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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.

Luke Collins
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  • I was thinking that I could maybe define a repeated root to be a root of $f$ such that when $f$ is written as $(x-\alpha)g(x)$, then $\alpha$ is a root of $g(x)$, although I don't know how circular this makes things. – Luke Collins Mar 17 '20 at 03:17
  • You probably need to define "root of multipliplicity r" recursiveiy in terms of r, – P. Lawrence Mar 17 '20 at 03:27
  • Be careful to ensure your definition of multiplicity is well-defined. See also what is multiplicity?. Alternatively the result follows immediately from the fact that $,\Bbb R[x],$ is a UFD and the $,x-a_i,$ are nonassociate primes – Bill Dubuque Mar 17 '20 at 15:00
  • @BillDubuque I am writing a set of notes for my students. They are still doing things at the high school level, but I like to make sure they are thinking mathematically and that when we do gloss over formalities, they are aware that we are doing so. This is the definition I provided. What do you think? – Luke Collins Mar 17 '20 at 15:53

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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 $f(x)$.

nilpotence
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