Factoring a quartic polynomial over the reals

  • Context: Undergrad 
  • Thread starter Thread starter Mr Davis 97
  • Start date Start date
  • Tags Tags
    Factoring Polynomial
Click For Summary

Discussion Overview

The discussion revolves around the factorization of the quartic polynomial ##x^4 + 1## over the reals. Participants explore the conditions under which such a polynomial can be factored, particularly in the absence of real roots.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions how to determine if the polynomial ##x^4 + 1## is factorable over the reals, noting its lack of real roots and suggesting it could only factor into two quadratic polynomials.
  • Another participant points out that since all coefficients are real, any complex roots must occur in conjugate pairs, proposing a method to express the polynomial as a product of two quadratics based on this structure.
  • A third participant references a previous discussion on the same problem, indicating that this is a recurring topic of interest.
  • Another participant asserts that every non-trivial polynomial with real coefficients can be factored into linear and quadratic terms, citing the relationship between complex roots and their conjugates.

Areas of Agreement / Disagreement

Participants express differing views on the factorization of the polynomial, with some focusing on the implications of complex roots and others emphasizing the general properties of polynomials with real coefficients. The discussion does not reach a consensus on the specific factorization of ##x^4 + 1##.

Contextual Notes

The discussion does not resolve the specific factorization steps or the implications of the absence of real roots on the factorization process.

Mr Davis 97
Messages
1,461
Reaction score
44
I have the simple quartic polynomial ##x^4+1##. How in general do I determine whether this is factorable over the reals or not? Since it has no real roots, it could only factor into two quadratic polynomials, but I am not sure what I can do to determine whether this is possible or not.
 
Mathematics news on Phys.org
All coefficients of the polynomial are real. Thus any roots with non-zero imaginary components come in conjugate pairs... Take advantage of this structure to split into 2 quadratics that you can multiply.

i.e.

for complex ##\lambda## (with non-zero imaginary component) we have

##(x - \lambda)(x - \bar{\lambda}) = x^2 - 2\Big(\text{real}(\lambda)\Big)x + \vert \lambda \vert^2 ##
 
You can factor every (non-trivial) polynomial with only real coefficients into linear and quadratic terms.
This is a direct consequence of the full factorization in the complex numbers and the conjugate pairs of complex roots.
 
  • Like
Likes   Reactions: Mr Davis 97

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 18 ·
Replies
18
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K