Roots of polynomials that are the product of 2 polynomials

Question

Given $p, q \in \mathbb{F}[X]$, $deg(p) \land deg(q) \geq 1$ with no roots in $\mathbb{F}$. How could we show that $p.q$ has no roots as well? Any hint would be appreciated.

Answer 1

The magic words are "division with remainder". If $a$ is a root of the product, then $(x-a)$ divides $p q,$ but we know that $x-a$ does not divide either $p$ or $q.$ So $p = s (x-a) + r_1, q = t (x-a) + r_2,$ $pq = (x-a) v +. r_1 r_2.$ $r_1 r_2 \neq 0$ (since $\mathbb{K}$ is a field and has no zero divisors.)

Answer 2

$p, q \in \mathbb{K}[X] \land \forall a \in \mathbb{K}:p,q \neq 0 \implies pq \neq 0$ because $ \mathbb{K} $ is a field and thus has no zero divisors (a product of nonzero elements can only be zero if at least one of them is zero).