Let $P(x)$ be a polynomial.
Let $a \neq b$.
Let $P(a) = 0$.
Let the multiplicity of $a$ be $m$.
Let $P(b) = 0$.
Let the multiplicity of $b$ be $n$.
Prove that $P(x) = (x-a)^m (x-b)^n Q(x)$ for some polynomial $Q(x)$ such that $Q(a) \neq 0$ and $Q(b) \neq 0$.
Can we prove this fact without using the unique factorization of a polynomial?
My attempt is here:
$P(x) = (x-a)^m R(x)$ for some polynomial $R(x)$ such that $R(a) \neq 0$.
Since $P(b) = 0$, $R(b) = 0$. So, $R(x) = (x-b) S(x)$ for some polynomial $S(x)$. $\cdots$