Determine which of the following polynomials has `(x+1)` a factor :
(i) `x^(3)+x^(2)+x+1`
(ii) `x^(4)+x^(3)+x^(2)+x+1`
(iii) `x^(4)+3x^(3)+3x^(2)+x+1`
(iv) `x^(3)-x^(2)-(2+sqrt(2))x+sqrt(2)`

Answer

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

### Step-by-step Solution:

1. **Evaluate the first polynomial:**
\[ P_1(x) = x^3 + x^2 + x + 1 \]
Substitute \(x = -1\):
\[ P_1(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0 \]
Since \(P_1(-1) = 0\), \((x + 1)\) is a factor of \(P_1(x)\).

2. **Evaluate the second polynomial:**
\[ P_2(x) = x^4 + x^3 + x^2 + x + 1 \]
Substitute \(x = -1\):
\[ P_2(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1 \]
Since \(P_2(-1) \neq 0\), \((x + 1)\) is not a factor of \(P_2(x)\).

3. **Evaluate the third polynomial:**
\[ P_3(x) = x^4 + 3x^3 + 3x^2 + x + 1 \]
Substitute \(x = -1\):
\[ P_3(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 = 1 - 3 + 3 - 1 + 1 = 1 \]
Since \(P_3(-1) \neq 0\), \((x + 1)\) is not a factor of \(P_3(x)\).

4. **Evaluate the fourth polynomial:**
\[ P_4(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2} \]
Substitute \(x = -1\):
\[ P_4(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + \sqrt{2} = -1 - 1 + 2 + \sqrt{2} + \sqrt{2} = -2 + 2 + 2\sqrt{2} = 2\sqrt{2} \]
Since \(P_4(-1) \neq 0\), \((x + 1)\) is not a factor of \(P_4(x)\).

### Conclusion:
The only polynomial that has \((x + 1)\) as a factor is:
- **(i) \(x^3 + x^2 + x + 1\)**