Irreducibility of f(x)=x^4+3x^3-9x^2+7x+27 – Abstract Algebra

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Question at hand is:

Is $x^4+3x^3-9x^2+7x+27$ irreducible in $\Bbb Q$ and/or $\Bbb Z$.

This is for an exam, reasoning is trivial, but no calculators in hand. Clearly, if there is a rational root, they are integers by Rational Root theorem and since $f$ is monic.

I am aware of

  1. Rational root theorem, which narrows down the options to $\pm1,\pm3,\pm9,\pm27$, and clearly, no roots.

  2. Eisenstein's Irreducibility Criteria, not helping here, thanks to $x$'s coefficient $7$

  3. Cohn's Irreducibility test: $12197$ is a prime, too large a number to prove that its a prime by hand.

  4. Descartes Rule of signs: at most 2 (or 0) positive/negative roots. Close enough.

None of which are helping me in any way since I can't use a calculator.

These are the solutions I tried:

  1. Alpha says all roots are complex. Made me search if there's some way to determine if all roots are complex, reaching nowhere.

  2. Check if there are any easy prime generation functions like Euler's, and if lucky 12197 falls in that list, the best I got is Euler's, $n^2+n+41, 1\le n<40$, and biggest such is $1601$, not helping.

Are there any better ways to determine if this polynomial is irreducible over $\Bbb Q$, without using calculators?

Best Answer

If you take the polynomial modulo 2 you get $x^4+x^3+x^2+x+1$. It doesn't have a root, so if it is decomposable, it is a product of two irreducible polynomials of degree 2. The only polynomial of degree 2 which doesn't have a root is $x^2+x+1$ and its square is $x^4+x^2+1$. It follows that your polynomial is irreducible modulo $2$ and therefore irreducible in over $\mathbb{Z}$.

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