To solve the problem of finding the value of [tex]\( k \)[/tex] such that the polynomial [tex]\( x^{100} + 2x^{99} + k \)[/tex] is divisible by [tex]\( x + 1 \)[/tex], we can utilize the Remainder Theorem. The Remainder Theorem states that if a polynomial [tex]\( f(x) \)[/tex] is divided by [tex]\( x - a \)[/tex], the remainder of this division is [tex]\( f(a) \)[/tex]. For the polynomial to be divisible by [tex]\( x + 1 \)[/tex], the remainder when [tex]\( x = -1 \)[/tex] should be zero.
Given polynomial:
[tex]\[ f(x) = x^{100} + 2x^{99} + k \][/tex]
Divisor:
[tex]\[ x + 1 \][/tex]
From the Remainder Theorem:
[tex]\[ f(-1) = 0 \][/tex]
We substitute [tex]\( x = -1 \)[/tex] in the polynomial:
[tex]\[ f(-1) = (-1)^{100} + 2(-1)^{99} + k \][/tex]
Calculating these terms individually,
[tex]\[ (-1)^{100} = 1 \][/tex]
[tex]\[ 2(-1)^{99} = 2(-1) = -2 \][/tex]
So, the expression becomes:
[tex]\[ f(-1) = 1 - 2 + k = 0 \][/tex]
Simplify it to solve for [tex]\( k \)[/tex]:
[tex]\[ 1 - 2 + k = 0 \][/tex]
[tex]\[ -1 + k = 0 \][/tex]
[tex]\[ k = 1 \][/tex]
Therefore, the value of [tex]\( k \)[/tex] is:
[tex]\[ \boxed{1} \][/tex]
