Factoring this thing with 6 degree

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Homework Help Overview

The discussion revolves around factoring a polynomial equation of degree six: u - u^6 - u^3 + 1 = 0. Participants explore the factorization of this polynomial and the methods involved in the process.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to identify roots and suggests potential factors, questioning the need for long division. Others discuss the validity of their factorization attempts and the possibility of shortcuts in the process.

Discussion Status

Participants are actively engaging in exploring different methods of factoring, with some guidance offered regarding the use of synthetic division and the implications of identified roots. There is recognition of the complexity of the quartic factor and its roots.

Contextual Notes

Participants note the absence of simple rules for factoring polynomials of this form and discuss the limitations of their current approaches, including the challenges posed by long division.

zeion
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Homework Statement



I need to factor this:

u - u^6 -u^3 +1 = 0


Homework Equations





The Attempt at a Solution



I know that (u +1) and (u-1) are roots.. but not sure what to do now without long division..
Do I put them into multiples like (u^3 + 1)(u^3 -1) ??
 
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(u^3 + 1)(u^3 -1) is wrong. What's the problem with long division?
 
It takes too long.. I was just wondering if there was simply rule I could follow when it's in this kind of form?
 
zeion said:
It takes too long.. I was just wondering if there was simply rule I could follow when it's in this kind of form?

No, no simple rule. You know (u-1) and (u+1) are factors, but you don't how many times they divide into your original polynomial. In this case, the answer for both is once. You just have to divide them out. If you want to minimize the work a bit you could divide by (u-1)*(u+1)=u^2-1. Saves you dividing by both separately.
 
Ah okay.. so there is no shortcut :(
 
There is a sort of shortcut, synthetic division. You have to actually do the division, but you don't have to write down all the x powers. You might have learned it already and forgotten it.
 
So I got (u^2-1)(-u^4-u^2-u-1) = -(u+1)(u-1)(u^4+u^2+u+1)
Does that mean there are roots only at u = -1 and u = 1?
 
zeion said:
So I got (u^2-1)(-u^4-u^2-u-1) = -(u+1)(u-1)(u^4+u^2+u+1)
Does that mean there are roots only at u = -1 and u = 1?

The quartic doesn't have any rational roots. It has complex roots, of course. But as far as a factoring exercise, I think you can just leave it there.
 
Oh okay, thanks!
 

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