Is $x^8+6x^7-15x^4-9x+12$ irreducible over $\mathbb{Q}$?

Question

As the title states, I'm trying to figure out whether $x^8+6x^7-15x^4-9x+12$ is irreducible in $\mathbb{Q}[X]$. So far I have been introduced to the Rational Root Test/Theorem which has given me the following potential candidates for a root: $\{\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\}$, none of which are actually a root.

The fact this is over $\mathbb{Q}$ makes me think maybe there is a rational solution, but I'm not sure how to figure it out. I've done some Googling already and found a lot of things I hadn't been introduced to yet (Eisenstein, for example).

Any help would be appreciated.

Answer 1

I suppose that Eisenstein's criterion with $p=3$ aplies here immediately.

For a more basic attempt that requires some effort:

Let $p(x)=x^8+6x^7-15x^4-9x+12$.

Suppose that $p(x)=(ax^2+bx+c)(dx^6+ex^5+fx^4+gx^3+hx^2+ix+j)$. By the distributive law you will have a system of linear equations that you will shot that it has no rational roots.

Then suppose that $p(x)=(ax^3+bx^2+cx+d)(ex^5+fx^4+gx^3+hx^2+ix+j)$ and do the same.

Then suppose that $p(x)=(ax^4+bx^3+cx^2+dx+e)(fx^4+gx^3+hx^2+ix+j)$ and do the same.

You have already checked that $p(x)$ has no rational roots hence these are all the cases you have to consider.

Answer 2

Eisenstein is the best solution. After this, one could apply the modular criterion with $p=17$. The proof that the polynomial is irreducible over the finite field $\Bbb F_{17}$ also uses that we can write down a system of linear equations from some assumed decomposition. However, the system is easier to solve over $\Bbb F_{17}$ than over $\Bbb Q$.