I need to show that $x^4+4x^3+6x^2+9x+11$ is irreducible in the integers.

First, I tried to apply Eisenstein's irreducibility criterion by shifting $x$ to $x+\alpha$. However, I can't think of any shift to apply that would fit the criterion.

Next, I tried using polynomial division. If it were reducible, this polynomial would have either a linear or a quadratic factor. It has no integer roots, so I tried to divide by a factor $x^2+ax+b$ and require that the remainder be zero; however, this yields two very difficult equations, I'm not sure how to prove that they have no integer solution.

Any hints?

  • 7
    $\begingroup$ Just a thought, but considering that $(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$, maybe we can say let $y = x+1$ and get $x^4 + 4x^3 + 6x^2 + 9x + 11 = (x+1)^4 + 5x + 10 = (x+1)^4 + 5(x+1) + 5 = y^4 + 5y + 5$, and then start from there? Just thought it'd be nice to have fewer terms if possible. $\endgroup$
    – 2012ssohn
    Feb 23 '14 at 18:14
  • $\begingroup$ Hint: Consider the polynomial modulo $2$; if it is reducible, it stays reducible modulo any number too. $\endgroup$ Feb 23 '14 at 18:18
  • 3
    $\begingroup$ @2012ssohn ... Actually, Eisenstein applies to $y^4+5y+5$. Huh. $\endgroup$
    – Andrea
    Feb 23 '14 at 18:21
  • $\begingroup$ @PeterKošinár I didn't know that, that's a very useful trick! $\endgroup$
    – Andrea
    Feb 23 '14 at 18:22

To push it from unanswered queue:

  1. Notice that given polynomial is monic, hence by Rational Root Theorem, all rational solutions are integer, namely $\pm1,\pm11$. Also, Descartes Rule of Signs assures that there are no positive roots, making our list $-1,-11$, and its now easy to verify both won't work.

From comments:

  1. if a polynomial is irreducible over a finite field, it is irreducible over $\mathbb Q$, too, and check if its irreducible in $\mod 2$, as Peter Košinár pointed out.

  2. As 2012sshon pointed out, let $y=x+1$ and as OP pointed out, then use Eisenstein's criteria.


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