Hey!!

Let $f = x^4−2x^2−1 \in \mathbb{Q}[x]$.

We have that $f(x+1)=(x+1)^4-2(x+1)^2-1=x^4+4x^3+6x^2+4x+1-2(x^2+2x+1)-1=x^4+4x^3+4x^2-2$

We have that $p=2$ divides all the coefficients $4,4,-2$ and $p^2=4$ does not divide the constant term $-2$.

So, the polynomial $f(x+1)$ is Eisenstein. This means that $f(x+1)$ is irreducible, or not?

Knowing that, can we conclude that $f(x)$ is irreducible as follows:

Suppose that $f(x)$ is reducible, then there are non-constant polynomials $g(x),h(x)\in \mathbb{Q}[x]$ such that $f(x)=g(x)h(x)$. Then $f(x+1)=g(x+1)h(x+1)$. Since the polynomials are non-constant that would mean that $f(x+1)$ is reducible, a contradiction.

So, $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Is this correct?