Answer
$3x^2 +2x+3+\frac{9-6x}{x^2-1}$.
Work Step by Step
We will divide the polynomial $3x^4+2x^3-8x+6$ by $x^2-1)$ using long division.
Rewrite as
$(3x^4+2x^3+0x^2-8x+6)\div(x^2+0x-1)$
Perform division:
$\begin{matrix}
& 3x^2 & +2x &+3 && & \leftarrow &Quotient\\
&-- &-- &--&--&--& \\
x^2+0x-1) &3x^4 &+2x^3&+0x^2&-8x&+6 & \\
& 3x^4 & +0x^3 &-3x^2 && & \leftarrow &3x^2(x^2+0x-1) \\
& -- & -- &-- & && \leftarrow &subtract \\
& 0 & 2x^3 & +3x^2 & -8x & \\
& & 2x^3 & +0x^2 & -2x & & \leftarrow & 2x(x^2+0x-1) \\
& & -- & -- & -- && \leftarrow & subtract \\
& & 0&3x^2 &-6x &+6 \\
& & & 3x^2& +0x &-3 & \leftarrow & 3(x^2+0x-1) \\
& & & -- & -- & --& \leftarrow & subtract \\
& & & 0 & -6x & +9& \leftarrow & Remainder
\end{matrix}$
The answer is
$\Rightarrow Quotient + \frac{Remainder}{Divisor}$
$\Rightarrow 3x^2 +2x+3+\frac{-6x+9}{x^2+0x-1}$
$\Rightarrow 3x^2 +2x+3+\frac{9-6x}{x^2-1}$.
