Question

use factor theorem to prove that (x+a) is a factor of (xn+an) for any odd positive integer n.​

Answer 1

Answer:

Let p(x) = x^n + a^n , where n is odd positive integer.

Take (x+a)= 0

=> x = -a

Consider:

p(-a) = (-a) ^n + (a) ^n

= -a^n + a^n

= 0

Since, n is odd.

By Factor theorem,

(x+a) is a factor of p(x) when n is odd positive integer.

Step-by-step explanation:

Answer 2

Answer:

༆ ɑnswer ࿐

let p(x) = x^n + a^n , Where n is odd positive integer

g(x) = x + a

= x + a = 0

= x = — a

p(—a) = (—a)^n + (a)^n

= —a^n + a^n

= 0

since is odd number

therefore by factor theorem m, x + a is a factor of p(x) Where n is odd positive integer...

hope it may help you

༆ Mɑrk me ɑs brɑin list ɑnswer ࿐