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Theorem. Let $P(z)$ be a polynomial of degree $n\geq 2$ and let $\Pi(z)=\prod\limits_{k=0}^{n-1}P^{k}(z)$. \Pi(z)=\prod\limits_{k=0}^{n-1}P^{(k)}(z)$where$P^{(k)}$is the$k$th derivative of$P$. Then either$\Pi(z)$has exactly one distinct zero root or$\Pi(z)$has at least$n+1$distinct zeroesroots. See the original paper by Sudbery. 2 added 64 characters in body The strongest result in this direction that I've heard of is Sudbery's theorem (which was originally conjectured by Popoviciu and Erdös). Theorem. Let$P(z)$be a polynomial of degree$n\geq 2$and let$\Pi(z)=\prod\limits_{k=0}^{n-1}P^{k}(z)$. Then either$\Pi(z)$has exactly one distinct zero or$\Pi(z)$has at least$n+1$distinct zeroes. See the original paper by Sudbery. 1 The strongest result in this direction that I've heard of is Sudbery's theorem. Theorem. Let$P(z)$be a polynomial of degree$n\geq 2$and let$\Pi(z)=\prod\limits_{k=0}^{n-1}P^{k}(z)$. Then either$\Pi(z)$has exactly one distinct zero or$\Pi(z)$has at least$n+1\$ distinct zeroes.