Why if $P(x) = (x-a)Q(x)$ where $Q(x)$ is linear, $P(x)$ must have $2$ roots over prime modulus?

Question

I am wondering how to formalize the logic behind the following answer:

4. Any polynomial of degree exactly 2 over arithmetic modulo a prime $p>2$ has either 0 roots or 2 roots. Answer: True. Any linear polynomial has one root as one can set it to zero and solve. If there is 1 root, then factoring gives another.

I am trying to formally justify why if there is 1 root, then factoring yields another. I know that if polynomial of degree $d$ $P(x)$ has a root $a$, then it can be written as $(x-a)Q(x)$ for some degree $d-1$ polynomial $Q(x)$. We could then set $Q(x) \equiv 0 \pmod p$ and solve for $x$; however, how do we formally show that there is a solution and that the solution is different than $a$?