# A polynomial whose roots form an arithmetic progression

Let $$f$$ be a fourth degree polynomial whose roots form an arithmetic progression. Prove that $$f'$$'s roots also form an arithmetic progression.
I didn' t make much progress, I just wrote $$f(x) =a(x-b-r) (x-b-2r)(x-b-3r)(x-b-4r)$$ and I tried to differentiate, which obviously doesn't help too much.

Clearly, the problem is unaffected by shifting the polynomial horizontally via $$x \mapsto x-a$$. Using this observations, we can WLOG that $$f(x)=k(x-3d)(x-d)(x+d)(x+3d)$$. Since $$f(x)=f(-x)$$, we have $$f’(x)=-f’(-x)$$. Note this implies $$f’(0)=0$$, and if $$\alpha$$ is a root of $$f’(x)$$ then $$-\alpha$$ is too. So the roots of $$f’(x)$$ are $$0, \pm \alpha$$ for some $$\alpha \in \mathbb{R}$$ which is clearly an arithmetic progression.