Solution 5.8.4.1 We wish to factor the polynomial s 3 +9s 2 +20s +12: Wedosoby nding the roots of the equation s 3 +9s 2 +20s +12=0: Wecan reformulate this as a root locus problem as follows. Divide both sides of the equation by s 3 +9s 2 +20s;; to obtain s 3 +9s 2 +20s s 3 +9s 2 +20s + 12 s 3 +9s 2 +20s =0;; or 1+ 12 s 3 +9s 2 +20s =0: Wenowfactor s 3 +9s 2 +20s. s 3 +9s 2 +20s = s(s 2 +9s + 20) = s(s +4)(s+5) Thus wehave 1+ 12 s(s + 4)(s+5) =0: Nowconsider the root locus problem 1+ K s(s + 4)(s+5) =0: The root locus is shown in Figure 1. The root locus shows the solutions to the equation s 3 +9s 2 +20s + K =0;; for all possible values of K. What wewantisthesolution for K =12. From the root locus we see that for K =12therewilleither be three real roots, or one real root and twocomplex roots. Further, weknowthat the real root will be greater than 5. So webegin by searching along the real axis to the left of s = ;5forthe pointwherethe gain is 12. We rst guess s = ;8. This is a pure guess, but once weevaluate K for s = ;8, wewill have information about our next guess. 1 Im(s) Re(s) -5 -4 Figure 1: Root Locus j Im(s) Re(s) -5 -4 -8 s + 5 s + 4 s Figure 2: Gain Calculation 2 Figure 2 shows howwecalculate K for s = ;8. We know that K s(s +4)(s+5) j s=;8 =1 6 ;180  : To separate the angle and magnitude information, wewrite the left hand side of the equation in polar form: K (jsj 6 s)(js +4j 6 s +4)(js+5j 6 s +5) = 1 6 ;180  : But, from Figure 2, we knowthat K (jsj 6 s)(js +4j 6 s +4)(js +5j 6 s +5) = K (jsj)(js+4j)(js+5j)( 6 s + 6 s +4+ 6 s +5) = K jsjjs+4jjs+5j 6 ;540  = K jsjjs+4jjs+5j 6 ;180  : Clearly, K jsjjs +4jjs +5j =1;; K = jsjjs +4jjs +5j: For s = ;8weget K = [jsjjs+4jjs+5j] j s=;8 = 8 4 3 = 96: The value of K wehavefound is too big. This tells us that the root we are looking for is between s = ;8 and s = ;5. So we next guess s = ;7, for whichweobtain: K = [jsjjs+4jjs+5j] j s=;7 = 7 3 2 = 42: The value is still too large, so weguesss = ;6, for whichweobtain: K = [jsjjs+4jjs+5j] j s=;6 = 6 2 1 = 12;; 3 whichisthegain wearelooking for, and wenowknow one of the roots of the polynomial. If wedivide out the known root, weobtain s 3 +9s 2 +20s +12 = (s +6)(s 2 +3s +2)) = (s +6)(s +1)(s +2) 4