Solution 4.6.2.1 The characteristic polynomial is p(s)=s 3 ;2s 2 ;5s +6: The initial Routh table is s 3 1 -5 0 s 2 -2 6 0 s 1 b 1 b 2 0 s 0 c 1 0 0 . Then b 1 = ;Det " 1 ;5 ;2 6 # ;2 = ;(6;10) ;2 = ;2;; b 2 = ;Det " 1 0 ;2 0 # ;2 = ;(0;0) 3 =0 c 1 = ;Det " ;2 6 ;2 0 # ;2 = ;(0 + 12) ;2 =6: The nal Routh table is s 3 1 ;5 0 s 2 ;2 6 0 s 1 ;2 0 0 s 0 6 0 0 . There sign changes, positivetonegativeaswegofrom row one to row2,and negativebacktopositiveaswegofromrow3torow4.Thus there are two roots in the righthalf plane. This is veri ed by the MATLAB dialogure: EDU>p = [1 -2 -5 6] p= 1 -2 -5 6 1 EDU>roots (p) ans = 3.00000000000000 -2.00000000000000 1.00000000000000 EDU> 2