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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.

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

2 Answers 2

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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.

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    $\begingroup$ 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? $\endgroup$
    – Mike
    Mar 29, 2021 at 12:21
  • $\begingroup$ Yes you are right, I forgot the degree while writing my solution $\endgroup$
    – 1123581321
    Mar 29, 2021 at 12:25
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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 $\Bbb Q$.

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

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