Question

If \( \frac{4x^3 + 16x + 7}{(x^2 + 4)^2} = \frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{(x^2 + 4)^2} \), then the number of non-zero values among \( A, B, C, D \) is:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( 4 \)

Hint:

Partial Fraction with Repeated Quadratic}
Multiply both sides by the denominator to eliminate fractions.
Expand and collect like terms.
Compare coefficients and solve the resulting equations.

Answer 1

The Correct Option is B

Let: \[ \frac{4x^3 + 16x + 7}{(x^2 + 4)^2} = \frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{(x^2 + 4)^2} \] Multiply both sides by \( (x^2 + 4)^2 \), equate coefficients: \[ 4x^3 + 16x + 7 = (Ax + B)(x^2 + 4) + (Cx + D) \] Expanding: \[ Ax^3 + 4Ax + Bx^2 + 4B + Cx + D = Ax^3 + Bx^2 + (4A + C)x + (4B + D) \] Compare: - \( A = 4 \) - \( B = 0 \) - \( 4A + C = 16 \Rightarrow C = 0 \) - \( 4B + D = 7 \Rightarrow D = 7 \) So non-zero values: \( A = 4, D = 7 \Rightarrow \text{2 values} \)