Solution:

$ \begin{aligned} \text { i) }\quad F(s) &=s^4+s^3+4 s^2+2 s+3\\ m(s) &=s^4+4 s^2+3 \\ n(s) &=s^3+2 s \end{aligned} $

$ \begin{array}{c|ccc} s^4 & 1 & 4 & 3 \\\\ \hline s^3 & 1 & 2 & - \\\\ \hline s^2 & 2 & 3 & - \\\\ \hline s^1 & 1 / 2 & - \\\\ \hline s^0 & 3 & \end{array} \\ $

All elements in the first column are +Ve \& non-zero. Hence given equation is Hurwitz's polynomial.

$ \begin{aligned} \\ \text { ii) } \quad F(s) &=(s+2)^3 \\\\ &=s^3+3(s)(2)^2+3\left(s^2(2)+(2)^3\right.\\\\ &=s^3+12 s+6 s^2+8 \\\\ &=s^3+6 s^2+12 s+8 \\\\ m(s) &=6 s^2+28 ; n(s)=s^3+12 s \\ \end{aligned} \\ $

$ \begin{array}{c|c} \\ s^3 & 1 & 12 \\\\ \hline s^2 & 6 & 8 \\\\ \hline s^1 & \frac{32}{3} & - \\\\ \hline s^0 & 8 \\ \end{array} \\ $

All elements in first coloumn are +ve \& non zero. Hence it is Hurwitz's polynomial.