# Roots of polynomial with real coefficients appear in conjugate pairs.

How to prove most simply that if a polyonmial $f$, has only real coefficients and $f(c)=0$, and $k$ is the complex conjugate of $c$, then $f(k)=0$?

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 en.wikipedia.org/wiki/Complex_conjugate_root_theorem – number7 Mar 18 '12 at 8:24 Did you try and conjugate $f(c)=0$? You get exactly $f(k)=0$, by using basic properties of the conjugation. – Beni Bogosel Mar 18 '12 at 12:15

Look at $\overline{f(c)}$ and use that conjugation is a homomorphism of $\mathbb{C}$. That is, $\overline{a+b} = \overline{a}+\overline{b}$ and $\overline{a\cdot b} = \overline{a} \cdot \overline{b}$.
You use the fact that the coefficients of $f$ are real to show that $$f(\overline c)=\overline{f(c)}.$$