Prove $x^3-3x+4$ is irreducible in $\mathbb{Q}[x]$ I want to prove $x^3-3x+4$ is irreducible in $\mathbb{Q}[x]$. Eisenstein's criterion doesn't apply here, so I think the simplest method would be to use the Rational Roots Test, right?
If I can use the rational roots test here, then since it is monic I simply check factors of the constant term:
\begin{align*}
(1)^3-3(1)+4&=2 \\
(2)^3-3(2)+4&=6 \\
(4)^3-3(4)+4&=56 \\
(-1)^3-3(-1)+4&=6 \\
(-2)^3-3(-2)+4&=2 \\
(-4)^3-3(-4)+4&=-48
\end{align*}
Therefore if $x^3-3x+4$ is reducible, it would have a degree 1 monomial factor $(x-a)$ for one of the $a=1,2,4,-1,-2,-4$ I tested above. However since none of these $a$ are roots, so then it does not have a degree 1 factor and is therefore irreducible. 
Unfortunately I can't find any reference to the Rational Roots Theorem in Artin's Algebra text (2nd edition), not even in the index (surprising!). So I'm referring to Wikipedia, and it does not specifically say -- does this test hold in $\mathbb{Q}[x]$? Hopefully it is applicable here and I have used it correctly...
Thanks!

Try showing that it has no roots in $\mathbb{Q}$, that is a good criteria for irreducibility.
Another trick, is that if it is irreducible, it is irreducible for all $x\in \mathbb{R}$ so let $x=\frac{1}{y}$ then try eisenstein on the new polynomial.
