`(x+1)` is a factor of `x^n+1` only if  

Options

A. n is an odd integer
B. n is an even integer
C. n is a positive integer
D. n is a negative integer

Answer

To determine when \( (x + 1) \) is a factor of \( x^n + 1 \), we can follow these steps:

### Step 1: Understand the Factor Condition
If \( (x + 1) \) is a factor of \( x^n + 1 \), then substituting \( x = -1 \) into the polynomial \( x^n + 1 \) should yield zero. This is based on the Factor Theorem, which states that if \( (x - c) \) is a factor of a polynomial \( f(x) \), then \( f(c) = 0 \).

### Step 2: Substitute \( x = -1 \)
Substituting \( x = -1 \) into the polynomial:
\[ f(-1) = (-1)^n + 1 \]

### Step 3: Set the Equation to Zero
For \( (x + 1) \) to be a factor, we need:
\[ (-1)^n + 1 = 0 \]

### Step 4: Solve the Equation
Rearranging the equation gives:
\[ (-1)^n = -1 \]
This means that \( n \) must be such that \( (-1)^n \) equals \(-1\).

### Step 5: Determine the Condition on \( n \)
The expression \( (-1)^n \) is equal to \(-1\) when \( n \) is an odd integer. Therefore, for \( (x + 1) \) to be a factor of \( x^n + 1 \), \( n \) must be an odd integer.

### Conclusion
Thus, the answer is that \( (x + 1) \) is a factor of \( x^n + 1 \) only if \( n \) is an odd integer.