# Show that the $n$th real polynomial has $n$ simple real roots

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.

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 $$x-\alpha_i|p_n\implies\prod_i(x-\alpha_i)|p_n$$.
• 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? – James2020 May 23 '20 at 20:20