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.
Best Answer
You need to pick some definition of repeated root. In this case, the approach you suggest in your comment is probably the right one. You could even go so far as to talk about the multiplicity being the power of $(x-\alpha)$ that divides $f(x)$.