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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.

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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.

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