Question

Given: $$ \frac{2x^2 + 1}{x^3 - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} $$ Find the value of $ 7A + 2B + C $

• A. 8
• B. 9
• C. 10
• D. 11

Hint:

In partial fractions, clear the denominator and compare coefficients to form a system of equations.

Answer 1

The Correct Option is B

Factor the denominator: \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \] So write: \[ \frac{2x^2 + 1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \] Multiply both sides by the denominator: \[ 2x^2 + 1 = A(x^2 + x + 1) + (Bx + C)(x - 1) \] Expand both parts:
1. \( A(x^2 + x + 1) = Ax^2 + Ax + A \)
2. \( (Bx + C)(x - 1) = Bx^2 - Bx + Cx - C = Bx^2 + (C - B)x - C \) Add both: \[ 2x^2 + 1 = (A + B)x^2 + (A + C - B)x + (A - C) \] Compare coefficients:
- \( x^2 \): \( A + B = 2 \)
- \( x \): \( A + C - B = 0 \)
- Constant: \( A - C = 1 \) Solve system:
From (1): \( A + B = 2 \)
From (2): \( A + C - B = 0 \Rightarrow C = B - A \)
From (3): \( A - C = 1 \Rightarrow C = A - 1 \) Equating both values of \( C \): \[ B - A = A - 1 \Rightarrow B = 2A - 1 \] Substitute into (1): \[ A + 2A - 1 = 2 \Rightarrow 3A = 3 \Rightarrow A = 1 \Rightarrow B = 2(1) - 1 = 1,\quad C = 1 - 1 = 0 \] Now compute: \[ 7A + 2B + C = 7(1) + 2(1) + 0 = \boxed{9} \]