If both x-1 and x-2 are factors of `x^(3)-3x^(2)+2x-4b`, then b must be

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to determine the value of \( b \) such that both \( x - 1 \) and \( x - 2 \) are factors of the polynomial \( x^3 - 3x^2 + 2x - 4b \).

### Step 1: Use the Factor Theorem
Since \( x - 1 \) and \( x - 2 \) are factors of the polynomial, we can substitute \( x = 1 \) and \( x = 2 \) into the polynomial and set it equal to zero.

### Step 2: Substitute \( x = 1 \)
Substituting \( x = 1 \):
\[ 1^3 - 3(1^2) + 2(1) - 4b = 0 \]
This simplifies to:
\[ 1 - 3 + 2 - 4b = 0 \]
\[ 0 - 4b = 0 \]
From this, we can deduce:
\[ 4b = 0 \implies b = 0 \]

### Step 3: Substitute \( x = 2 \)
Now, we will check if \( b = 0 \) satisfies the condition by substituting \( x = 2 \):
\[ 2^3 - 3(2^2) + 2(2) - 4b = 0 \]
Substituting \( b = 0 \):
\[ 8 - 3(4) + 4 - 0 = 0 \]
This simplifies to:
\[ 8 - 12 + 4 = 0 \]
\[ 0 = 0 \]
This confirms that \( b = 0 \) is indeed a valid solution.

### Conclusion
Thus, the value of \( b \) must be \( 0 \).