If $Q(x)$, $R(x)$, and $P(x):=Q(x)/R(x)$ are all polynomials, then is it true that $\deg(P)=\deg(Q)-\deg(R)$?

Question

Let $Q(x)$ and $R(x)$ be polynomials. Suppose I know that $$ P(x):=\frac{Q(x)}{R(x)} $$ is also a polynomial (and not just some rational function).

Is it always true that $\deg(P(X))=\deg(Q(x))-\deg(R(x))$?

I was thinking this would be useful, since I might want to know the degree of $P(x)$, but it might be the case that $P(x)$ is complicated so I might not have an easy way to write $P(x)$ as a polynomial and look at its largest degree. But of course the degree of $Q(x)$ and $R(x)$ will always be obvious, so if this fact is true I can always calculate $\deg(P(x))$.

Any hints or tips would be appreciated. I am in particular looking for a proof.

Answer 1

Supposing that $P$ is indeed a polynomial, you can multiply both sides by $R$ to get $P(x)R(x) = Q(x)$. The lhs the product of two polynomials $\implies$ it is also a polynomial. We get $\deg(PR) = \deg(P) + \deg(R) = \deg(Q) \implies \deg(P) = \deg(Q) - \deg(R)$.

Answer 2

Note that your relation is equivalent to $P(x) \cdot R(x) = Q(x)$, and that the degree of a product of polynomials is the sum of their degrees. So $\deg P(x) + \deg R(x) = \deg Q(x)$, which is equivalent to what you wanted to hold.