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

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.

• Yet the answer is immediate by Eisenstein criterion. Mar 29, 2021 at 12:12
• By the rational roots theorem, this polynomial is irreducible over $\mathbb{Q}$. You have checked all possible rational roots. Mar 29, 2021 at 12:13
• @ncmathsadist So does that mean each root of this polynomial is irrational? (I graphed it and it does have 4 roots). Mar 29, 2021 at 12:15
• @ncmathsadist: The rational roots theorem does no say it is irreducible, except for a quadratic or cubic polynomial. Mar 29, 2021 at 12:15
• For instance, $(x^2-2)(x^2-3)$ has no rational roots, but it is not irreducible in $\Bbb Q[X]$. Mar 29, 2021 at 12:18

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.

• But for your 2nd case, $p(x)$ has degree 8, not 6; why can't $p(x)$ be the product of 2 degree-4 polynomials. Or am I missing something here?
– Mike
Mar 29, 2021 at 12:21
• Yes you are right, I forgot the degree while writing my solution Mar 29, 2021 at 12:25

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 $$\Bbb Q$$.

• How did you find $17$? Trial and error? ) Mar 29, 2021 at 12:42
• @AlexeyBurdin Yes, because for some small $p$ there is an obvious root, e.g., for $p=3$ or $p=5$. Mar 29, 2021 at 12:49