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