IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q?

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Discussion Overview

The discussion revolves around the irreducibility of the polynomial f(x) = x^4 + x^3 - x^2 - 2x - 2 over the rational numbers Q. Participants explore methods for determining irreducibility, including checking for linear factors and considering polynomial factorizations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant, Niall, claims to have shown that f(x) has no linear factors by testing roots ±1 and ±2, concluding that they are not roots.
  • Niall proposes a factorization approach using F(x) = (x^2 + ax + b)(x^2 + cx + d) but expresses difficulty in finding suitable values for a, b, c, and d.
  • Another participant notes that there are limited solutions to the equation bd = -2, which may affect the factorization.
  • A later reply from Niall indicates that they have concluded f(x) is reducible, although the reasoning is not detailed.
  • Another participant asks about checking for linear factors in the polynomial F(x) = x^3 - 5 by testing roots ±5 and ±1, seeking clarification on whether this is sufficient.
  • One participant explains that the only limit on polynomial factorization is that the degrees of the factors must sum to the degree of the polynomial, noting that cubics can factor into a linear and a quadratic polynomial.
  • There is a discussion about finding the multiplicative inverse of an element in a field defined by cuberoot(5), with participants exploring methods for solving the equation for the inverse.
  • One participant suggests using linear algebra to compute the minimal polynomial, while another expresses uncertainty about the method due to it not being covered in their course.
  • Participants discuss alternative approaches to finding the inverse, including writing down an indeterminate field element and solving the corresponding equation.

Areas of Agreement / Disagreement

Participants express differing views on the irreducibility of f(x), with Niall concluding it is reducible while others have not confirmed this. The discussion on the polynomial F(x) = x^3 - 5 also indicates uncertainty about the sufficiency of testing specific roots. Overall, multiple competing views remain on the methods and conclusions regarding polynomial irreducibility.

Contextual Notes

Participants' discussions include various assumptions about polynomial factorization and the nature of roots, which may not be fully resolved. The conversation also touches on methods not yet covered in the participants' coursework, indicating a potential gap in knowledge or technique.

Niall101
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IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

Attempt:

Ive shown that this has no linear factors as +-1 +-2 are not roots.

Then I let F(x)=(x^2+ax+b)(x^2+cx+d)

And I end up with
a +c=1
d+ac+b=-1
ad+bc=-2
bd=-2

But i can seam to show why no a b c d exist (or Exist)

Is there an easier approach?

I have an exam in 12 hours so I hope soeone can help!

Thanks in advance!

Niall
 
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There aren't many solutions to bd = -2...
 


hey thanks for your reply I managed to get it. It is reducable. :)
 


Sorry for another question. If i have F(x)=x^3-5 Is it sufficiant to show it have no linear factors by plugging in +-5 +-1 and showing not equal to zero? I was thinking because the only factors could be linear or of degree 3. Thanks again.
 


The only limit on the ways that polynomials can factor is that the degrees of the factors add up to the degree of the polynomial. In particular, there are cubics that factor into a linear and a quadratic.
 


Over Q tho are +-5 +-1 The only possible roots of x^3-5?

cool thanks so much!

Let A = cuberoot(5),
and let
F = a + bA + cA in R
a; b; c in Q
Find the multiplicative inverse of the element 2 +A +A^2 in the field F.

For this I Know There is an element in the field which multiplied by this give the Multiplicitive identity. Is there a way of doing this by taking powers of 2 +A +A^2?

I really appreciate the help!

Edit: I like your site with the comics made me smile
 
Last edited:


The method you suggest sounds like you're planning on using linear algebra to compute the minimal polynomial of your field element. That will work, but I doubt it's the most efficient way to do it by hand.
 


Yeah i think that's what i was trying to do. Have to do this in an exam later. But We haven't covered minimal polynomials yet so he probably wants a different method. Could you point me in the right direction?
 


What about the direct approach? Write down an indeterminate field element, and solve the equation that says your field element times the general field element equals 1.
 
  • #10


great thanks I got it now. Thanks for your help!
 

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