We have,
$\begin{aligned}
& \frac{4 x^2+5 x^4+7}{\left(x^2+1\right)\left(x^4+x^2+1\right)}=\frac{A x+B}{x^2+1}+\frac{C x^3+D x^2+E x+F}{x^4+x^2+1} \\
& \Rightarrow 4 x^2+5 x^4+7=(A x+B)\left(x^4+x^2+1\right) \\
& +\left(C x^3+D x^2+E x+F\right)\left(x^2+1\right) \\
& \Rightarrow 5 x^4+4 x^2+7=A x^5+A x^3+A x+B x^4+B x^2 \\
& +B+C x^5+C x^3+D x^4+D x^2+E x^3+E x+F x^2+F \\
& \Rightarrow 5 x^4+4 x^2+7=(A+C) x^5+(B+D) x^4 \\
& +(A+C+E) x^3+(B+D+F) x^2+(A+E) x+(B+F) \\
\end{aligned}$
On comparing, we get
$\begin{aligned}
& A+C=0 \\
& B+D=5
\end{aligned}$
$\begin{array}{r}
A+C+E=0 \\
B+D+F=4 \\
A+E=0 \\
B+F=7
\end{array}$
On solving, we get
$\begin{gathered}
A=C=E=0, B=8 . D=-3, F=-1 \\
\therefore B+2(D+F+E)-C \cdot A=8+2(-3-1+0)-0 \times 0 \\
=8-8=0
\end{gathered}$