Solution 5.8.4.2 Wewish to factor the polynomial s 3 +22s 2 +152s +320 Using root locus techniques. Wedosoby nding the roots of the equation s 3 +22s 2 +152s +320=0: Wecan reformulate this as a root locus problem as follows. Divide both sides of the equation by s 3 +22s 2 ;; to obtain s 3 +22s 2 s 3 +22s 2 + 152s +320 s 3 +22s 2 =0;; or 1+ 152(s+320=152) s 3 +22s 2 =0: Wecaneasily factor s 3 +22s 2 ,toobtain 1+ 152(s+320=152) s 2 (s +22) =0: Nowconsider the root locus problem 1+ K(s +320=152) s 2 (s +22) =0: The root locus is shown in Figure 1. The root locus shows the solutions to the equation s 3 +22s 2 + K(s +320=152) = 0;; for all possible values of K. What wewantisthesolution for K =152. From the root locus weseethat for K =152 there will either be three real roots, or one real root and twocomplex roots. Further, weknowthat the real roots lie between s = ;22 and s = ;152=320. So webegin by searching along the real axis to the left of s = ;152=320 for the point where the gain is 152. We rstguess s = ;6. This is a pure guess, but once we evaluate K for s = ;6, wewillhave information about our next guess. Figure 2 shows howwecalculate K for s = ;6. 1 Im(s) Re(s) -22 -(152/320) 2 Figure 1: Root Locus Im(s) Re(s) -22 -(152/320) 2 s + 22 s = 152/320 s Figure 2: Gain Calculation 2 For s = ;6we get K = [ jsjjsjjs+22j js+320=152j ] j s=;6 = 147:9 The value of K wehavefound is too small. This tells us that the root we are looking for is between s = ;6ands = ;22. So wenext guess s = ;8, for whichweobtain: K = [ jsjjsjjs+22j js+320=152j ] j s=;8 = 152 Wehavenowfound one of the roots of the poi;;ynomial. To nd the others, wedivide out the term s +8and apply the quadratic formula. s 3 +22s 2 +152s+320 s +8 = s 2 +14s 2 +40 = (s +4)(s+10) Thus s 3 +22s 2 +152s +320 = (s +4)(s + 8)(s +10) 3