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 ?
Thanks in advance !
 A: 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 $\,\overline{ab}\,$.
