If frac{x^2-7 x+2}{x^4+3 x^2+4}=frac{A x+B}{x | vedclass

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(B) Given $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$  ...$(i)$Factorize the denominator: $x^4+3 x^2+4 = (x^2+2)^2

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$2 a+b$

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(B) Given $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$  ...$(i)$Factorize the denominator: $x^4+3 x^2+4 = (x^2+2)^2 - x^2 = (x^2+x+2)(x^2-x+2)$.Thus,$\frac{x^2-7 x+2}{(x^2+x+2)(x^2-x+2)} = \frac{Px+Q}{x^2+x+2} + \frac{Rx+S}{x^2-x+2}$.Equating numerators: $x^2-7x+2 = (Px+Q)(x^2-x+2) + (Rx+S)(x^2+x+2)$.Expanding the right side: $x^2-7x+2 = x^3(P+R) + x^2(-P+Q+R+S) + x(2P-Q+2R+S) + (2Q+2S)$.Comparing coefficients:$P+R=0$$-P+Q+R+S=1$$2P-Q+2R+S=-7$$2Q+2S=2 \Rightarrow Q+S=1$Solving these,we get $P=0, Q=4, R=0, S=-3$.So,$\frac{x^2-7x+2}{x^4+3x^2+4} = \frac{4}{x^2+x+2} + \frac{-3}{x^2-x+2}$.Comparing with $(i)$,we have $a=1, b=-1$ (since $a>b$),$A=0, B=4, C=0, D=-3$.Then $B+D = 4-3 = 1$.Checking options: $2a+b = 2(1) + (-1) = 1$.Therefore,$B+D = 2a+b$.

If $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$ and $a>b$ then $B+D=$

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