Complex zeros of polynomial with no real zeros

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    Complex Polynomial
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Discussion Overview

The discussion revolves around finding the complex zeros of the polynomial x^6+x^4+x^3+x^2+1. Participants explore methods of factoring the polynomial and seek to identify the complex solutions, noting the absence of real solutions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the polynomial x^6+x^4+x^3+x^2+1 and notes it has no real solutions, proposing an initial factorization into (x^2-x+1)(x^4+x^3+x^2+x+1).
  • Another participant acknowledges the factorization and suggests considering what the fourth-degree polynomial is a factor of.
  • One participant expresses certainty that x^4+x^3+x^2+x+1 also lacks real solutions and contemplates a further factorization approach using a quadratic form.
  • Another participant hints at a connection to a fifth-order polynomial, prompting further exploration.
  • A later reply identifies that x^4+x^3+x^2+x+1 can be expressed as (x^5 - 1)/(x - 1) using the geometric sum formula, suggesting this could lead to finding the four complex roots.

Areas of Agreement / Disagreement

Participants generally agree that the original polynomial and the fourth-degree polynomial do not have real solutions. However, there is no consensus on the methods of factorization or the exact nature of the complex solutions.

Contextual Notes

Some participants express confusion regarding the factorization steps and the connections between the polynomials, indicating potential gaps in understanding or missing assumptions about the relationships between the expressions.

bobus
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I have to find the complex zeros of the following polynomial:
x^6+x^4+x^3+x^2+1
This evidently doesn't have any real solution so I tried to facto it with long division and by guessing I came up with:
(x^2-x+1)(x^4+x^3+x^2+x+1)

How can I factor the 4th degree polynomial now? Or how can I factor the original polynomial from beginning?

Thanks!
 
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welcome to pf!

hi bobus! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
bobus said:
This evidently doesn't have any real solution so I tried to facto it with long division and by guessing I came up with:
(x^2-x+1)(x^4+x^3+x^2+x+1)

How can I factor the 4th degree polynomial now?

that was very good factoring! :biggrin:

ok, you have x4 + x3 + x2 + x + 1 …

hmm :confused: … let's do a bit of lateral thinking :rolleyes:

what is it a factor of ? :wink:
 
Ok I'm pretty sure that not even X^4+x^3+x^2+x+1 has any real solution. So I guess there are 4 more complex solutions. So I was thinking about (ax^2+bx+c)(dx^2+ex+f) and trying to come up with some values for a, b, c, d, e and f but it seems to don't work...
Any idea??
 
no, you're looking down, look up! :rolleyes:

what fifth-order expression is it a factor of? :smile:
 
x^5+x^4+x^3+x^2+x
How does this help me?
 
now subtract something :wink:
 
hmmm? I guess I'm a little confused :(
 
x^4 + x^3 +x^2 + x + 1 = (x^5 - 1)/(x - 1) (Geometric sum formula.)

You should be able to get the four roots you need.
 

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