Let f(x) be a polynominal with real coefficients such that `f(x)=f'(x) times f'''(x).` If f(x)=0 is satisfied x=1,2,3 only, then the value of f'(1)f'(2)f'(3) is

Options

A. (a)positive
B. (b)negative
C. (c)0
D. (d)Inadequate data

Answer

To solve the problem, we start with the given polynomial function \( f(x) \) and the condition that \( f(x) = f'(x) \cdot f'''(x) \). We also know that \( f(x) = 0 \) has roots at \( x = 1, 2, 3 \).

### Step 1: Determine the degree of the polynomial
Since \( f(x) = 0 \) has roots at \( x = 1, 2, 3 \) and is satisfied only at these points, we can express \( f(x) \) as:
\[ f(x) = k(x - 1)(x - 2)(x - 3)(x - r) \]
where \( k \) is a constant and \( r \) is a root that we need to determine. The polynomial must have a degree of at least 4 since it has 3 distinct roots and one of them must be repeated (as indicated by the problem).

### Step 2: Analyze the condition \( f(x) = f'(x) \cdot f'''(x) \)
The degree of \( f'(x) \) will be \( 3 \) (since the degree of \( f(x) \) is \( 4 \)), and the degree of \( f'''(x) \) will be \( 1 \). Therefore, the degree of \( f'(x) \cdot f'''(x) \) will be \( 3 + 1 = 4 \). This matches the degree of \( f(x) \), which confirms our polynomial structure.

### Step 3: Identify the repeated root
Given that \( f(x) \) has roots at \( x = 1, 2, 3 \) and one of these roots must be repeated, we can assume without loss of generality that \( x = 1 \) is the repeated root. Thus, we can rewrite \( f(x) \) as:
\[ f(x) = k(x - 1)^2(x - 2)(x - 3) \]

### Step 4: Calculate the derivatives
Next, we need to find \( f'(x) \):
Using the product rule:
\[ f'(x) = k \left[ 2(x - 1)(x - 2)(x - 3) + (x - 1)^2 \left( (x - 2) + (x - 3) \right) \right] \]

Now we will evaluate \( f'(1) \), \( f'(2) \), and \( f'(3) \).

### Step 5: Evaluate \( f'(1) \), \( f'(2) \), and \( f'(3) \)
1. **Calculate \( f'(1) \)**:
\[ f'(1) = k \left[ 2(1 - 1)(1 - 2)(1 - 3) + (1 - 1)^2 \left( (1 - 2) + (1 - 3) \right) \right] = 0 \]

2. **Calculate \( f'(2) \)**:
\[ f'(2) = k \left[ 2(2 - 1)(2 - 2)(2 - 3) + (2 - 1)^2 \left( (2 - 2) + (2 - 3) \right) \right] = 0 \]

3. **Calculate \( f'(3) \)**:
\[ f'(3) = k \left[ 2(3 - 1)(3 - 2)(3 - 3) + (3 - 1)^2 \left( (3 - 2) + (3 - 3) \right) \right] = 0 \]

### Step 6: Compute the product \( f'(1)f'(2)f'(3) \)
Since \( f'(1) = 0 \), \( f'(2) = 0 \), and \( f'(3) = 0 \):
\[ f'(1)f'(2)f'(3) = 0 \cdot 0 \cdot 0 = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]