Are These Polynomials Irreducible Over Q?

gtfitzpatrick
Messages
372
Reaction score
0

Homework Statement



determine whether the following polynomials are irreducible over Q,

i)f(x) = x^5+25x^4+15x^2+20
ii)f(x) = x^3+2x^2+3x+5
iii)f(x) = x^3+4x^2+3x+2
iv)f(x) = x^4+x^3+x^2+x+1


Homework Equations





The Attempt at a Solution



By eisensteins criterion let f(x) = a_n x^n+a_{n-1} x^{n-1}+...a_0
if there exists p, a prime such that p does not divide a_n , p divides a_{n-1},...,p divides a_0 and p^2 does not divide a_0 then f(x) is irreducible over Q

So i) if p=5 => it is irreducible over Q

but not sure how to go about the others...
 
Physics news on Phys.org
I think ii) and iii) are both reducible but iv) is irreducible as4-1 =3, a prime
 
Since ii and iii are cubics, if they were reducible, they would have to have at least one linear factor, so at least one rational root. By the "rational root theorem", any rational root to ii would have to be \pm 1 or \pm 5. Check whether any of those is a root. Similarly, any rational root to iii would have to be \pm 1 or \pm 2.

(Clearly neither ii nor iii has a positive root so you really only have two values to check in each problem.)
 
For (iv), the roots are easy to compute since

(x^5-1)=(x-1)(x^4+x^3+x^2+x+1)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top