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 $Q$ 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?

  • $\begingroup$ Do you know about Eisenstein's criterion? $\endgroup$ – Alex Wertheim Dec 4 '14 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 Dec 4 '14 at 5:57
  • $\begingroup$ Since this one is reducible, Eisenstein won't help. $\endgroup$ – Robert Israel Dec 4 '14 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. $\endgroup$ – Alex Wertheim Dec 4 '14 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 Dec 4 '14 at 6:18

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.

| cite | improve this answer | |

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.

| cite | improve this answer | |
  • $\begingroup$ 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 Dec 4 '14 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.