$x^3+nx+2$ is irreducible over ${\mathbb Z}$ for $n\neq 1,\ -3\ -5$

Asked Aug 04 '14 at 19:26

Active Aug 04 '14 at 19:44

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I want to show that $$x^3+nx +2 $$ is irreducible over ${\bf Z}$ for $n\neq 1,\ -3\ -5$

By Eisenstein, if $n$ is even then it is irreducible. How can we solve odd case ?

[add] Note that factorization is one of these : $$ (x\pm 2)(x^2+ax\pm 1),\ (x\pm 1)(x^2+ax\pm 2) $$

Since $x^2$-term do not exist we can determine $a$. So $n$

edited Aug 04 '14 at 19:44

asked Aug 04 '14 at 19:26

HK Lee

1

Since this is a cubic it has at most three real roots. You should look at the formula (or methods for solving) cubics and see what you can say about $n$ when you apply them. – postmortes Aug 04 '14 at 19:29
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Hint: Any integer root must be one of $\pm 1,\pm 2$. – Thomas Andrews Aug 04 '14 at 19:32
I will add solution from helps. – HK Lee Aug 04 '14 at 19:34
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Compute $f(x)$ for $x=-2,-1,1,2$. – Hagen von Eitzen Aug 04 '14 at 21:25

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