If `x^(3) + 2x^(2) + ax + b` is exactly divisible by `x^(2) - 1`, then the value of a and b are respectively
(a)1 and 2
(b)1 and 0
(c)`-1` and -2
(d)0 and 1

Options

A. 1 and 2
B. 1 and 0
C. `-1` and -2
D. 0 and 1

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the polynomial \( x^3 + 2x^2 + ax + b \) is exactly divisible by \( x^2 - 1 \).

### Step-by-Step Solution:

1. **Understand the divisibility condition**:
Since \( x^2 - 1 = (x - 1)(x + 1) \), the polynomial \( x^3 + 2x^2 + ax + b \) must equal zero for both \( x = 1 \) and \( x = -1 \).

2. **Substitute \( x = 1 \)**:
\[ P(1) = 1^3 + 2(1^2) + a(1) + b = 1 + 2 + a + b = 0 \]
This simplifies to:
\[ 3 + a + b = 0 \quad \Rightarrow \quad a + b = -3 \quad \text{(Equation 1)} \]

3. **Substitute \( x = -1 \)**:
\[ P(-1) = (-1)^3 + 2(-1)^2 + a(-1) + b = -1 + 2 - a + b = 0 \]
This simplifies to:
\[ 1 - a + b = 0 \quad \Rightarrow \quad -a + b = -1 \quad \Rightarrow \quad b - a = 1 \quad \text{(Equation 2)} \]

4. **Solve the system of equations**:
We have two equations:
- \( a + b = -3 \) (Equation 1)
- \( b - a = 1 \) (Equation 2)

From Equation 2, we can express \( b \) in terms of \( a \):
\[ b = a + 1 \]
Substitute this expression for \( b \) into Equation 1:
\[ a + (a + 1) = -3 \]
This simplifies to:
\[ 2a + 1 = -3 \]
\[ 2a = -4 \quad \Rightarrow \quad a = -2 \]

5. **Find \( b \)**:
Substitute \( a = -2 \) back into the expression for \( b \):
\[ b = -2 + 1 = -1 \]

6. **Final values**:
Therefore, the values of \( a \) and \( b \) are:
\[ a = -2, \quad b = -1 \]

### Conclusion:
The values of \( a \) and \( b \) are respectively \( -2 \) and \( -1 \). This corresponds to option (c) \(-1\) and \(-2\).