Solution:

$ N(s)=2 s^4+7 s^3+11 s^2+12 s+4 $

$ \begin{array}{c|ccc} s^4 & 2 & 11 & 4 \\\\ \hline s^3 & 7 & 12 & - \\\\ \hline s^2 & \frac{53}{7} & 4 & - \\\\ \hline s^1 & \frac{440}{53} & - & - \\\\ \hline s^0 & 4 & & \\ \end{array} \\ $

$N(s)$ is Hurwitz's polynomial

$ D(s)=s^4+5 s^3+9 s^2+11 s+6 \\ $

$ \begin{array}{c|ccc} s^4 & 1 & 9 & 6 \\\\ \hline s^3 & 5 & 11 & - \\\\ \hline s^2 & \frac{34}{s} & 6 & - \\\\ \hline s^1 & \frac{112}{17} & - & - \\\\ \hline s^0 & 6 & & \\\\ \end{array} \\ $

D(S) is Hurwitz's polynomial Determination of poles

$ \quad D(S)=0 \\ $

$ \therefore s^4+5 s^3+9 s^2+11 s+6=0 \\ $

Using synthetic division

$ \begin{array}{c|ccccc} -1 & 1 & 5 & 9 & 11 & 6 \\\\ & & -1 & -4 & -5 & -6 \\\\ \hline & 1 & 4 & 5 & 6 & 0 \\\\ & & & & \\ \end{array} \\ $

$\therefore \quad s^4+5 s^3+9 s^2+11 s+6=(s+1)\left(s^3+4 s^2+5 s+6)\right.$ $\quad=(s+1)(s+3)(s+0.5+11.32)(s+0.5j)$

poles are at

$ -1,-3,-0.5-j 1.32 \&+0.5-j 1.32 \\ $

i.e roots are complex & real & -ve Hence residuals are not required

checking of $A(w)$ function, where

$ \begin{aligned} \\ A(w)=m_1 m_2-n_1 n_2 \\\\ m_1=&\left(2 s^4+11 s^2+4\right) \quad m_2=\left(s^4+9 s^2+6\right) \\\\ n_1=&\left(7 s^3+12 s\right) \quad n_2=\left(5 s^3+11 s\right) \\\\ A(s)=&\left(2 s^4+11 s^2+4\right)\left(s^4+9 s^2+6\right)-(7 s^3+12s + 5 s^3+11 s) \\\\ =&(2 s^8+18 s^6+12 s^4+11 s^6+9 s^4 s^4+66 s^2+4 s^4)+(36 s^2+24)-(35 s^6+77 s^4+60 s^4+132 s^2) \\\\ =& 2 s^8-6 s^6-22 s^4-30 s^2+24 \\ \end{aligned} \\ $

$ \begin{aligned} \\ &\text { put } s=j w^4 \\\\ &\therefore A(w)=2(j w)^8-6(j w)^6-22(j \omega)^4-30\left(j \omega^2)+24\right. \\\\ &=2 w^8+6 w^6-22 w^4+30 w^2+24 \\\\ &\text { it is clear that for all values } \\\\ &\text { of } w \text { A(w) } \geqslant 0 \\\\ &\text { Hence all canditions are } \\\\ &\text { satisfird } \\\\ &\therefore \text { it is P.R.F. } \\ \end{aligned} \\ $