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I have already shown that any polynomial $P\in\mathbb{R}[x]$ satisfies $\overline{P(\overline{x})}=P(x),\forall x\in\mathbb{C}$

My question is, given a polynnomial $P\in\mathbb{C}[x]$, how I can verify whether $\overline{P(\overline{x})}=P(x),\forall x\in\mathbb{C}\Rightarrow P\in\mathbb{R}[x]$ is always true?

So, does there exist a polynomial $P\in\mathbb{C}[x]\setminus\mathbb{R}[x]$ such that $\forall x\in\mathbb{C}:\overline{P(\overline{x})}=P(x)$?

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Let $P(x)=\sum_{k=0}^na_kx^k$. The hypothesis applied for $x$ real gives $\overline{\sum_{k=0}^na_kx^k}=\sum_{k=0}^na_kx^k,$ hence $\sum_{k=0}^n(\overline{a_k}-a_k)x^k=0.$ The LHS is a polynomial with complex entries, which vanishes on the real numbers: it's the null polynomial, hence $a_k\in\Bbb R$ for $0\leq k\leq n$ and $P\in\Bbb R[x]$.

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    Right (we can sa$y$ 'zeros' of a pol$y$nomial, or 'roots'.2012-10-14