If the partial fraction decomposition of frac | vedclass

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(C) Given the partial fraction decomposition:$\frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{A}{x^2+1} + \frac{B}{(x^2+1)^2} + \frac{C}{(x^2+1)^3}$Multiplying

Vedclass · Accepted Answer

$25$

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(C) Given the partial fraction decomposition:$\frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{A}{x^2+1} + \frac{B}{(x^2+1)^2} + \frac{C}{(x^2+1)^3}$Multiplying both sides by $(x^2+1)^3$,we get:$x^4+24x^2+28 = A(x^2+1)^2 + B(x^2+1) + C$$x^4+24x^2+28 = A(x^4+2x^2+1) + B(x^2+1) + C$$x^4+24x^2+28 = Ax^4 + (2A+B)x^2 + (A+B+C)$Comparing the coefficients of $x^4$,$x^2$,and the constant term:$1 = A$$24 = 2A+B$ $\Rightarrow 24 = 2(1)+B$ $\Rightarrow B = 22$$28 = A+B+C$ $\Rightarrow 28 = 1+22+C$ $\Rightarrow C = 5$Finally,calculating $B-2A+C$:$B-2A+C = 22 - 2(1) + 5 = 22 - 2 + 5 = 25$

If the partial fraction decomposition of $\frac{x^4+24x^2+28}{(x^2+1)^3}$ is $\frac{A}{x^2+1}+\frac{B}{(x^2+1)^2}+\frac{C}{(x^2+1)^3}$,then $B-2A+C=$

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