Roots of the derivative of a nice polynomial function (with explicit roots)

Question

Let $n\in\mathbb{N}$ and consider the polynomial function $f:\mathbb{R}\to\mathbb{R}$ given by $$ f(x):=x^2(x^2-a_1^2)^2(x^2-a_2^2)^2...(x^2-a_n^2)^2, $$ where $a_1,...,a_n\in\mathbb{R}$, all different, i.e. $a_i\neq a_j$ for all $i\neq j$. It is very clear that all roots of $f(x)$ are exactly located at $x=0,\pm a_1,...,\pm a_n$. Is there any chance to "easily" obtain where are located the roots of $\tfrac{d}{dx}f(x)$? Up to now, by explicitly differentiating I've seen that $\tfrac{d}{dx}f(x)$ has $2n+1$ roots located at $x=0,\pm a_1,...,\pm a_n$, but I am still missing the remaining $2n$ other roots (the most complicated ones, because there seem to be a combination of factors). Is there any chance (or clever way) to do it? At least to understand more or less where they should be located? Or maybe a reason why I shouldn't expect to get these roots so easily?

Answer 1

Writing $a_0=0$ and $a_{-i}=-a_i$, the given polynomial is $$p(x)=x^2\prod_{i=1}^n(x^2-a_i^2)^2=q(x)^2$$ where $q(x)=\prod_{i=-n}^n(x-a_i)$ is a polynomial of degree $2n+1$.

Therefore $p'(x)=2q(x)q'(x)$ has the same roots as $q(x)$ and $q'(x)$ combined. Those of $q(x)$ are the $2n+1$ mentioned in the question. The remaining ones are the roots of $q'(x)$. There are $2n$ of them and they lie in between the roots of $q(x)$, that is, $a_i<b_i<a_{i+1}$. For a proof of this, see Kyle's answer in Relation between real roots of a polynomial and real roots of its derivative.