`x+1` is a factor of the polynomial

Options

A. `x^(3)+x^(2)-x+1`
B. `x^(3)+x^(2) +x+1`
C. `x^(4)+x^(3)+x^(2)+1`
D. `x^(4)+3x^(3)+3x^(2)+x+1`

Explanation

To determine which polynomial has \( x + 1 \) as a factor, we can use the Factor Theorem. According to the theorem, if \( x + 1 \) is a factor of a polynomial \( P(x) \), then \( P(-1) = 0 \).

Let's evaluate each polynomial at \( x = -1 \):

### Step 1: Evaluate the first polynomial \( P_1(x) = x^3 + x^2 - x + 1 \)

1. Substitute \( x = -1 \):
\[ P_1(-1) = (-1)^3 + (-1)^2 - (-1) + 1 \]
\[ = -1 + 1 + 1 + 1 \]
\[ = 2 \]

2. Since \( P_1(-1) = 2 \neq 0 \), \( x + 1 \) is **not** a factor of \( P_1(x) \).

### Step 2: Evaluate the second polynomial \( P_2(x) = x^3 + x^2 + x + 1 \)

1. Substitute \( x = -1 \):
\[ P_2(-1) = (-1)^3 + (-1)^2 + (-1) + 1 \]
\[ = -1 + 1 - 1 + 1 \]
\[ = 0 \]

2. Since \( P_2(-1) = 0 \), \( x + 1 \) **is** a factor of \( P_2(x) \).

### Step 3: Evaluate the third polynomial \( P_3(x) = x^4 + x^3 + x^2 + 1 \)

1. Substitute \( x = -1 \):
\[ P_3(-1) = (-1)^4 + (-1)^3 + (-1)^2 + 1 \]
\[ = 1 - 1 + 1 + 1 \]
\[ = 2 \]

2. Since \( P_3(-1) = 2 \neq 0 \), \( x + 1 \) is **not** a factor of \( P_3(x) \).

### Step 4: Evaluate the fourth polynomial \( P_4(x) = x^4 + 3x^3 + 3x^2 + x + 1 \)

1. Substitute \( x = -1 \):
\[ P_4(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 \]
\[ = 1 - 3 + 3 - 1 + 1 \]
\[ = 1 \]

2. Since \( P_4(-1) = 1 \neq 0 \), \( x + 1 \) is **not** a factor of \( P_4(x) \).

### Conclusion

The only polynomial for which \( x + 1 \) is a factor is \( P_2(x) = x^3 + x^2 + x + 1 \).

### Final Answer

The correct option is **B**: \( x^3 + x^2 + x + 1 \).

---