Question:medium

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

Show Hint

In partial fractions, the sum of coefficients for terms of the highest degree must match the original numerator's degree.
  • 2
  • 1
  • -1
  • 3
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
This question involves resolving a proper rational fraction into partial fractions.
The denominator contains a repeated linear factor \( (x+2)^2 \) and a simple linear factor \( (x+3) \).
We need to find the constants \( A, B, \) and \( C \) and calculate their sum.
Step 2: Key Formula or Approach:
Multiply both sides by the common denominator \( (x+2)^2(x+3) \) to obtain the identity:
\[ x + 4 = A(x+3) + B(x+2)(x+3) + C(x+2)^2 \]
We can solve for the constants by substituting specific values for \( x \) or equating coefficients.
Step 3: Detailed Explanation:

Finding A: Let \( x = -2 \).
\[ -2 + 4 = A(-2 + 3) + B(0) + C(0) \]
\[ 2 = A(1) \implies A = 2 \]

Finding C: Let \( x = -3 \).
\[ -3 + 4 = A(0) + B(0) + C(-3 + 2)^2 \]
\[ 1 = C(-1)^2 \implies 1 = C(1) \implies C = 1 \]

Finding B: Let's compare the coefficient of \( x^2 \) on both sides.
On the LHS, the coefficient of \( x^2 \) is 0.
On the RHS, \( x^2 \) terms come from \( B(x^2 + \dots) \) and \( C(x^2 + \dots) \).
\[ 0 = B + C \]
Substitute \( C = 1 \):
\[ 0 = B + 1 \implies B = -1 \]

Calculating A + B + C:
\[ A + B + C = 2 + (-1) + 1 = 2 \]

Step 4: Final Answer:
The value of \( A + B + C \) is 2.
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