Question

Using factor theorem show that x+a is a factor of x^n+a^n when n is any odd positive integer.

Answer 1

Factor theorem says that if (x - x1) is a factor of  polynomial P(x), then P(x1) = 0.

$\ If\ (x\ +\ a)\ is\ a\ factor\ of\ P(x)\ =\ [\ x^n\ +\ a^n\ ]\ then\ P(\ -a\ )\ =\ 0\ \\ \\ P(\ -a\ )\ =\ (-a)^n\ +\ a^n\ =\ (-1)^n\ a^n\ +\ a^n\ = \\ \\. \ \ \ \ \ - a^n\ +\ a^n\ =\ 0,\ as\ (-1)^n\ =\ -1,\ if\ n\ =\ an\ odd\ integer.$

Answer 2

Step-by-step explanation:

༆ 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 ࿐