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I have this polynomial:

$f(x)=x^4+x^3-4x^2-5x-5$. How can I find out if this polynomial is irreducible over the field $\mathbb{Q}$ of rational numbers? I know about mod $p$ irreducibility test but it fails in this case. In general how do you find out if a polynomial is irreducible or prove that it is reducible?

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    $\begingroup$ Do you know about Eisenstein's criterion? $\endgroup$ Commented Dec 4, 2014 at 5:52
  • $\begingroup$ @AWertheim I have read about that theorem but I was unable to use it in this case. It says I have to find a prime p which does not divide $a_4$ (1), divides $a_3, a_2, a_1$ and $p^2$ does not divide $a_0$. How can I apply that theorem here? $\endgroup$
    – khajvah
    Commented Dec 4, 2014 at 5:57
  • $\begingroup$ Since this one is reducible, Eisenstein won't help. $\endgroup$ Commented Dec 4, 2014 at 5:57
  • $\begingroup$ @khajvah: as Robert Israel notes, your polynomial is reducible. But it is a general technique for showing a polynomial is irreducible over $\mathbb{Q}$. Note that you need not always consider $f(x)$ - you can apply Eisenstein's criterion equally well to $f(x+k), k\in \mathbb{Z}$, which sometimes yields the desired conclusion. $\endgroup$ Commented Dec 4, 2014 at 6:12
  • $\begingroup$ @AWertheim So in conclusion, we might be able to prove that a polynomial is irreducible but it is impossible to prove that it is reducible without actually finding the factors ? $\endgroup$
    – khajvah
    Commented Dec 4, 2014 at 6:18

2 Answers 2

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Being a quartic, this polynomial is reducible if and only if it has a linear or quadratic factor with integer coefficients.

A linear factor implies an integer root. The only possible roots are $1,-1,5,-5$. Checking, none of them works. Note that for $x=\pm5$ you don't need to do all the calculations since it is easy to see that then $$x^4+x^3-4x^2-5x-5\equiv-5\pmod{25}$$ and so LHS${}\ne0$.

So try $$x^4+x^3-4x^2-5x-5=(x^2+ax+b)(x^2+cx+d)\ .$$ Expanding and equating coefficients, $$a+c=1\ ,\quad ac+b+d=-4\ ,\quad ad+bc=-5\ ,\quad bd=-5\ .$$ The last equation gives four possibilities for $b$ and $d$, (in fact only two, as we may assume by symmetry that $b=\pm1$ and $d=\pm5$) and then it's easy to find the other coefficients:

  • $b=-1$, $d=5$, $a=(b+5)/(b-d)=-\frac23$, didn't work;
  • $b=1$, $d=-5$, $a=(b+5)/(b-d)=1$, $c=1-a=0$,

and we then check that $$x^4+x^3-4x^2-5x-5=(x^2+x+1)(x^2-5)\ .$$

Note that if our second attempt above had failed, this would be enough to conclude that the polynomial is irreducible.

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There are algorithms, e.g. Kronecker's, that will find factors over $\mathbb Q$ if they exist. These are implemented in the various Computer Algebra systems, so in practice you would just ask Maple or Wolfram Alpha.

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  • $\begingroup$ I got this in my test and was unable to solve. I doubt my professor wanted us to google the result. Is mod p irreducibility test reliable? $\endgroup$
    – khajvah
    Commented Dec 4, 2014 at 6:00

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