8. In the expansion of $\left(4 x^{2}-2 x-5\right)\left(1+\frac{1}{x^{2}}\right)^{5}$, the constant term is $\qquad$
8. 15 .
$$
\text { Original expression }=\left(4 x^{2}-2 x-5\right)\left(1+\frac{5}{x^{2}}+\frac{10}{x^{4}}+\cdots\right) \text {. }
$$

Therefore, the constant term in the expanded form is $(-5) \times 1+4 \times 5=15$.