Determining whether $2x ^ 3 - x^2 + 1$ irreducible in $\mathbb{Q}[x]$? I'm checking if $f(x) = 2x ^ 3 - x^2 + 1$ is irreducible in $\mathbb{Q}[x]$.
We have a polynomial of degree three so if it has no roots it is irreducible. But as we are dealing with $\mathbb{Q}[x]$ it is not easy to check the roots, apart from the trivial options such as $0, 1$.
Eisenstein criterion's can't be applied. I tried subbing in some values such as $(x + 1)$ but found that Eisenstein still was not applicable.
So what other options do I have?
 A: Let $f(x) = 2x^3 - x^2 + 1$.  Note that, since $\deg(f) \leq 3$, it is irreducible over the rationals $\iff$ $f$ has no rational roots.  From here, we can apply the rational root theorem.  This tells us that the set of possible roots is $\displaystyle \left\{ \pm 1, \ \pm \frac{1}{2} \right\}$, and one can check that these all fail to satisfy $f(\alpha) = 0$.  Therefore, it will not split into a product of two smaller-degree polynomials with coefficients in $\mathbb{Q}$.

Alternative method: It is a theorem that if a polynomial in $\mathbb{Z}[x]$ is irreducible modulo a prime $p$, then it is also irreducible over $\mathbb{Q}$.  Notice that $f$ reduces as $f(x) = 2x^3 + 2x^2 + 1 \pmod{3}$.  As above, it suffices to show that $f$ has no roots in $\mathbb{Z}_3$, which is easier given that we only have $3$ things to check.  Evaluated modulo $3$, we have $f(0) \equiv 1$, $ \ \ f(1) \equiv 2$, and $f(2) \equiv 1$.  We see that $f$ has no roots modulo $3$, so it is irreducible over $\mathbb{Z}_3$, and thus it is irreducible over $\mathbb{Q}$.
