If frac{x^2}{2 x^4+7 x^2+6}=frac{A x+B}{x^2+a | vedclass

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(D) Given the partial fraction decomposition: $\frac{x^2}{2 x^4+7 x^2+6}=\frac{A x+B}{x^2+a}+\frac{C x+D}{a x^2+3}$.First, factor the denominator: $2

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$4$

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(D) Given the partial fraction decomposition: $\frac{x^2}{2 x^4+7 x^2+6}=\frac{A x+B}{x^2+a}+\frac{C x+D}{a x^2+3}$.First, factor the denominator: $2 x^4+7 x^2+6 = (x^2+2)(2 x^2+3)$.Comparing this with the given form, we identify $a=2$.Now, express the fraction as: $\frac{x^2}{(x^2+2)(2 x^2+3)} = \frac{P}{x^2+2} + \frac{Q}{2 x^2+3}$.Using partial fractions: $x^2 = P(2 x^2+3) + Q(x^2+2)$.For $x^2 = -2$: $-2 = P(-4+3) \implies -2 = -P \implies P = 2$.For $x^2 = -3/2$: $-3/2 = Q(-3/2+2) \implies -3/2 = Q(1/2) \implies Q = -3$.Thus, $\frac{x^2}{2 x^4+7 x^2+6} = \frac{2}{x^2+2} - \frac{3}{2 x^2+3}$.Comparing with $\frac{A x+B}{x^2+2} + \frac{C x+D}{2 x^2+3}$, we get $A=0, B=2, C=0, D=-3$.Calculating $A+B+C-2 D = 0+2+0-2(-3) = 2+6 = 8$.Since $a=2$, $4a = 4(2) = 8$.Therefore, $A+B+C-2 D = 4 a$.

If $\frac{x^2}{2 x^4+7 x^2+6}=\frac{A x+B}{x^2+a}+\frac{C x+D}{a x^2+3}$, then find the value of $A+B+C-2 D$. (in $a$)

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