Let $p( z)$ be a polynomial with real coefficients, and let $\alpha\in\mathbb{C}.$ Prove that $p(\alpha) = 0$ if and only in $p(\bar{\alpha})=0$ 
Let $p( z)$ be a polynomial with real coefficients, and let $\alpha\in\mathbb{C}.$ Prove that $p(\alpha) = 0$ if and only in $p(\bar{\alpha})=0$.

I have just started a linear algebra class and I have no idea how to even start this problem.
 A: Write $p(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots +a_{1}z+a_{0}$, where $a_{i}\in\mathbb{R}$ for all $0\leq i\leq n$. Then, note that we have:
$$
\begin{aligned}
p(\alpha)=0&\iff a_{n}\alpha^{n}+a_{n-1}\alpha^{n-1}+\cdots +a_{1}\alpha+a_{0}=0\\
&\iff\overline{a_{n}\alpha^{n}+a_{n-1}\alpha^{n-1}+\cdots +a_{1}\alpha+a_{0}}=0\\
&\iff\overline{a_{n}\alpha^{n}}+\overline{a_{n-1}\alpha^{n-1}}+\cdots+\overline{a_{1}\alpha}+\overline{a_{0}}=0\\
&\iff a_{n}\overline{\alpha}^{n}+a_{n-1}\overline{\alpha}^{n-1}+\cdots+a_{1}\overline{\alpha}+a_{0}=0\\
&\iff p(\overline{\alpha})=0,
\end{aligned}
$$
since for all complex $\beta,\gamma$ we have $\overline{\beta\gamma}=\overline{\beta}\overline{\gamma}$ and $\overline{\beta+\gamma}=\overline{\beta}+\overline{\gamma}$, and $\overline{r}=r$ for all $r\in\mathbb{R}$. 
A: Let $\alpha$ be a complex number that is not real. And let $Q(x)=(x-\alpha)(x-\overline\alpha)$. Clearly $Q(x)$ is a polynomial with real coefficients.
Suppose that there is a polynomial $R(x)$ with $R(\alpha)=0$. Then $Q(x)$ divides $R(x)$.
Otherwise we have $R(x)=P(x)Q(x)+S(x)$, where $S(x)$ is a nonzero polynomial of degree at most $1$ with $S(\alpha)=0$, none exist, this is a contradiction.
A: write:
$$
f(\alpha) = \Re(f(\alpha))+i\Im(f(\alpha))
$$
when $f$ has a zero, both real and imaginary parts must be zero, ie $\Re(f(\alpha))=0$ and $\Im(f(\alpha))=0$
now notice that if $f$ has real coefficients then
$$
f(\bar \alpha)=\overline {f(\alpha)} =\Re(f(\alpha)) -i\Im(f(\alpha))
$$
