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

1 Answer
Apr 25, 2018

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.