If `x^3+3x^2-9x+c` is of the form `(x-alpha)^2(x-beta)` then `c` is equal to

Options

A. 27
B. -27
C. 5
D. -5

Explanation

To solve the problem, we need to find the value of \( c \) such that the polynomial \( x^3 + 3x^2 - 9x + c \) can be expressed in the form \( (x - \alpha)^2(x - \beta) \).

### Step-by-Step Solution:

1. **Understanding the Form**:
Since the polynomial is of the form \( (x - \alpha)^2(x - \beta) \), it implies that \( \alpha \) is a double root. Therefore, the polynomial can be expressed as:
\[ f(x) = (x - \alpha)^2(x - \beta) \]

2. **Finding the Derivative**:
Since \( \alpha \) is a double root, we know that:
\[ f'(\alpha) = 0 \]
We first compute the derivative of the polynomial \( f(x) = x^3 + 3x^2 - 9x + c \):
\[ f'(x) = 3x^2 + 6x - 9 \]

3. **Setting the Derivative to Zero**:
Now, we set the derivative equal to zero to find \( \alpha \):
\[ 3\alpha^2 + 6\alpha - 9 = 0 \]
Dividing the entire equation by 3 gives:
\[ \alpha^2 + 2\alpha - 3 = 0 \]

4. **Factoring the Quadratic**:
We can factor this quadratic equation:
\[ (\alpha - 1)(\alpha + 3) = 0 \]
Thus, the possible values for \( \alpha \) are:
\[ \alpha = 1 \quad \text{or} \quad \alpha = -3 \]

5. **Finding \( c \) for Each \( \alpha \)**:
We will now substitute each value of \( \alpha \) back into the original polynomial to find \( c \).

- **Case 1**: \( \alpha = 1 \)
\[ f(1) = 1^3 + 3(1^2) - 9(1) + c = 0 \]
Simplifying:
\[ 1 + 3 - 9 + c = 0 \implies c - 5 = 0 \implies c = 4 \]

- **Case 2**: \( \alpha = -3 \)
\[ f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + c = 0 \]
Simplifying:
\[ -27 + 27 + 27 + c = 0 \implies c + 27 = 0 \implies c = -27 \]

6. **Final Answer**:
The values of \( c \) that satisfy the condition are:
\[ c = 4 \quad \text{or} \quad c = -27 \]