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## Irreductible polynomial with paired roots

Suppose $f \in \Bbb{Q}[X]$ is an irreductible polynomial which has a complex root $a$ such that $-a$ is also a root for $f$. Prove that for any other root $b$ of $f$, $-b$ is also a root for $f$.

Proof: We may assume WLOG that the leading coefficient of $f$ is $1$. Then $f$ is the minimal polynomial of $a$ and $-a$. Of course $a \neq 0$, since $f$ is irreductible. Moreover, since $\pm f(-x)$ is the minimal polynomial for $-a$ we get that $\pm f(-x)=f(x)$. Because $0$ is not a root for $f$ we must have $f(x)=f(-x)$, which means that for any root $b$ of $f$, we have that $-b$ is also a root of $f$.