If the middle term in the expansion of ${\left(\frac{\mathrm{x}}{2}+2\right)}^8$is 1120, then the product of possible real values of x is ____.

Middle term is ${\left(\frac{\mathrm{n}}{2}+1\right)}^{\text{th }}\text{, i.e., }(4+1{)}^{\text{th }}\text{, i.e., }{\mathrm{T}}_5$
$\therefore {\mathrm{T}}_5{=}^8{\mathrm{C}}_4{\left(\frac{\mathrm{x}}{2}\right)}^4\times 2^4=1120$
or    ${\mathrm{x}}^4=\frac{8\times 7\times 6\times 5}{1\times 2\times 3\times 4}{\mathrm{x}}^4=1120$
$\begin{array}{l}\Rightarrow {\mathrm{x}}^4=\frac{1120}{70}=16\\ \Rightarrow \left({\mathrm{x}}^2+4\right)\left({\mathrm{x}}^2-4\right)=0\end{array}$
Therefore, $\mathrm{x}=\pm 2\text{ only as }\mathrm{x}\in \mathrm{R}$