If `(x ^(2) + x - 2)` is the HCF of the expressions `(x - 1) (2x ^(2) + ax +2) and (x +2) (3x ^(2) + bx +1),` then find the valued of a and b.

Options

A. `a = 3, b = 4`
B. `a = 5, b =- 4`
C. `a =- 3, b = 5`
D. `a = 5, b = 6`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given that \( (x^2 + x - 2) \) is the highest common factor (HCF) of the two expressions \( (x - 1)(2x^2 + ax + 2) \) and \( (x + 2)(3x^2 + bx + 1) \).

### Step 1: Factor the HCF
First, we factor the expression \( x^2 + x - 2 \):
\[ x^2 + x - 2 = (x - 1)(x + 2) \]

### Step 2: Set up the expressions
The given expressions are:
1. \( (x - 1)(2x^2 + ax + 2) \)
2. \( (x + 2)(3x^2 + bx + 1) \)

### Step 3: Identify the roots of the HCF
The roots of \( (x - 1)(x + 2) = 0 \) are \( x = 1 \) and \( x = -2 \). Since \( (x^2 + x - 2) \) is the HCF, both expressions must equal zero at these roots.

### Step 4: Substitute \( x = 1 \)
Substituting \( x = 1 \) into both expressions:
1. For the first expression:
\[ (1 - 1)(2(1)^2 + a(1) + 2) = 0 \quad \text{(This is zero, so we need to check the second expression)} \]

2. For the second expression:
\[ (1 + 2)(3(1)^2 + b(1) + 1) = 3(3 + b + 1) = 3(4 + b) = 0 \]
This implies \( 4 + b = 0 \) or \( b = -4 \).

### Step 5: Substitute \( x = -2 \)
Now substitute \( x = -2 \) into both expressions:
1. For the first expression:
\[ (-2 - 1)(2(-2)^2 + a(-2) + 2) = -3(8 - 2a + 2) = -3(10 - 2a) = 0 \]
This implies \( 10 - 2a = 0 \) or \( 2a = 10 \) leading to \( a = 5 \).

2. For the second expression:
\[ (-2 + 2)(3(-2)^2 + b(-2) + 1) = 0 \quad \text{(This is zero, so we don't need to check further)} \]

### Conclusion
Thus, we have found:
\[ a = 5 \quad \text{and} \quad b = -4 \]

### Final Answer
The values of \( a \) and \( b \) are:
\[ \boxed{5} \quad \text{and} \quad \boxed{-4} \]