and For a fourth-order polynomial , all the coefficients must satisfy , and (When this is derived you do not know all coefficients should be positive, and you add .) In general the Routh stability criterion states a polynomial has all roots in the open left half-plane if and only if all first-column elements of the Routh array have the same sign. All coefficients being positive (or all negative) is necessary for all roots to be located in the open left half-plane. That is why here is fixed to 1, which is positive. When this is assumed, we can remove from fourth-order polynomial, and conditions for fifth- and sixth-order can be simplified. For fifth-order we only need to check that and for sixth-order we only need to check and this is further optimised in Liénard–Chipart criterion. Indeed, some coefficients being positive is not independent with principal minors being positive, like check can be removed for third-order polynomial. Higher-order example A tabular method can be used to determine the stability when the roots of a higher order characteristic polynomial are difficult to obtain. For an nth-degree polynomial the table has n + 1 rows and the following structure: where the elements and can be computed as follows: When completed, the number of sign changes in the first column will be the number of non-negative roots. In the first column, there are two sign changes (0.75 → −3, and −3 → 3), thus there are two non-negative roots where the system is unstable. The characteristic equation of a servo system is given by: for stability, all the elements in the first column of the Routh array must be positive. So the conditions that must be satisfied for stability of the given system as follows: We see that if then Is satisfied. We have the following table : there are two sign changes. The system is unstable, since it has two right-half-plane poles and two left-half-plane poles. The system cannot have jω poles since a row of zeros did not appear in the Routh table. Sometimes the presence of poles on the imaginary axis creates a situation of marginal stability. In that case the coefficients of the "Routh array" in a whole row become zero and thus further solution of the polynomial for finding changes in sign is not possible. Then another approach comes into play. The row of polynomial which is just