If frac{2x^4-3x^2+4}{(x^2+1)(x^2+2)} = a + fr | vedclass

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(D) Let $u = x^2$. Then the expression is $\frac{2u^2-3u+4}{(u+1)(u+2)} = a + \frac{px+q}{u+1} + \frac{mx+n}{u+2}$.Performing polynomial division on t

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$\frac{n}{q}$ is undefined

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(D) Let $u = x^2$. Then the expression is $\frac{2u^2-3u+4}{(u+1)(u+2)} = a + \frac{px+q}{u+1} + \frac{mx+n}{u+2}$.Performing polynomial division on the left side: $\frac{2u^2-3u+4}{u^2+3u+2} = 2 + \frac{-9u}{u^2+3u+2} = 2 + \frac{-9u}{(u+1)(u+2)}$.Using partial fractions for $\frac{-9u}{(u+1)(u+2)} = \frac{A}{u+1} + \frac{B}{u+2}$.$-9u = A(u+2) + B(u+1)$.For $u = -1$,$-9(-1) = A(1) \implies A = 9$.For $u = -2$,$-9(-2) = B(-1) \implies B = -18$.So,$\frac{2x^4-3x^2+4}{(x^2+1)(x^2+2)} = 2 + \frac{9}{x^2+1} - \frac{18}{x^2+2}$.Comparing this with $a + \frac{px+q}{x^2+1} + \frac{mx+n}{x^2+2}$,we get $a = 2$,$p = 0$,$q = 9$,$m = 0$,$n = -18$.Thus,$\frac{n}{q} = \frac{-18}{9} = -2$.

If $\frac{2x^4-3x^2+4}{(x^2+1)(x^2+2)} = a + \frac{px+q}{x^2+1} + \frac{mx+n}{x^2+2}$,then $\frac{n}{q} =$

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