Question

if (x+1)is a factor of(x^n+1)then n is necessarily ?
(a).an odd integer. (b).an even integer. (c).a negative integer.
(d).a positive integer.

Answer 1

Answer:

n is an odd integer

option (a) is correct

Step-by-step explanation:

Formula used:

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+..........ab^{n-2}+b^{n-1})$

Now,

$x^n+1$

$x^n+1^n$

$x^n-(-1)^n$

$=(x-(-1))(x^{n-1}+x^{n-2}(-1)+..........x(-1)^{n-2}+(-1)^{n-1})$

$=(x+1)(x^{n-1}+x^{n-2}(-1)+..........x(-1)^{n-2}+(-1)^{n-1})$ which is divisible by (x+1)

But it is possible only when n is an odd integer.

Answer 2

Step-by-step explanation:

only is a ln odd integer

hope it helps you

thank you so much