# Is this statement about roots of polynomials well-known?

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 ?

• May 25, 2022 at 20:40
• @JeanMarie Thank you ! Do we have something interesting about the second part !? May 25, 2022 at 21:15
• 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... May 25, 2022 at 21:50
• Warning: A complex line is a plan respect to $\mathbb R$ (dimensión $1$ over $\mathbb C$ but $2$ over $\mathbb R$. May 25, 2022 at 22:10
• @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, 2022 at 0:42

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

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