Show that $3x^5-4x^3-6x^2+6$ is irreducible over $\mathbb{Q}[x]$

Question

Can’t seem to apply Eisenstein’s criterion. It doesn’t make sense to reduce modulo 2 since the constant term disappears… What are some other possible strategies to apply here?

Answer 1

I do not have enough reputation to comment, and so I will leave an answer.

V.Ch. is correct, the Eisenstein criterion will give you irreducibility. We apply the criterion with the prime $2$ because it divides every coefficient and doesn't divide the constant coefficient more than once.

From your question it seems that perhaps you do not know the Eisenstein criterion. In this case, another method is to reduce modulo $5$. If $\alpha$ is a root of $3x^5 - 4x^3 - 6x^2 + 6 \equiv 3x^5 + x^3 - x^2 + 1 \; (\textrm{mod } 5)$ then since $\alpha^5 \equiv \alpha \; (\textrm{mod } 5)$, we have $\alpha^3 - \alpha^2 + 3\alpha + 1 \equiv 0 \; (\textrm{mod } 5)$. You can verify that the polynomial $x^3 - x^2 + 3x + 1$ has no roots by plugging in integers modulo $5$.