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Here is the statement :

Let $P$ be a non constant polynomial of $\mathbb{C}[X]$ which has at least two distinct roots. If $P''$ divides $P$ hence all the roots of $P$ belong to the same (real) line.

The ingredients used are : $P$ has in fact only simple roots, the Gauss-Lucas theorem and the extreme points of a convex set.

It seems slightly similar to Jensen's theorem.

I was wondering if it was a well-known fact about the roots of a non constant complex polynomial ? If so, are there any references about it ?

Thanks in advance !

Maman
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    Connected : https://math.stackexchange.com/q/1454206/305862 and https://math.stackexchange.com/q/2839020/305862 – Jean Marie May 25 '22 at 20:40
  • @JeanMarie Thank you ! Do we have something interesting about the second part !? – Maman May 25 '22 at 21:15
  • What are you calling "the second part" ? – Jean Marie May 25 '22 at 21:21
  • @JeanMarie the fact that all the roots of $P$ belong to the same line. – Maman May 25 '22 at 21:34
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    According to the answer given to the second reference I gave, it is the line joining roots $a$ and $b$... Strange formulation but not contradictory... – Jean Marie May 25 '22 at 21:50
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    Warning: A complex line is a plan respect to $\mathbb R$ (dimensión $1$ over $\mathbb C$ but $2$ over $\mathbb R$. – Piquito May 25 '22 at 22:10
  • @Piquito Indeed you are right ! – Maman May 25 '22 at 22:51
  • @JeanMarie Ok it seems easier than using theorems like Gauss-Lucas or extreme points in a convex set... – Maman May 25 '22 at 23:29
  • @Maman The answer that I previously posted was wrong, so I removed it. The mistake was actually in my other answer it relied upon, which I fixed since. In summary, your $P$ has two distinct roots, and the conclusion there is that it must have all distinct roots (otherwise it would have to be a perfect power). It is not immediately obvious how/whether that helps with your question here, but at least the question makes a lot more sense now. Sorry for the noise. – dxiv May 27 '22 at 05:49
  • @dxiv Thank you for clarifying the reasoning ! That means that $P$ will be of the form $c\prod (x-\lambda_i)$. And for the roots to be on the same (real) line, how would we conclude ? Maybe it is a this point that we have to use Gauss-Lucas theorem or something like that ? – Maman May 27 '22 at 10:30
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    @Maman Right, it should work along that line, though I was looking/hoping for a more direct and possibly constructive proof, which I still believe must exist. – dxiv May 28 '22 at 00:42

1 Answers1

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My answer here proves that any polynomial of degree $\,n\ge 3\,$ which satisfies $\,P''(x) \mid P(x)\,$ and has at least two distinct roots must have all $\,n\,$ roots distinct.

Let $\,P(x) = \lambda(x-a)(x-b)P''(x)\,$ with $\,a \ne b\,$ and $\,c_i \big|_{i=1,2,\dots,n-2}\,$ the roots of $\,P''\,$, all distinct and different from $\,a, b\,$. If $\,d_j \big|_{j=1,2,\dots,k \,\le\,n-2}\,$ are the extreme points of the convex hull $\,C_{P''} =\text{Conv}(c_i)\,$ then $\,\text{Conv}(d_i) = C_{P''}\,$ by the Krein–Milman theorem.

Let $\,C_{P'}\,$ be the convex hull of the roots of $\,P'\,$, and $\,C_P = \text{Conv}(a, b, c_i)\,$, then $\,C_{P''} \subseteq C_{P'} \subseteq C_P\,$ by the Gauss-Lucas theorem. Root $\,d_1\,$ of $\,P'' \mid P\,$ is in both $\,C_P\,$ and $\,C_{P''}\,$, so it must be in $\,C_{P'}\,$. However, $\,d_1\,$ cannot be an extreme point of $\,C_{P'}\,$ since those are among the roots of $\,P'\,$, and $\,P'\,$ has no roots in common with $\,P\,$ because all roots are simple. Therefore $\,d_1\,$ cannot be an extreme point of $\,C_P\,$, either, since an extreme point of a convex set is an extreme point of any convex subset that contains it.

By symmetry, none of the $\,d_j\,$ can be extreme points of $\,C_P\,$, either. Since $\,C_P\,$ contains $\,a,b\,$ and, by convexity, the entire segment $\,\overline{ab}\,$, it follows that the only extreme points are $\,a,b\,$, and all other roots $\,c_i\,$ must lie within segment $\,\overline{ab}\,$.

dxiv
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  • As I just commented on the linked answer, the claim at the top is false, and $P(x) = x^3 - x$ is a counterexample. – Ravi Fernando May 27 '22 at 01:11
  • @RaviFernando Thanks for catching that. Revised and reworked. – dxiv May 28 '22 at 00:46
  • Thank you ! Do you know if there are any references about this statement ? – Maman May 28 '22 at 12:12
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    @Maman I did run a cursory search on polynomials with collinear roots, but came up empty. The case of the cubic was answered on MSE under When are the roots of a polynomial of degree 3 aligned?, but I didn't find much about the general case. Polynomials with $,P'' \mid P,$ form just a small subset of those with all roots on a line. For example, in the case of a cubic, $,P'' \mid P,$ iff one root is the midpoint of the others. – dxiv May 28 '22 at 17:47
  • @Maman On second thought, polynomials with collinear roots may not be "interesting" enough to merit much study on their own, since they directly relate to polynomials with all-real roots and their ranges, which is a more general and better studied problem. – dxiv May 29 '22 at 08:16
  • @dxiv Maybe you'll find that claim interesting. If we suppose that the two distinct roots of $P$ are real then under the same conditions, we can deduce that all the roots of $P$ are in fact real (and simple) ! – Maman May 29 '22 at 14:00
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    @Maman Right. Another direct consequence is that, if $,P,$ is a real polynomial with distinct roots on a line, then that line must be either the real axis or parallel to the imaginary axis. This follows because the roots of $,P^{(n-2)},$ must be on the same line, and the roots of a quadratic with real coefficients are either real or complex conjugates (and, in fact, it's enough for the leading three coefficients of $,P,$ to be real). – dxiv May 29 '22 at 22:52