Question

using factor theram show that x+1 is a factor of x power19+1

Answer 1

Factor theorem-

$f(x) = q(x).d$
where f(x) is given expression,
q(x) is quotient
d is divisor

Here,
$f(x) = {x}^{19} + 1 \\ q(x) = x + 1$
Putting x=(-1)
$f(x) = { (- 1)}^{19} + 1 = - 1 + 1 = 0 \\ q(x) = - 1 + 1 = 0 \\ q(x).d = 0.d = 0 = f(x)$
So q(x) = x+1 is a factor of f(x) = x^19 +1