Roots of polynomial with real coefficients appear in conjugate pairs.

Question

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 $f(k)=0$?

Answer 1

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 $\overline{a\cdot b} = \overline{a} \cdot \overline{b}$.

Answer 2

You use the fact that the coefficients of $f$ are real to show that $$ f(\overline c)=\overline{f(c)}. $$