Using Eisenstein criterion show that $x^3+x^2-2x-1$ is irreducible? [duplicate]

You can use the reducite criterium.

I.e., consider the polynomial over the field $\mathbb{F}_2$.I.e., we get $X^3 + X^2 +1$. If this reduced polynomial is irreducible over $\mathbb{F}_2$, it is also irreducible over $\mathbb{Q}$ (this theorem is known as the reducite criterium).

This polynomial is irreducible over $\mathbb{F}_2$ (it has no roots, this is easily seen).

Hence, by the reducite criterium it follows that $X^3 + X^2 - 2X -1$ is irreducible in $\mathbb{Q}[X]$.

We don't need to pull the Eisenstein shift out of a hat like magic. I explain how to find it below.

Hint $\, \bmod\color{#c00}7\!:\,\ f(x) \equiv (x\!-\!2)^{3} \ $ is a prime power.

So Eisenstein works on $\,g(x) = f(x\!+\!2) \equiv x^3\ $ by $\,g(0) = f(2)\equiv 7\not\equiv 0\pmod{\!49}$

Remark $\ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $\,p,\,$ powers of a prime, viz. $\,\equiv x^n,\,$ and products of primes always factor uniquely. The same works for its shift $\,(x-c)^n,\,$ so we seek primes $\,p\,$ such that, mod $\,p,\,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $\,p\,$ that can yield such powers are those dividing the discriminant (here by Alpha = $\,\color{#c00}7^2).\,$ Indeed, if $\,f\equiv a (x-c)^n,\,\ n> 1\,$ then $\,f\,$ and $\,f'\,$ have a common root $\,x\equiv c,\,$ hence their resultant $\, R(f,f')\equiv 0.\,$ But this is, up to sign, the discriminant of $\,f\,$ (presumed monic).