Irreducibility of $8x^3 -6x +1$ in $\mathbb{Q}[x]$

Question

I want to use Gauss' Theorem for that. So I have to show that $f(x) = 8x^3 -6x +1$ is irreducible in $\mathbb{Z}[x]$ which is equivalent to it not having any roots in $\mathbb{Z}$. Since any integral root would have to divide $1$, one only has to check $\pm 1$. But $f(\pm 1) \ne 0$. So $f$ is irreducible in $\mathbb{Z}[x]$ and then, by Gauss, irreducible in $\mathbb{Q}[x]$.

Is that correct?

Answer 1

As almagest pointed out in comments, your argument would seemingly apply to the reducible polynomial $4x^2-4x+1$ as well. For non-monic polynomials, you need to check for rational roots, whose denominators divide the lead coefficient. So in your case, you would need to evaluate $f(x)$ for $x=\pm1$, $\pm1/2$, $\pm1/4$, and $\pm1/8$.

For a different approach, note that

$$f(x+1)=8(x+1)^3-6(x+1)+1=8x^3+24x^2+18x+3$$

This allows you to use Eisenstein's criterion (with the prime $p=3$) to show the polynomial is irreducible.

Answer 2

Set $u=2x$ and rewrite the polynomial as $u^3-3u+1$. The original polynomial is reducible over $\mathbf Q$ if and only if this polynomial is. Having degree $3$, it is reducible if and only if it has a rational root. Now the rational root theorem says this root can be only $\pm 1$. Furthermore, this root must be positive, hence you only have to test $1$, which is not a root. Hence the original polynomial is irreducible.