Question

Please help! Show that x + 1 is a factor of x^n+1 for all odd powers n . Show also that it cannot be a factor for any even n.

Answer 1

if n is odd , n = 2k+1
show that x+1 divides evenly into x^(2k+1) + 1

Answer 2

it seems to be true
(3^3 +1 )/ (3 + 1) is an integer

Answer 3

I think there is a simple argument
let f(x) = x^n +1 , where n is odd.

Now if n is odd, (-1)^n =(-1)^(2k+1) = (-1)^(2k)* (-1) = -1

It is true that
f(-1) = (-1)^n + 1 = -1 + 1 = 0 since n is odd.
So -1 is a root of f(x)
so (x - (-1)) must be a factor of x^n+1
but that is the same as x+1

Answer 4

so x^n+1 = (x+1) ( some polynomial) , if n is odd

Answer 5

shortened proof:
(x+1) is a factor of f(x)=x^n+1 if and only if f(-1) = 0
but (-1)^n +1 = 0 when n is odd and not when n is even.

Answer 6

this is the factor theorem i am using

Answer 7

Quick question: is x^n+1 equal to x^(n+1) or rather (x^n) + 1

Answer 8

the latter

Answer 9

because x+1 is not a factor of x^(n+1)