Let f(x) be a polynomial of degree 5 such that $\mathrm{f}\left(\left|\mathrm{x}\right|\right)=0$  has 8 real distinct. Then number of real roots of f(x) = 0 is _______

$\begin{array}{l}\mathrm{Since} \mathrm{f}\left(\left|\mathrm{x}\right|\right)=0 \mathrm{has} 8 \mathrm{roots}\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right) =0 \mathrm{has} 4 \mathrm{real} \mathrm{positive} \mathrm{roots} \mathrm{but} \mathrm{since} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is}\\ \mathrm{polynomial} \mathrm{of} \mathrm{degree} 5\\ \Rightarrow \mathrm{fifth} \mathrm{root} \mathrm{is} \mathrm{also} \mathrm{real}.\\ \therefore \mathrm{Number} \mathrm{of} \mathrm{real} \mathrm{roots} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right)=0 \mathrm{is} 5\end{array}$