Question

If $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$ then $a-b+c =$

• A. 0
• B. 1
• C. 3
• D. 2

Hint:

For partial fractions, using a combination of methods is often fastest. The "cover-up" method (substituting the root of a linear factor) is very quick for finding the coefficient corresponding to that factor. After that, comparing coefficients of the highest and lowest powers of $x$ is usually enough to find the remaining constants.

Answer 1

The Correct Option is C

Step 1: Partial Fraction Decomposition: The given identity is: \[ \frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2} \] Multiply both sides by \( (x+1)(x^2+2) \): \[ x+3 = a(x^2+2) + (bx+c)(x+1) \]
Step 2: Find 'a': Substitute \( x = -1 \) (the root of \( x+1 \)) into the equation: \[ -1 + 3 = a((-1)^2 + 2) + 0 \] \[ 2 = a(3) \implies a = \frac{2}{3} \]
Step 3: Find 'b' and 'c' by comparing coefficients: Expand the RHS: \[ x+3 = ax^2 + 2a + bx^2 + bx + cx + c \] \[ x+3 = (a+b)x^2 + (b+c)x + (2a+c) \] Comparing coefficient of \( x^2 \): \[ a+b = 0 \implies b = -a = -\frac{2}{3} \] Comparing constant term: \[ 2a + c = 3 \implies \frac{4}{3} + c = 3 \implies c = 3 - \frac{4}{3} = \frac{5}{3} \]
Step 4: Calculate \( a-b+c \): \[ a - b + c = \frac{2}{3} - \left(-\frac{2}{3}\right) + \frac{5}{3} \] \[ = \frac{2}{3} + \frac{2}{3} + \frac{5}{3} = \frac{9}{3} = 3 \]