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?

Nobody seems to have posted the synthetic division route; I'll include it here for completeness.
We first monicize the divisor (i.e., set things up such that the divisor has leading coefficient $1$); thus, consider
$$\frac{-\frac12x^3+x^2-\frac34x+\frac54}{x^2+\frac54}$$
and set up the array
$$\begin{array}{r|cc|cc}
 &-\frac12&1&-\frac34&\frac54\\\hline
0&\times&  &  &\times\\ 
-\frac54&\times&\times&  & \\\hline
 &  &  &  & 
\end{array}$$
Note that the coefficients in the leftmost column are the negated coefficients of the terms with degree less than the degree of the polynomial. The second dividing line (can't do dashed lines in an array environment, sorry) indicates that we expect the quotient to be linear ($3-2=1$). The $\times$ entries indicate the sections of the array that shouldn't be filled.
This generalized synthetic division proceeds like this:
$$\begin{array}{r|cc|cc}
 &-\frac12&1&-\frac34&\frac54\\\hline
0&\times&  &  &\times\\ 
-\frac54&\times&\times&  & \\\hline
 &-\frac12&  &  & 
\end{array}$$
(bring down leading coefficient)
$$\begin{array}{r|cc|cc}
 &-\frac12&1&-\frac34&\frac54\\\hline
0&\times&0&  &\times\\ 
-\frac54&\times&\times&\frac58& \\\hline
 &-\frac12&  &  & 
\end{array}$$
(multiply the bottom of the first column by the coefficients in the leftmost column, and fill the array diagonally)
$$\begin{array}{r|cc|cc}
 &-\frac12&1&-\frac34&\frac54\\\hline
0&\times&0&  &\times\\ 
-\frac54&\times&\times&\frac58& \\\hline
 &-\frac12&1&  & 
\end{array}$$
(add numbers in second column)
$$\begin{array}{r|cc|cc}
 &-\frac12&1&-\frac34&\frac54\\\hline
0&\times&0&0&\times\\ 
-\frac54&\times&\times&\frac58&-\frac54\\\hline
 &-\frac12&1&  & 
\end{array}$$
(multiply bottom of the second column by the coefficients in the leftmost column and fill diagonally)
$$\begin{array}{r|cc|cc}
 &-\frac12&1&-\frac34&\frac54\\\hline
0&\times&0&0&\times\\ 
-\frac54&\times&\times&\frac58&-\frac54\\\hline
 &-\frac12&1&-\frac18&0
\end{array}$$
(add numbers in third and fourth columns)
We thus have the result
$$\frac{-\frac12x^3+x^2-\frac34x+\frac54}{x^2+\frac54}=-\frac12 x+1+\frac{-\frac18x+0}{x^2+\frac54}=-\frac12 x+1+\frac{-\frac12x}{4x^2+5}$$
