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: Understanding the Concept
This problem requires decomposing a rational function into its partial fractions. The denominator is already factored into a linear term and an irreducible quadratic term.
Step 2: Key Formula or Approach
To find the constants $a, b,$ and $c$, we first combine the fractions on the right side over a common denominator and then equate the numerators. \[ x+3 = a(x^2+2) + (bx+c)(x+1) \] We can find the constants by substituting convenient values of $x$ (like the root of the linear factor) and by comparing the coefficients of the powers of $x$.
Step 3: Detailed Explanation
We start with the identity: \[ x+3 = a(x^2+2) + (bx+c)(x+1) \] Finding 'a' (Cover-up method): To find 'a', we can substitute the root of its denominator, which is $x=-1$. \[ (-1)+3 = a((-1)^2+2) + (b(-1)+c)(-1+1) \] \[ 2 = a(1+2) + (-b+c)(0) \] \[ 2 = 3a \implies a = \frac{2}{3} \] Finding 'b' and 'c' (Comparing coefficients): Expand the right side of the identity: \[ x+3 = a(x^2+2) + bx(x+1) + c(x+1) \] \[ x+3 = ax^2 + 2a + bx^2 + bx + cx + c \] Group the terms by powers of $x$: \[ 0x^2 + 1x + 3 = (a+b)x^2 + (b+c)x + (2a+c) \] Now, equate the coefficients of corresponding powers of $x$ on both sides. Coefficient of $x^2$: \[ 0 = a+b \implies b = -a \] Since $a = 2/3$, we get $b = -2/3$. Coefficient of $x$: \[ 1 = b+c \implies c = 1-b \] Since $b = -2/3$, we get $c = 1 - (-2/3) = 1 + 2/3 = 5/3$. We can check our result with the constant term: Constant term: $3 = 2a+c$. $2(\frac{2}{3}) + \frac{5}{3} = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3$. This matches, so our values are correct. Calculate the final expression: We need to find the value of $a-b+c$. \[ a-b+c = \left(\frac{2}{3}\right) - \left(-\frac{2}{3}\right) + \left(\frac{5}{3}\right) \] \[ = \frac{2}{3} + \frac{2}{3} + \frac{5}{3} = \frac{2+2+5}{3} = \frac{9}{3} = 3 \] Step 4: Final Answer
The value of $a-b+c$ is 3.