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

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(D) Given the partial fraction decomposition: $\frac{3 x^2+ax+3}{(2 x+3)(x^2+2)}=\frac{3}{2 x+3}+\frac{Bx+C}{x^2+2}$ Equating the numerators after tak

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$2$

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(D) Given the partial fraction decomposition: $\frac{3 x^2+ax+3}{(2 x+3)(x^2+2)}=\frac{3}{2 x+3}+\frac{Bx+C}{x^2+2}$ Equating the numerators after taking the common denominator: $3 x^2+ax+3 = 3(x^2+2) + (Bx+C)(2x+3)$ $3 x^2+ax+3 = 3x^2 + 6 + 2Bx^2 + 3Bx + 2Cx + 3C$ $3 x^2+ax+3 = (3+2B)x^2 + (3B+2C)x + (6+3C)$ Comparing the coefficients of $x^2$, $x$, and the constant term: $1) \ 3+2B = 3 \implies 2B = 0 \implies B = 0$ $2) \ 3B+2C = a \implies 3(0)+2C = a \implies 2C = a \implies C = \frac{a}{2}$ $3) \ 6+3C = 3 \implies 3C = -3 \implies C = -1$ Substituting $C = -1$ into $C = \frac{a}{2}$: $-1 = \frac{a}{2} \implies a = -2$ Finally, calculating $a(B+C)$: $a(B+C) = -2(0 + (-1)) = -2(-1) = 2$ Thus, the correct option is $D$.

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

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