As I was reading this question, I saw Ethan's answer. However, perhaps this is very obvious, but why does the degree of the polynomial be at most $2$? I get that the polynomial must be irreducible but does that force the degree to be at most $2$?
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Let $P(X) \in \mathbb R[X]$ be the minimal polynomial of $\alpha$.
If $\alpha \in \mathbb R$ then $X-\alpha |P(X)$. Since $X-\alpha \in \mathbb R[X]$ has $\alpha$ as a root, it follows that $P(X)=X-\alpha$.
if $\alpha \notin \mathbb R$ then $X-\alpha |P(X)$. Moreover, $P(\bar{\alpha})=0$ means $X-\bar{\alpha} |P(X)$. From here, since $X-\alpha$ and $X-\bar{\alpha}$ are relatively prime we get that $(X-\alpha)(X-\bar{\alpha})|P(X)$.
Now, it is easy to see that $(X-\alpha)(X-\bar{\alpha}) \in \mathbb R[X]$ from where it follows that $$P(X)= (X-\alpha)(X-\bar{\alpha}) $$
N. S.
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We know that, over $\mathbb{C}$, we can factor any polynomial entirely into linear terms (this is the fundamental theorem of algebra). Moreover, one can show that whenever $f$ has real coefficients, then $z$ and $\overline{z}$ (the complex conjugate) must both be roots of $f$.
Now, given $f \in \mathbb{R}[x]$, we factor it into linear terms over $\mathbb{C}$. We look at each root $\alpha$ in turn. If it is real, then $(x-\alpha)$ is a factor of $f$ over $\mathbb{R}$ as well. If it has a complex part, then $\overline{\alpha}$ must be a root too, and then $(x - \alpha)(x - \overline{\alpha})$ is a quadratic with real coefficients.
Since these are the only two cases which can occur, we have factored $f$ into linear and quadratic parts.
I hope this helps ^_^
Chris Grossack
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