Dividing by a monic polynomial

Division of polynomials $P_1$ by $P_2$ should behave like division of integers, that is you would like to find quotient polynomial $Q$ and remainder polynomial $R$ such that:

$$P_1 = P_2 \times Q + R$$    AND $$\deg(R) < \deg(P_2).$$

This is NOT always possible when $P_1$, $P_2$, $Q$, and $R$ must be polynomials over the integers. For example, what quotient and remainder would you suggest for $P_1$ $=$ $x^2$, and $P_2$ $=$ $3x$?

In the special case where $P_2$ has leading (highest) coefficient either 1 or -1 (that is, $P_2$ is monic), and $P_2$ has degree no greater than $P_1$, then division is possible. Here's a recipe:

1. Set the remainder $R$ initially equal to $P_1$, and the quotient $Q$ initially equal to 0.
2. While the degree of $R$ is no less than the degree of $P_2$ do the following steps:
   1. Construct a monomial $M$ (a polynomial with one term) $m$ by raising $x$ (or whatever variable you're using) to the exponent equal to the degree of $R$ minus the degree of $P_2$, and then multiplying this power of $x$ by the leading coefficient of $R$ times the leading coefficient of $P_2$ (either 1 or -1).
   2. Recalculate $Q$ by adding $M$ to it.
   3. Recalculate $R$ by subtracting ($M$ $\times$ $P_2$) from it. This new remainder will have lower degree than the old one.

Implement the function monDiv specified in Polynomial.h. For $P_1$ or order $n$, your function should have complexity $O(n^3)$.