Solving Polynomial Equations with Exponents of 40

  • Thread starter Thread starter heartyface
  • Start date Start date
  • Tags Tags
    Polynomials
Click For Summary

Homework Help Overview

The problem involves a polynomial equation of the form x^4 - x^3 + x^2 - x + 1 = 0 and seeks to determine the value of another expression, x^40 - x^30 + x^20 - x^10 + x^0. The subject area pertains to polynomial equations and their properties, particularly in relation to roots and exponents.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods such as symmetry, factoring, and the potential use of roots of unity. Some express frustration over the complexity of the problem and the lack of shortcuts. There are questions about how to connect the original polynomial to the expression being evaluated.

Discussion Status

The discussion is ongoing, with participants exploring different avenues for solving the problem. Some have suggested using roots of unity, while others are considering evaluating the roots of the polynomial directly. There is no explicit consensus on a method yet, but several lines of reasoning are being examined.

Contextual Notes

Participants note the challenge of finding roots and the implications of polynomial properties, such as evaluating p(-1) to check for roots. There is an acknowledgment of the complexity involved in the problem, which may require more extensive calculations.

heartyface
Messages
28
Reaction score
0
Polynomials help~~

Heh, so I posted this thread in the wrong category so I'm reposting it! =)
Hello. So here was this problem I came across:

If x^4-x^3+x^2-x^1+x^0=0, what is the numerical value of x^40-x^30+x^20-x^10+x^0?
I did try doing many stuffs (symmetry) & factoring, but I think none of these steps helped.
Enlighten the youngster, gracias.
 
Physics news on Phys.org


If 2X=0, what is 20X?
 


@adaptation -_- very funny, if it was a simple problem like that I wouldn't have posted it.
Btw, the answer's 1.
Hmm... but How...?
 


I found no shortcut in doing this :rolleyes: ,
maybe then the only way to do this is find the roots and put them in the second equation ,
Yea lengthy it is
 

Attachments

  • ca2 copy.png
    ca2 copy.png
    5.6 KB · Views: 558


There could be a work around , using roots of unity , you need to think about it .
( sum of roots of unity and their properties )
 


our original polynomial is p(x) = x4-x3+x2-x+1.

note that p(-1) = 1 + 1 + 1 + 1 + 1 = 5, so -1 is not a root of p(x).

now consider q(x) = (x+1)p(x) = x5 + 1.

Mod note: rest of solution removed[/color]
 
Last edited by a moderator:

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
9
Views
3K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K