I have the following polynomial $f(x)=x^4-x^2+2x+1$.
How do you suggest me check for its irreducibility over the rationals?
The possible rational roots of this polynomial are $\pm 1$ but I checked and they don't nullify the polynomial.
Eisenstein criterion doesn't seem to work. I thought also to equate similar powers in the following decomposition: $$x^4-x^2+2x+1=(x^2+ax+b)(x^2+cx+d)$$
Is my only chance is using Ferrari formula? https://en.wikipedia.org/wiki/Quartic_equation
On
A bit easier is to verify that $f$ is irreducible over $\Bbb F_5$. It has no root there, and assuming $f=(x^2+ax+b)(x^2+cx+d)$ gives quickly a contradiction by comparing the coefficients and solving the equations over $\Bbb F_5$. By the modular criterion $f$ is irreducible over $\Bbb Z$, and hence also over $\Bbb Q$ by Gauss' Lemma.
A slightly more pedestrian approach. By inspection the polynomial $f$ has exactly two real roots: $\alpha \in (-2,-1)$ and $\beta \in (-1,0)$. If $f$ factors as two quadratics over $\mathbb Q$ then it factors over $\mathbb Z$ (Gauss’ lemma). Now both real roots must come from the same factor. Given that $\alpha + \beta$ and $\alpha \beta$ are then integers, and considering the intervals in which the roots are located, this leaves no other option than $x^2+2x+1$ as possible factor. But $f$ is not divisible by $x^2 + 2x + 1$ (e.g. it has no root at $-1$) and therefore it has no quadratic factor at all.