Question

If \[ \frac{x+4}{(x+2)^2(x+3)} = \frac{A}{(x+2)^2} + \frac{B}{x+2} + \frac{C}{x+3}, \] then \(A+B+C=\)

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

Hint:

For partial fractions, multiply by the complete denominator first, then substitute values that make factors zero.

Answer 1

The Correct Option is A

Concept: In partial fractions, when the denominator contains repeated linear factors, we write separate terms for each power.

Step 1: Start with the given expression. \[ \frac{x+4}{(x+2)^2(x+3)} = \frac{A}{(x+2)^2} + \frac{B}{x+2} + \frac{C}{x+3} \]

Step 2: Multiply both sides by \((x+2)^2(x+3)\). \[ x+4=A(x+3)+B(x+2)(x+3)+C(x+2)^2 \]

Step 3: Put \(x=-3\) to find \(C\). \[ -3+4=C(-3+2)^2 \] \[ 1=C(1) \] \[ C=1 \]

Step 4: Put \(x=-2\) to find \(A\). \[ -2+4=A(-2+3) \] \[ 2=A \] \[ A=2 \]

Step 5: Find \(B\) by comparing coefficients or substituting \(x=0\). \[ 0+4=A(3)+B(2)(3)+C(2)^2 \] \[ 4=2(3)+6B+1(4) \] \[ 4=6+6B+4 \] \[ 6B=-6 \] \[ B=-1 \]

Step 6: Now calculate \(A+B+C\). \[ A+B+C=2+(-1)+1 \] \[ A+B+C=2 \] Therefore, \[ \boxed{2} \]