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!
 A: The rational roots test does indeed work in $\mathbb{Q}[x]$.  If a polynomial has degree less than or equal to $3$, then it is irreducible over the rationals if and only if it has a rational root (Prove why).  
Wikipedia states that the rational roots test holds for polynomials with integer coefficients.  Note that any polynomial in $\mathbb{Q}[x]$ can be transformed into a polynomial in $\mathbb{Z}[x]$ by simply multiplying through by a constant, which will not affect irreducibility.  In other words, $f(x)$ is irreducible if and only if $c \cdot f(x)$ is irreducible for any nonzero $c \in \mathbb{Q}$.
A: If one really is in an Eisenstein mood (and one should not be), let $y=x+1$. 
Then $x^3-3x+4=(y-1)^3-3(y-1)+4=y^3-3y^2+6$. and we can use the Irreducibility Criterion with $p=3$.  
A: 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.
A: 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 $mod 5$.
