Answer
$2x^2-x-3 +\frac{x+1}{3x^2-1}$.
Work Step by Step
The given expression is
$(6x^4-3x^3-11x^2+2x+4)\div(3x^2-1)$
Rewrite the given expression as
$(6x^4-3x^3-11x^2+2x+4)\div(3x^2+0x-1)$
$\begin{matrix}
& 2x^2 & -x &-3 & && \leftarrow &Quotient\\
&--&-- &-- &--&--& \\
3x^2+0x-1)& 6x^4&-3x^3&-11x^2&+2x&+4 & \\
& 6x^4 & +0x^3 & -2x^2 & && \leftarrow &2x^2(3x^2+0x-1) \\
& -- & -- & --& & & \leftarrow &subtract \\
& 0 & -3x^3 & -9x^2 & +2x && \\
& & -3x^3 & -0x^2 &+ x & & \leftarrow & -x(3x^2+0x-1) \\
& & -- & -- & && \leftarrow & subtract \\
& & 0&-9x^2 &+x &+4 \\
& & & -9x^2& -0x &+3 & \leftarrow & -3(3x^2+0x-1) \\
& & & -- & -- &--& \leftarrow & subtract \\
& & & 0 & x & +1& \leftarrow & Remainder
\end{matrix}$
The answer is
$\Rightarrow Quotient +\frac{Remainder}{Divisor}$.
$\Rightarrow 2x^2-x-3 +\frac{x+1}{3x^2-1}$.
