Solution: 5.8.4.4 Weare trying to factor the polynomial s 4 +17s 3 +80s 2 +100s Wedosoby nding the roots of the equation s 4 +17s 3 +80s 2 +100s =0: Clearly wehavearootats =0,sowereally only need to factor s 3 +17s 2 +80s +100=0: Wecan reformulate this as a root locus problem as follows. Divide both sides of the equation by s 3 +17s 2 ;; to obtain s 3 +17s 2 s 3 +17s 2 + 80s +100 s 3 +17s 2 =0;; or 1+ 80(s +1:25) s 3 +17s 2 =0: Wecaneasily factor s 3 +17s 2 ,toobtain 1+ 80(s +1:25) s 2 (s +17) =0: Nowconsider the root locus problem 1+ K(s +1:25) s 2 (s +17) =0: The root locus is shown in Figure 1. Wenowsolve the equation K = jsj 2 js +17j js +1:25j for some values of s along the real axis. We are looking for values of s for which K =80. In the process we nd there is a break-in pointaround s = ;3and a break-out pointnears = ;8. Wealsoidentify three values of s for which K =80,asshown in Table 1 shows the results of the search. Thus s 3 +17s 2 +80s +100=(s + 2)(s+5)(s +3)(s +10) 1 Re(s) -1.25 2 X -17 Im(s) Figure 1: Root Locus s -1 -2 -3 -4 -5 -6 -8 -10 -12 K 64 80 72 75.6 80 83.4 85.3 80 67 Table 1: SearchfroK =80 2