Question

How do we prove that `(x+a)` is a factor of `x^n+a^n` for all odd positive integer n?

Answer 1

Please see below.

Explanation:

According to factor theorem, if `x-k` is a factor of a polynomial function `f(x)`, then `f(k)=0`

Now to test for `x+a` being a factor of `f(x)=x^n+a^n`,

as `x+a=x-(-a)`, we must check for `f(-a)`

Now `f(x)=x^n+a^n`

Hence, `f(-a)=(-a)^n+a^n`

Observe if `n` is even then `f(-a)=(-a)^n+a^n=a^n+a^n=2a^n!=0` and hence `x+a` is not a factor of `x^n+a^n` if `n` is even

but if `n` is odd then `f(-a)=(-a)^n+a^n=-a^n+a^n=0`

and hence `x+a` is a factor of `x^n+a^n` if `n` is odd.