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!

To prove that this polynomial is irreducible you can prove that it hasn't roots in $\mathbb{Q}$. Suppose for absurd that $\frac {a}{b} $ is a root of this polynomial then we have $$ a^3-3b^2a+4b^3=0$$ with $a$ and $b\neq 0$ integers. Therefore $$ a^3=b^2(3a-4b)$$ with $$3a-4b=b^t\cdot k^{3z}$$ and $k,z,t$ integers with $t\equiv 1 \pmod 3$. We obtain that $b|3a-4b$ therefore $3a=cb$ with $c$ integer. From previous equation we obtain $$\frac {c^3\cdot b^3}{27}-cb^3+4b^3=0$$ and dividing for $b^3$ we have $$ c^3=27(c-4)$$ but this equation hasn't solution in $mod 5$.
