Roots of product of two polynomials is the union of the roots of each polynomial

Question

I'm trying to prove this lemma: The roots of $P(x)*Q(x)$ is the union of the roots of $P(x)$ and $Q(x)$ for all $x$.

It's trivially true, which is why I find it hard to prove. Let $r(x) = P(x)*Q(x)$. When you completely factor $r(x)$, you will get a root in the set of roots of either $P(x)$, $Q(x)$ or both. But I'm not sure how to prove that more rigorously.

Any help would be appreciated!

Answer 1

This is actually true for any pair of real-valued functions $f, g$ with the same domain in $\mathbb{R}$. (In fact, I think it's true for any two maps from a given set into an integral domain.)

Let $R$ denote the set of roots of $f$, $S$ the set of roots of $g$, and $T$ the set of roots of $fg$. We must show that $R \cup S = T$, and to do this we can show that both $R \cup S \subseteq T$ and $R \cup S \supseteq T$.

To show the containment $\subseteq$, suppose $r \in R \cup S$. Then, $r \in R$ or $r \in S$; if it is in $R$, then by definition $f(r) = 0$ and so $$(fg)(r) = f(r)g(r) = 0 \cdot g(r) = 0,$$ that is, $r \in T$; by symmetry $r \in S$ leads to the same conclusion.

Can you see how to handle the reverse containment $\supseteq$?

The first step is: Suppose $r \in T$, so that $(fg)(r) = 0$.)