# Is x^3 + 2x + 4 irreducible?

• Apr 19th 2009, 04:56 AM
Louise
Is x^3 + 2x + 4 irreducible?
Hi, could you help me with the following question

Is x^3 + 2x + 4 irreducible in the ring Q[x]?

(Q=rationals)

Thanks.
• Apr 19th 2009, 05:18 AM
peteryellow
yes, it is you can try to prove it as following.

By Rational root theorem it is easily seen that it has no rational root. then try to show that
$x^3+2*x+4 = (x^2+bx+c)(x^2+dx+e)$ and then get a contardiction.
• Apr 19th 2009, 05:43 AM
Louise
Thanks, is it also right that x^4 -4x^3+17x-17 is not irreducible by Eisenstein's criterion and reverse Eisenstein's criterion?
• Apr 19th 2009, 05:54 AM
peteryellow
This is irreducible, you can try to prove as I mentioned before. But you cannot use Eisenstein directly. You can try to find f(x+1), f(x-1) and so on and then see if you can use Eisenstein.
• Apr 19th 2009, 06:38 AM
Louise
Thanks again, i found f(x+1) =x^4 +6x^2+9x-3 and so it's irreducible by Eisenstein's criterion with p=3.

Do you have any idea on how to do the following;

Let F be a field. Show that any element in the field F(x) can be written uniquely as f(x)/g(x) with f(x) and g(x) coprime polynomials and g(x) monic.

The hint suggests looking at the proof of any rational number can be written as m/n with m and n comprime and n positive.
• Apr 19th 2009, 10:29 AM
Gamma
Regarding the first polynomial

Just a quick note, showing a polynomial of degree 2 or 3 has no roots in a particular field is enoughto say it is irreducible over that field. In particular you ruled them all out over the rationals with rational root test, so you are done. No need to worry about another possible factorization.
• Apr 19th 2009, 11:57 AM
peteryellow
Quote:

Originally Posted by peteryellow
yes, it is you can try to prove it as following.

By Rational root theorem it is easily seen that it has no rational root. then try to show that
$x^3+2*x+4 = (x^2+bx+c)(x^2+dx+e)$ and then get a contardiction.

I am vey sorry it is my mistake this should have been
$x^3+2*x+4 = (x^2+bx+c)(x+e)$ and then get a contradiction.

And it is enough to check if the polynomial has rational roots or not since deg(f)=3. Sorry again.
• Apr 19th 2009, 12:05 PM
Gamma
Right
Oh, don't apologize, to be honest I didn't even notice the degrees didn't add up.

I hope that you mean your contradiction will be that e cannot be in $\mathbb{Q}$ as you have shown there are not roots, and therefore cannot have a linear term, thus is necessarily irreducible if it has no linear term factors because it has degree 3. I just don't want you to multiply anything out and start solving systems of equations or anything, lol. Just want to make sure we are on the same page.