Answer
$(4x^4+6x^3+3x-1)\div(2x^2+1)=2x^2+3x-1+\frac{0}{2x^2+1}$
Work Step by Step
$\begin{array}{lllll}
& 2x^{2} & +3x & -1 & \\
& -- & -- & -- & -\\
2x^2+1\ \ \ \ \ ) & 4x^{4} & +6x^{3} & +3x & -1\\
& -4x^{4} & -2x^{2} & & \\
& -- & -- & & \\
& & 6x^{3} & -2x^2 & +3x \\
& & -6x^{3} &+0x^2 &-3x \\
& & -- & -- & \\
& & & -2x^2 & 0x&-1\\
& & & 2x^2 &0x& +1\\
& & & -- & --\\
& & & & 0
\end{array} $
$(4x^4+6x^3+3x-1)\div(2x^2+1)=2x^2+3x-1+\frac{0}{2x^2+1}$
