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$p(x) = x^7+20x-6$.

If someone can give me the idea I would appreciate.

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    Are you familiar with Eisenstein's criterion?2017-02-21

1 Answers 1

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$2$ is a prime that divides both $6$ and $20$.

$2$ is a prime that does not divide $1$.

$4=2^2$ does not divide $6$.

Thus by Eisenstein's criterion $p(x)=1x^7 + 20x-6~$ is irreducible over $\mathbb{Q}[x]$.

Since $\gcd(1,6,20)=1~$ and $~p(x)$ is irreducible over $\mathbb{Q}[x]$, then $p(x)$ is also irreducible over $\mathbb{Z}[x]$.

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    Your first answer (Eisenstein) is good, but your second (concerning roots) is faulty. There are many polynomials that are reducible over $\mathbb{Q}$ but have no rational roots: $(x^2 + 1)^2$, for example.2017-02-21
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    Good point. Rational root theorem only guarantees no linear factors over $\mathbb{Q}$. I will edit.2017-02-21