If `(x+2) and (x-2)` are factors of `ax^4 + 2x - 3x^2 + bx - 4 ` , then the value of a + b is

Options

A. `-1`
B. 1
C. 7
D. `-8`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that \( (x + 2) \) and \( (x - 2) \) are factors of the polynomial \( ax^4 + 2x - 3x^2 + bx - 4 \).

### Step-by-Step Solution:

1. **Define the Polynomial**:
Let \( f(x) = ax^4 - 3x^2 + bx + 2x - 4 \).

2. **Use Factor Theorem**:
Since \( (x + 2) \) and \( (x - 2) \) are factors, we know that:
- \( f(-2) = 0 \)
- \( f(2) = 0 \)

3. **Calculate \( f(-2) \)**:
\[ f(-2) = a(-2)^4 - 3(-2)^2 + b(-2) + 2(-2) - 4 \]
\[ = 16a - 3(4) - 2b - 4 - 4 \]
\[ = 16a - 12 - 2b - 4 - 4 \]
\[ = 16a - 2b - 20 \]
Setting this equal to zero gives us our first equation:
\[ 16a - 2b - 20 = 0 \quad \text{(Equation 1)} \]

4. **Calculate \( f(2) \)**:
\[ f(2) = a(2)^4 - 3(2)^2 + b(2) + 2(2) - 4 \]
\[ = 16a - 3(4) + 2b + 4 - 4 \]
\[ = 16a - 12 + 2b + 4 - 4 \]
\[ = 16a + 2b - 12 \]
Setting this equal to zero gives us our second equation:
\[ 16a + 2b - 12 = 0 \quad \text{(Equation 2)} \]

5. **Solve the System of Equations**:
Now we have two equations:
- \( 16a - 2b = 20 \) (from Equation 1)
- \( 16a + 2b = 12 \) (from Equation 2)

We can add these two equations:
\[ (16a - 2b) + (16a + 2b) = 20 + 12 \]
\[ 32a = 32 \]
\[ a = 1 \]

6. **Substitute \( a \) back to find \( b \)**:
Substitute \( a = 1 \) into Equation 1:
\[ 16(1) - 2b = 20 \]
\[ 16 - 2b = 20 \]
\[ -2b = 20 - 16 \]
\[ -2b = 4 \]
\[ b = -2 \]

7. **Find \( a + b \)**:
Now we can find \( a + b \):
\[ a + b = 1 + (-2) = -1 \]

### Final Answer:
Thus, the value of \( a + b \) is \( -1 \).