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 in \(A,B,C,D\) is

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

Hint:

In partial fraction comparison, multiply by the common denominator first and then compare coefficients of equal powers of \(x\).

Answer 1

The Correct Option is B

Step 1: Multiply out and match coefficients.
\(4x^3+16x+7=(Ax+B)(x^2+4)+Cx+D=Ax^3+Bx^2+(4A+C)x+(4B+D)\). Comparing: \(A=4, B=0, 4A+C=16\Rightarrow C=0, 4B+D=7\Rightarrow D=7\).

Step 2: Count non-zero values.
Non-zero among \(\{A,B,C,D\}=\{4,0,0,7\}\) are \(A\) and \(D\), so count \(=2\). \[ \boxed{2} \]