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 \( f(x) = x^3 + 3x^2 - 9x + c \) can be expressed in the form \( (x - \alpha)^2 (x - \beta) \).

### Step 1: Set up the equation
Given:
\[ f(x) = x^3 + 3x^2 - 9x + c \]
We want to express this polynomial in the form:
\[ f(x) = (x - \alpha)^2 (x - \beta) \]

### Step 2: Expand the desired form
Expanding \( (x - \alpha)^2 (x - \beta) \):
\[ (x - \alpha)^2 = x^2 - 2\alpha x + \alpha^2 \]
Now multiplying by \( (x - \beta) \):
\[ (x^2 - 2\alpha x + \alpha^2)(x - \beta) = x^3 - \beta x^2 - 2\alpha x^2 + 2\alpha \beta x + \alpha^2 x - \alpha^2 \beta \]
Combining like terms:
\[ = x^3 + (-\beta - 2\alpha)x^2 + (2\alpha \beta + \alpha^2)x - \alpha^2 \beta \]

### Step 3: Compare coefficients
Now we compare coefficients with \( f(x) = x^3 + 3x^2 - 9x + c \):
- Coefficient of \( x^2 \):
\[ -\beta - 2\alpha = 3 \quad (1) \]
- Coefficient of \( x \):
\[ 2\alpha \beta + \alpha^2 = -9 \quad (2) \]
- Constant term:
\[ -\alpha^2 \beta = c \quad (3) \]

### Step 4: Solve the equations
From equation (1):
\[ \beta = -2\alpha - 3 \]
Substituting \( \beta \) into equation (2):
\[ 2\alpha(-2\alpha - 3) + \alpha^2 = -9 \]
Expanding:
\[ -4\alpha^2 - 6\alpha + \alpha^2 = -9 \]
Combining like terms:
\[ -3\alpha^2 - 6\alpha + 9 = 0 \]
Dividing by -3:
\[ \alpha^2 + 2\alpha - 3 = 0 \]
Factoring:
\[ (\alpha + 3)(\alpha - 1) = 0 \]
Thus, \( \alpha = -3 \) or \( \alpha = 1 \).

### Step 5: Find corresponding \( \beta \) and \( c \)
**Case 1**: If \( \alpha = -3 \):
\[ \beta = -2(-3) - 3 = 6 - 3 = 3 \]
Substituting into equation (3):
\[ c = -(-3)^2(3) = -9 \cdot 3 = -27 \]

**Case 2**: If \( \alpha = 1 \):
\[ \beta = -2(1) - 3 = -2 - 3 = -5 \]
Substituting into equation (3):
\[ c = -(1)^2(-5) = -1 \cdot -5 = 5 \]

### Conclusion
The possible values of \( c \) are \( -27 \) and \( 5 \).

### Final Answer
Thus, \( c \) can be either \( -27 \) or \( 5 \).