Question

If \(\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1}+ \frac{Bx+C}{2x^2+3}\), then A - B + C=

• A. 2
• B. 1
• C. 3
• D. 6

Answer 1

The Correct Option is A

We are given that

$$ \frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3} $$

Step 1: Eliminating the Denominator
Multiplying both sides by $(x+1)(2x^2+3)$, we have:

$$ 3x+2 = A(2x^2+3) + (Bx+C)(x+1) $$

Step 2: Expanding the Right-Hand Side
Expanding the terms on the right-hand side:

$$ 3x+2 = 2Ax^2 + 3A + Bx^2 + Bx + Cx + C $$

Combining like terms:

$$ 3x+2 = (2A+B)x^2 + (B+C)x + (3A+C) $$

Step 3: Comparing Coefficients
By comparing the coefficients of $x^2$, $x$, and the constant terms, we obtain the following system of equations:

$$ \begin{aligned} 2A + B &= 0, \\ B + C &= 3, \\ 3A + C &= 2. \end{aligned} $$

Step 4: Solving the System of Equations
From the first equation, $B = -2A$. Substituting this into the second equation:

$$ -2A + C = 3. $$

We now solve the two equations:

$$ \begin{aligned} -2A + C &= 3, \\ 3A + C &= 2. \end{aligned} $$

Subtracting the first equation from the second:

$$ (3A + C) - (-2A + C) = 2 - 3, $$

$$ 3A + 2A = -1 \quad \Rightarrow \quad 5A = -1 \quad \Rightarrow \quad A = -\frac{1}{5}. $$

Using $A = -\frac{1}{5}$, we find $B$:

$$ B = -2A = -2\left(-\frac{1}{5}\right) = \frac{2}{5}. $$

Using $B = \frac{2}{5}$ in $B + C = 3$, we find $C$:

$$ \frac{2}{5} + C = 3 \quad \Rightarrow \quad C = 3 - \frac{2}{5} = \frac{15}{5} - \frac{2}{5} = \frac{13}{5}. $$

Step 5: Calculating $A - B + C$
We now compute $A - B + C$:

$$ A - B + C = -\frac{1}{5} - \frac{2}{5} + \frac{13}{5} = \frac{-1 - 2 + 13}{5} = \frac{10}{5} = 2. $$

Final Answer:
The final answer is ${2}$.