# If $f$ has 3 zeros, then $\exists x_0: f''(x_0)=0$

I'm having some trouble proving the following:

Let $$f$$ be a polynomial with exactly 3 distinct real zeros. Prove that there exists $$x_0 \in \mathbb R$$ such that $$f''(x_0)=0$$.

Visually I understand why this would be true, but I don't really know how to prove this rigorously.

I'm thinking that we can use the fact that if $$f$$ is a polynomial, then $$f'$$ is also a polynomial and so will be $$f''$$. So, because $$f''$$ is a polynomial it's continuous and then maybe use the Intermediate value theorem to prove that there is a point $$x_0$$ such that $$f''(x_0) = 0$$, But I'm not so sure how to do it.

How can I prove this?

It does not even need to be a polynomial. Just assume $$f$$ to be twice differentiable (which, of course, applies to every polynomial). Let $$y_0 < y_1 < y_2 \in \mathbb{R}$$ be the zeros of $$f$$. Then $$f(y_0) = f(y_1) = f(y_2) = 0$$. So by Rolle's theorem, there exist $$z_0 \in (y_0, y_1)$$ and $$z_1 \in (y_1, y_2)$$ such that $$f'(z_0) = f'(z_1) = 0$$. Then apply Rolle's theorem again to get $$x_0 \in (z_0, z_1)$$ such that $$f''(x_0) = 0$$.
Let $$f$$ have three roots $$\alpha\lt \beta \lt \gamma$$, then clearly $$f(\alpha)=f(\beta)=f(\gamma)=0$$
By Rolle's theorem, there exist $$y_1,y_2$$ such that $$y_1\in (\alpha, \beta)$$ and $$y_2\in (\beta, \gamma)$$ such that $$f'(y_1)=f'(y_2)=0$$. Again apply Rolle's theorem on $$[y_1,y_2]$$.
You may write $$f(x)=(x-a)(x-b)(x-c)g(x)$$ , where $$a and $$g$$ has no real roots so it retains it's sign. Now $$f''(a)$$, $$f''(c)$$ have different signs and as $$f''$$, is continous you get the result.