<s>[INST] How do you divide a polynomial by a binomial of the form $ax^2+b$, where $a$ and $b$ are greater than one? I came across a question that asked me to divide $-2x^3+4x^2-3x+5$ by $4x^2+5$. Can anyone help me?
 [/INST] The answer in no way depends on whether $a$ and $b$ are more than $1$ or less; the only fact about those that you need is that the polynomial you're dividing by is not zero.  You have this:
$$
\begin{array}{ccccccccccccccccc}
 \\
4x^2+5 & \big) & -2x^3 & + & 4x^2 & - & 3x & + & 5 \\
       &       &  
\end{array}
$$
So ask what you need to multiply $4$ by to get $-2$.  In other words, divide $-2$ by $4$.  You get $-2/4=-1/2$.  So write
$$
\begin{array}{ccccccccccccccccc}
& & \frac{-1}{2} x \\
\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & - & 3x & + & 5 \\
       &       &  
\end{array}
$$
Then multiply:
$$
\begin{array}{ccccccccccccccccc}
& & \frac{-1}{2} x \\
\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & - & 3x & + & 5 \\
       &       &  -2x^3 & & & -& \frac 52 x
\end{array}
$$
Then subtract:
$$
\begin{array}{ccccccccccccccccc}
& & \frac{-1}{2} x \\
\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & - & 3x & + & 5 \\
       &       &  -2x^3 & & & -& \frac 52 x \\  \hline
& & & &  4x^2 & - & \frac 12 x & + 5 \\  \hline
\end{array}
$$
The ask what you have to multiply $4x^2$ by to get $4x^2$, then multiply, then subtract.....
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