If frac{3x+2}{(x+1)(2x^2+3)} = frac{A}{x+1} + | vedclass

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(B) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ Multiplying both sides by $(x+1)(2x^2

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$2$

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(B) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ Multiplying both sides by $(x+1)(2x^2+3)$, we get: $3x+2 = A(2x^2+3) + (Bx+C)(x+1)$ Setting $x = -1$: $3(-1)+2 = A(2(-1)^2+3) + 0 \Rightarrow -1 = A(5) \Rightarrow A = -\frac{1}{5}$ Expanding the right side: $3x+2 = 2Ax^2 + 3A + Bx^2 + Bx + Cx + C = (2A+B)x^2 + (B+C)x + (3A+C)$ Comparing coefficients: For $x^2$: $2A+B = 0 \Rightarrow B = -2A = -2(-\frac{1}{5}) = \frac{2}{5}$ For $x$: $B+C = 3 \Rightarrow C = 3 - B = 3 - \frac{2}{5} = \frac{13}{5}$ Calculating $A+C-B$: $-\frac{1}{5} + \frac{13}{5} - \frac{2}{5} = \frac{13-2-1}{5} = \frac{10}{5} = 2$

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

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