Solution: 5.8.4.3 Weare trying to factor the polynomial s 4 +11s 3 +41s 2 +61s +30 Wedosoby nding the roots of the equation s 4 +11s 3 +41s 2 +61s +30=0: Wecan reformulate this as a root locus problem as follows. Divide both sides of the equation by s 4 +11s 3 +41s 2 ;; to obtain s 4 +11s 3 +41s 2 s 4 +11s 3 +41s 2 + 61s +30 s 4 +11s 3 +41s 2 =0;; or 1+ 61(s+30=61) s 3 +22s 2 +41s 2 =0: Wecaneasily factor s 4 +11s 3 +41s 2 ,toobtain 1+ 61(s+30=61) s 2 (s +5:5;j3:279)(s+5:5+j3:279) =0: Nowconsider the root locus problem 1+ K(s +30=61) s 2 (s +5:5;j3:279)(s+5:5+j3:279) =0: The root locus is shown in Figure 1. The root locus was determined by searching along the real axis to the left to the left of the zero at [s = ;30=61 looking for break-in and break-out points. Tab. 1 shows the results of the search. s -9 -5 -4.5 -4 -3.5 -2.5 -2 -1.5 -1 K 268 61.01 59.37 59.3 60.07 61.47 61 59.7 61 Table 1: SearchFor Break-in and Break-out Points From the table we see that there is a break-in around s = ;4forK = 59:3, a break-out around s = ;2:5forK =61:47 and a break-n around s = ;1:5forK =59:7. Since wearelooking for a gain of 61, wemerely havetosearchinthe vicinityofthe break-in in break-out points. A little work shows that the factorization of the polynomial is s 4 +11s 3 +41s 2 +61s+30=(s +1)(s+2)(s +3)(s +5) 1 j Im(s) Re(s) (-30/61) 2 -5.5 j3.279 Figure 1: Root Locus 2