begin{aligned} & text{If } frac{4x^2+5x^4+7}{ | vedclass

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(A) Given the partial fraction decomposition: $\frac{5x^4+4x^2+7}{(x^2+1)(x^4+x^2+1)} = \frac{Ax+B}{x^2+1} + \frac{Cx^3+Dx^2+Ex+F}{x^4+x^2+1}$ Multipl

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(A) Given the partial fraction decomposition: $\frac{5x^4+4x^2+7}{(x^2+1)(x^4+x^2+1)} = \frac{Ax+B}{x^2+1} + \frac{Cx^3+Dx^2+Ex+F}{x^4+x^2+1}$ Multiplying both sides by the denominator $(x^2+1)(x^4+x^2+1)$, we get: $5x^4+4x^2+7 = (Ax+B)(x^4+x^2+1) + (Cx^3+Dx^2+Ex+F)(x^2+1)$ Expanding the right side: $5x^4+4x^2+7 = Ax^5+Ax^3+Ax + Bx^4+Bx^2+B + Cx^5+Cx^3+Dx^4+Dx^2+Ex^3+Ex+Fx^2+F$ Grouping the coefficients of like powers of $x$: $5x^4+4x^2+7 = (A+C)x^5 + (B+D)x^4 + (A+C+E)x^3 + (B+D+F)x^2 + (A+E)x + (B+F)$ Comparing coefficients on both sides: $1) A+C = 0$ $2) B+D = 5$ $3) A+C+E = 0$ $4) B+D+F = 4$ $5) A+E = 0$ $6) B+F = 7$ From $(1)$ and $(3)$, since $A+C=0$, we get $E=0$. From $(5)$, $A+E=0 \implies A=0$, which implies $C=0$. From $(2)$, $B+D=5$. From $(4)$, $(B+D)+F=4 \implies 5+F=4 \implies F=-1$. From $(6)$, $B+F=7 \implies B-1=7 \implies B=8$. Then $D = 5-B = 5-8 = -3$. Thus, $A=0, B=8, C=0, D=-3, E=0, F=-1$. Calculating the expression: $B+2(D+F+E)-C \cdot A = 8 + 2(-3-1+0) - 0 \cdot 0 = 8 + 2(-4) - 0 = 8-8 = 0$.

$\begin{aligned} & \text{If } \frac{4x^2+5x^4+7}{(x^2+1)(x^4+x^2+1)} = \frac{Ax+B}{x^2+1} \\ & + \frac{Cx^3+Dx^2+Ex+F}{x^4+x^2+1}, \text{ then } \\ & B+2(D+F+E)-C \cdot A = \end{aligned}$

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