If `x+a` is a factor of `x^4-a^2x^2+3x-6a ,` then `a=`

Options

A. 0
B. 1
C. 9
D. `- 9`

Answer

To solve the problem, we need to determine the value of \( a \) such that \( x + a \) is a factor of the polynomial \( P(x) = x^4 - a^2 x^2 + 3x - 6a \).

### Step-by-Step Solution:

1. **Understanding the Factor Condition**:
Since \( x + a \) is a factor of \( P(x) \), it means that if we substitute \( x = -a \) into the polynomial \( P(x) \), the result should be zero:
\[ P(-a) = 0 \]

2. **Substituting \( x = -a \)**:
We substitute \( -a \) into the polynomial:
\[ P(-a) = (-a)^4 - a^2(-a)^2 + 3(-a) - 6a \]

3. **Calculating Each Term**:
- The first term:
\[ (-a)^4 = a^4 \]
- The second term:
\[ -a^2(-a)^2 = -a^2(a^2) = -a^4 \]
- The third term:
\[ 3(-a) = -3a \]
- The fourth term:
\[ -6a = -6a \]

4. **Combining the Terms**:
Now, we combine all the terms:
\[ P(-a) = a^4 - a^4 - 3a - 6a \]
Simplifying this gives:
\[ P(-a) = 0 - 9a = -9a \]

5. **Setting the Polynomial Equal to Zero**:
For \( P(-a) \) to equal zero:
\[ -9a = 0 \]

6. **Solving for \( a \)**:
Dividing both sides by -9, we find:
\[ a = 0 \]

### Conclusion:
Thus, the value of \( a \) is \( 0 \).