If for two real constants a and b the expression ` ax ^(3) + 3x^(2) - 8 x + b ` is exactly divisible by (x + 2) and (x - 2) then

Options

A. ` a = 2, b = 12`
B. ` a = 12 , b = 2 `
C. ` a = 2, b = - 12`
D. ` a = - 2, b = 12`

Answer

To solve the problem, we need to find the values of the constants \( a \) and \( b \) such that the polynomial \( ax^3 + 3x^2 - 8x + b \) is exactly divisible by both \( (x + 2) \) and \( (x - 2) \).

### Step-by-Step Solution:

1. **Understand the Remainder Theorem**: According to the Remainder Theorem, if a polynomial \( f(x) \) is divisible by \( (x - c) \), then \( f(c) = 0 \). Therefore, if our polynomial is divisible by \( (x + 2) \) and \( (x - 2) \), we need to evaluate the polynomial at \( x = -2 \) and \( x = 2 \) and set both results to zero.

2. **Evaluate at \( x = -2 \)**:
\[ f(-2) = a(-2)^3 + 3(-2)^2 - 8(-2) + b = 0 \]
Simplifying this:
\[ f(-2) = a(-8) + 3(4) + 16 + b = 0 \]
\[ -8a + 12 + 16 + b = 0 \]
\[ -8a + b + 28 = 0 \quad \text{(Equation 1)} \]

3. **Evaluate at \( x = 2 \)**:
\[ f(2) = a(2)^3 + 3(2)^2 - 8(2) + b = 0 \]
Simplifying this:
\[ f(2) = a(8) + 3(4) - 16 + b = 0 \]
\[ 8a + 12 - 16 + b = 0 \]
\[ 8a + b - 4 = 0 \quad \text{(Equation 2)} \]

4. **Set up the system of equations**:
From Equation 1:
\[ b = 8a - 28 \]
From Equation 2:
\[ b = -8a + 4 \]

5. **Equate the two expressions for \( b \)**:
\[ 8a - 28 = -8a + 4 \]
\[ 16a = 32 \]
\[ a = 2 \]

6. **Substitute \( a \) back to find \( b \)**:
Using \( a = 2 \) in Equation 1:
\[ b = 8(2) - 28 = 16 - 28 = -12 \]

7. **Final answer**:
The values of the constants are:
\[ a = 2, \quad b = -12 \]