If frac{x^4+24 x^2+28}{left(x^2+1right)^3}=fr | vedclass

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Question

(C) Given $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3} = \frac{A x+B}{x^2+1} + \frac{C x+D}{\left(x^2+1\right)^2} + \frac{E x+F}{\left(x^2+1\right)^3}$

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(C) Given $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3} = \frac{A x+B}{x^2+1} + \frac{C x+D}{\left(x^2+1\right)^2} + \frac{E x+F}{\left(x^2+1\right)^3}$.Multiplying both sides by $(x^2+1)^3$,we get:$x^4+24 x^2+28 = (A x+B)(x^2+1)^2 + (C x+D)(x^2+1) + (E x+F)$.Expanding the right side:$x^4+24 x^2+28 = (A x+B)(x^4+2x^2+1) + (C x^3+D x^2+C x+D) + E x+F$.$x^4+24 x^2+28 = A x^5 + B x^4 + 2A x^3 + 2B x^2 + A x + B + C x^3 + D x^2 + C x + D + E x + F$.Grouping the powers of $x$:$x^4+24 x^2+28 = A x^5 + B x^4 + (2A+C) x^3 + (2B+D) x^2 + (A+C+E) x + (B+D+F)$.Comparing coefficients on both sides:$A = 0$$B = 1$$2A+C = 0 \Rightarrow C = 0$$2B+D = 24$ $\Rightarrow 2(1)+D = 24$ $\Rightarrow D = 22$$A+C+E = 0$ $\Rightarrow 0+0+E = 0$ $\Rightarrow E = 0$$B+D+F = 28$ $\Rightarrow 1+22+F = 28$ $\Rightarrow F = 5$.Therefore,$A+B+C+D+E+F = 0+1+0+22+0+5 = 28$.

If $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3}$,then the value of $A+B+C+D+E+F=$

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