Solution: 5.8.4.5 Weare trying to factor the polynomial s 4 +23s 3 +174s 2 +472s +320 Wedosoby nding the roots of the equation s 4 +23s 3 +174s 2 +472s+320=0 Wecan reformulate this as a root locus problem as follows. Divide both sides of the equation by s 4 +23s 3 +174s 2 ;; to obtain s 4 +23s 3 +174s 2 s 4 +23s 3 +174s 2 + 472s +320 s 4 +23s 3 +174s 2 =0;; or 1+ 472(s+(320=472)) s 4 +23s 3 +174s 2 =0: Wecaneasily factor s 4 +23s 3 +174s 2 ,toobtain 1+ 472(s+320=472) s 2 (s +11:5;j6:4614)(s+11:5+j6:4614) =0: Nowconsider the root locus problem 1+ K(s +320=472) s 2 (s +11:5;j6:4614)(s+11:5+j6:4614) =0: The root locus is shown in Figure 1. The root locus was determined by searching along the real axis on the interval [;12;;0] looking for break-in and break-out points, and for values of K =472.Tab. 1 shows the results of the search. s -12 -10 -9 -8 -6 -5 -4 -3 -2 -1 K 534.2 472 467.2 472 487 485.9 472 441.9 399.4 472 Table 1: SerarchFor Break-in and Break-out Points From the table weseethat there is a break-in around s = ;2and another break-in pointaround s = ;9. From the table it is clear that the factorization of the polynomial is s 4 +23s 3 +174s 2 +472s+320=(s +1)(s + 4)(s +8)(s +10) 1 Im(s) Re(s) (-320/472) 2 -11.5 6.4614 Figure 1: Root locus 2