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 both sides by \((x^2+4)^2\).
\[ 4x^3+16x+7 = (Ax+B)(x^2+4)+(Cx+D) \]

Step 2: Expand the right-hand side.
\[ (Ax+B)(x^2+4)+(Cx+D) \] \[ =Ax^3+Bx^2+4Ax+4B+Cx+D \] \[ =Ax^3+Bx^2+(4A+C)x+(4B+D) \]

Step 3: Compare coefficients.
Compare with \[ 4x^3+0x^2+16x+7. \] So, \[ A=4 \] \[ B=0 \] \[ 4A+C=16 \] \[ 4B+D=7 \]

Step 4: Find \(C\) and \(D\).
Since \[ A=4, \] we get \[ 4(4)+C=16 \] \[ 16+C=16 \] \[ C=0 \] Also, since \[ B=0, \] \[ 4(0)+D=7 \] \[ D=7 \] Thus, \[ A=4,\quad B=0,\quad C=0,\quad D=7. \]

Step 5: Count the non-zero values.
The non-zero values are \[ A=4 \] and \[ D=7. \] Hence, the number of non-zero values is \[ 2. \]

Step 6: Final conclusion.
Therefore, \[ \boxed{2} \]