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

Apr 25, 2018

#### Explanation:

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

Now to test for $x + a$ being a factor of $f \left(x\right) = {x}^{n} + {a}^{n}$,

as $x + a = x - \left(- a\right)$, we must check for $f \left(- a\right)$

Now $f \left(x\right) = {x}^{n} + {a}^{n}$

Hence, $f \left(- a\right) = {\left(- a\right)}^{n} + {a}^{n}$

Observe if $n$ is even then $f \left(- a\right) = {\left(- a\right)}^{n} + {a}^{n} = {a}^{n} + {a}^{n} = 2 {a}^{n} \ne 0$ and hence $x + a$ is not a factor of ${x}^{n} + {a}^{n}$ if $n$ is even

but if $n$ is odd then $f \left(- a\right) = {\left(- a\right)}^{n} + {a}^{n} = - {a}^{n} + {a}^{n} = 0$

and hence $x + a$ is a factor of ${x}^{n} + {a}^{n}$ if $n$ is odd.