Can you factor the following two polynomials?

  • Context: MHB 
  • Thread starter Thread starter DrLiangMath
  • Start date Start date
  • Tags Tags
    Polynomials
Click For Summary

Discussion Overview

The discussion centers around the factorization of two polynomials, specifically \(x^4 + 4\) and \(x^4 + 3y^2 + 2y^4 + 4x^2 + 5y^2 + 3\), over the integers. Participants explore various methods and approaches to factor these expressions, including potential solutions and alternative perspectives.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests a manipulation of \(x^4 + 4\) by rewriting it as \(x^4 + 4x^2 - 4x^2 + 4\) to facilitate factorization.
  • Another participant argues that \(x^4 + 4\) cannot be factored over the integers or real numbers, asserting that it can only be expressed in terms of linear factors.
  • A different participant provides an alternative approach by discussing the imaginary solutions of \(x^4 + 4 = 0\), noting that the solutions form conjugate pairs and lead to specific quadratic factors.

Areas of Agreement / Disagreement

Participants express disagreement regarding the factorability of \(x^4 + 4\). While some propose methods for factorization, others maintain that it cannot be factored over the integers or real numbers, indicating a lack of consensus.

Contextual Notes

The discussion includes various assumptions about the nature of factorization and the types of numbers considered (integers vs. reals vs. complex). The mathematical steps involved in the proposed factorizations are not fully resolved, leaving some ambiguity in the approaches discussed.

DrLiangMath
Messages
21
Reaction score
0
Can you factor the following polynomials over integers?

[math] x^4 + 4[/math]

[math] x^4 + 3 ~x^2~y^2 + 2 ~y^4 + 4 ~x^2 + 5 ~y^2 + 3[/math]

If not, you can get help from the following free math tutoring YouTube channel "Math Tutoring by Dr. Liang"

https://www.youtube.com/channel/UCWvb3TYCbleZjfzz8HEDcQQ
 
Mathematics news on Phys.org
Hint: [math]x^4 +4 = x^4 + 4x^2 - 4x^2 + 4[/math]

-Dan
 
(up)
 
I don't see how that helps. $x^4+ 4$ obviously cannot be factored over the integer, or even over the real numbers.
 
HallsofIvy said:
I don't see how that helps. $x^4+ 4$ obviously cannot be factored over the integer, or even over the real numbers.
As linear factors, yes. But:
[math]x^4 + 4 = x^4 + (4 x^2 - 4 x^2) + 4 = (x^4 + 4 x^2 + 4) - 4 x^2 = (x^2 + 2)^2 - 4 x^2 = (x^2 + 2 + 2 x)(x^2 + 2 - 2 x)[/math]

-Dan
 
Just to mention an alternative approach, $$x^4+4=0$$ has 4 imaginary solutions that form conjugate pairs.
The solutions are $$x=\pm 1\pm i$$

If we then put the conjugate pairs together, we get $$(x-(1+i))(x-(1-i))=x^2-2x+2$$ and $$(x-(-1+i))(x-(-1-i))=x^2+2x+2$$ just like topsquark found.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
9
Views
9K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K