`x^100 + 2x^99 + k`, is divisible by (x + 1), the value of k?

Options

A. 1
B. `-3`
C. 2
D. `-2`

Answer

To find the value of \( k \) such that the polynomial \( x^{100} + 2x^{99} + k \) is divisible by \( (x + 1) \), we can use the Remainder Theorem. According to this theorem, if a polynomial \( f(x) \) is divisible by \( (x - a) \), then \( f(a) = 0 \).

### Step-by-Step Solution:

1. **Identify the polynomial**:
We have the polynomial \( f(x) = x^{100} + 2x^{99} + k \).

2. **Determine the value of \( x \)**:
Since we want to check for divisibility by \( (x + 1) \), we set \( x = -1 \).

3. **Substitute \( x = -1 \) into the polynomial**:
\[ f(-1) = (-1)^{100} + 2(-1)^{99} + k \]

4. **Calculate the powers of -1**:
- \( (-1)^{100} = 1 \) (since 100 is even)
- \( (-1)^{99} = -1 \) (since 99 is odd)

5. **Substitute these values back into the polynomial**:
\[ f(-1) = 1 + 2(-1) + k \]
\[ f(-1) = 1 - 2 + k \]
\[ f(-1) = -1 + k \]

6. **Set the polynomial equal to zero for divisibility**:
For \( f(-1) \) to equal zero:
\[ -1 + k = 0 \]

7. **Solve for \( k \)**:
\[ k = 1 \]

### Final Answer:
The value of \( k \) is \( 1 \).