Question

If \( (x-a) \) is a factor of \( x^3 - 3x^2a + 2a^2x + b \), then the value of \( b \) is :

• A. 1
• B. 2
• C. 3
• D. 0

Hint:

Factor Theorem: If \( (x-k) \) is a factor of \( P(x) \), then \( P(k) = 0 \). Here, \( (x-a) \) is a factor, so \(k=a\). Let \( P(x) = x^3 - 3x^2a + 2a^2x + b \). Set \( P(a) = 0 \): \( a^3 - 3a^2(a) + 2a^2(a) + b = 0 \) \( a^3 - 3a^3 + 2a^3 + b = 0 \) \( (1-3+2)a^3 + b = 0 \) \( 0 \cdot a^3 + b = 0 \) \( 0 + b = 0 \) \( b = 0 \)

Answer 1

The Correct Option is D

Concept: This problem uses the Factor Theorem. The Factor Theorem states that if \( (x-k) \) is a factor of a polynomial \( P(x) \), then \( P(k) = 0 \). Conversely, if \( P(k) = 0 \), then \( (x-k) \) is a factor of \( P(x) \).

Step 1: Apply the Factor Theorem We are given that \( (x-a) \) is a factor of the polynomial \( P(x) = x^3 - 3x^2a + 2a^2x + b \). According to the Factor Theorem, if \( (x-a) \) is a factor, then \( P(a) = 0 \).

Step 2: Substitute \( x = a \) into the polynomial \( P(x) \) Replace every \( x \) in \( P(x) \) with \( a \): \[ P(a) = (a)^3 - 3(a)^2a + 2a^2(a) + b \]

Step 3: Simplify the expression for \( P(a) \) \[ P(a) = a^3 - 3(a^2 \cdot a) + 2(a^2 \cdot a) + b \] \[ P(a) = a^3 - 3a^3 + 2a^3 + b \] Combine the terms with \(a^3\): \[ P(a) = (1 - 3 + 2)a^3 + b \] \[ P(a) = ( -2 + 2)a^3 + b \] \[ P(a) = (0)a^3 + b \] \[ P(a) = 0 + b \] \[ P(a) = b \]

Step 4: Set \( P(a) = 0 \) and solve for \( b \) Since \( (x-a) \) is a factor, we know \( P(a) = 0 \). From Step 3, we found \( P(a) = b \). Therefore, we must have: \[ b = 0 \] The value of \( b \) is 0.