & & -24 & & \color{red}{\leftarrow \small{-8(x^{2}+3) } } & & \\
& & & & -- & -- & -- & & & & \\
& & & & & & 18 & & \color{red}{ \small{remainder}} & & \\
& & & & & & & & & &
\end{array} $
$(4x^{4}+3x^{3}+4x^{2}+9x-6)\div(x^{2}+3)=$
= $4x^2+3x-8+\dfrac{18}{x^2+3}$
