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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 !

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    $\begingroup$ Connected : math.stackexchange.com/q/1454206/305862 and math.stackexchange.com/q/2839020/305862 $\endgroup$
    – Jean Marie
    May 25, 2022 at 20:40
  • $\begingroup$ @JeanMarie Thank you ! Do we have something interesting about the second part !? $\endgroup$
    – Maman
    May 25, 2022 at 21:15
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    $\begingroup$ 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... $\endgroup$
    – Jean Marie
    May 25, 2022 at 21:50
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    $\begingroup$ Warning: A complex line is a plan respect to $\mathbb R$ (dimensión $1$ over $\mathbb C$ but $2$ over $\mathbb R$. $\endgroup$
    – Piquito
    May 25, 2022 at 22:10
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    $\begingroup$ @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. $\endgroup$
    – dxiv
    May 28, 2022 at 0:42

1 Answer 1

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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 $\,\overline{ab}\,$.

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  • $\begingroup$ As I just commented on the linked answer, the claim at the top is false, and $P(x) = x^3 - x$ is a counterexample. $\endgroup$ May 27, 2022 at 1:11
  • $\begingroup$ @RaviFernando Thanks for catching that. Revised and reworked. $\endgroup$
    – dxiv
    May 28, 2022 at 0:46
  • $\begingroup$ Thank you ! Do you know if there are any references about this statement ? $\endgroup$
    – Maman
    May 28, 2022 at 12:12
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    $\begingroup$ @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. $\endgroup$
    – dxiv
    May 28, 2022 at 17:47
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    $\begingroup$ @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). $\endgroup$
    – dxiv
    May 29, 2022 at 22:52

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