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Let $p_n$ be a real polynomial of exactly degree $n$. Suppose that $p_n$ has atleast $n-1$ distinct real roots of odd multiplicity. It is claimed that $p_n$ has $n$ simple real roots. How do you verify this claim?

Here is what I thought: if those roots were of multiplicity $\geq 3$, then $p_n$ will have atleast $n+1$ roots counting multiplicities (I hope I do math correctly). Also, there can't be a non-real root, in which case that conjugated one would also be a root, and so in total $n+1$ roots. Something like that.

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Your proof is fine. A slight variant is to divide out $x-\alpha$ for $\alpha$ each of the given roots, to obtain a degree $1$ quotient that can only have real coefficients. But your approach is much simpler. For starters, it doesn't require us to verify $x-\alpha_i|p_n\implies\prod_i(x-\alpha_i)|p_n$.

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  • $\begingroup$ If $\alpha_1,\dots, \alpha_m$ are different real roots of odd multiplicity of $p_n$, then $m\geq n-1$ by assumption. How do we conclude that $m=n$? I can't figure out how to prove it formally and completely. What I wrote above was my intuition, and I find it difficult to translate it into formal proof. Could you please show a way for me? $\endgroup$ – James2020 May 23 '20 at 20:20
  • $\begingroup$ @James I'd say what you wrote is already formal enough. $\endgroup$ – J.G. May 23 '20 at 20:21

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