Say I know that a polynomial $f\in\mathbb R[x]$ of degree $n$ has roots $\alpha_1,\dots,\alpha_n\in\mathbb R$. Can I show that $$f(x)=(x-\alpha_1)\cdots(x-\alpha_n)$$ just as a consequence of the factor theorem? I want to avoid the fundamental theorem of algebra, since I just want to demonstrate to my students that every polynomial is the product of its roots (if it has $n$ roots), and they haven't seen complex numbers yet.
The issue lies in dealing with repeated roots, since if we have that the $\alpha_i$ are all distinct, then by the factor theorem, we have $f(x)=(x-\alpha_1)s(x)$ for some $s\in\mathbb R[x]$ with $\deg(s)=n-1$, and since $(\alpha_2-\alpha_1)\neq 0$, we must have $s(\alpha_2)=0$, so we can apply the theorem again to get $f(x)=(x-\alpha_1)(x-\alpha_2)t(x)$, and so on until we factorise $f$ completely.
But if some of the $\alpha_i$ are repeated, this throws a spanner into the works. Indeed, what does it even mean to have a ``repeated root'' if I'm not allowed to assume that $f$ equals $\prod_i(x-\alpha_i)$? Note that my students haven't seen derivatives either, so I can't say that an $n$th order root is also a zero of the first $n-1$ derivatives or something like that.