maths factorise x^5-1?

Hi,
. _________________
1)1 . 0 . 0 . 0 . 0 . -1
.___1__1__1_.1__1_
. 1 . 1 . 1 . .1. 1. . 0

x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1) <==ANSWER

I hope that helps!! :-)

a^5 - b^5 => (a-b)(a^4+a^3 * b+a^2 * b^2 + a * b^3 + b^4)
so
x^5 - 1 = x^5 - 1^5
=> (x-1)(x^4 + x^3 * 1^1 + x^2 * 1^2 + x^1 * 1^3 + 1^4)
=> (x-1)(x^4 + x^3 + x^2 + x + 1)

_____________
Bakhtiar Kasi algebric identity u choose was wrong..its not binomial identity..its factorzing id.....careful when choosing appr. Id..its not for expanding its for factorizing...

y+1) (y-1)

Such a polynomial can be factorized by using remainder theorem.
here P(x) = x ^ 5 -1
P(1) = 1 ^ 5 -1= 1-1 = 0 so x - 1 is a factor of P(x)
Now divide P(x) by x - 1 to get the other factor. On dividing x^5 -1 by x - 1, the quotient is x ^ 4+x ^ 3+x ^ 2 + x + 1, hence the factors of x ^ 5 - 1 are
( x - 1 ).( x ^ 4 + x ^ 3 + x ^ 2 + x +1)

It depends where you are working. In R? In C?
It depends also how much you want to factorize.

A classical factorization is : x^5-1=(x-1) (x^4+x^3+x^2+x+1)
More generally, x^n-1=(x-1) (x^(n-1)+x^(n-2)+...+x+1)

In C, you have this classical factorization using the roots of 1 :
x^5-1=(x-1) (x-e^(2i Pi/5)) (x-e^(4i Pi/5)) (x-e^(6i Pi/5)) (x-e^(9i Pi/5))
=(x-1) (x-e^(2i Pi/5)) (x-e^(-2i Pi/5)) (x-e^(4i Pi/5)) (x-e^(-4i Pi/5))

Putting together conjugate zero of x^5-1, you obtain :
x^5-1=(x-1) (x^2-2cos(2Pi/5)X+1) (x^2-2cos(4Pi/5)X+1)

But, we don't know yet cos(2Pi/5) and cos(4Pi/5).
We know that : (x^2-2cos(2Pi/5)X+1) (x^2-2cos(4Pi/5)X+1)=x^4+x^3+x^2+x+1
We are looking for a and b such that :
(x^2+ax+1) (x^2+bx+1)=x^4+x^3+x^2+x+1
Solving this equation in a and b, you found that :
x^5-1=(x-1) (x^2+(1+sqrt(5))/2 x+1) (x^2+(1-sqrt(5))/2x+1)
where sqrt(5) is the square root of 5

Easiest factorization is (x-1) (x^4+x^3+x^2+x+1)

(x - 1)(x^4 + x^3 + x^2 + x + 1)

Check

x^5 + x^4 + x^3 + x^2 + x - x^4 - x^3 - x^2 - 1
x^5 - 1

I suppose we can find more roots out of this, but that would require alot more thinking ....

do you mean (x^5) - 1 or x^(5-1)
if first, then its already factorised,
if second, then same i guess..

unless first is something like {(x-1)(x+1)}^2 (x-1)

x^5 - 1
= x^5 - 1^5
= (x - 1)(x^4 + x^3 + x^2 + x + 1^4)
= (x - 1)(x^4 + x^3 + x^2 + x + 1)

(x^5/2 -1)(x^5/2 +1)