are exactly n positive zeros; in fact, it is well-known that for a polynomial of degree n , the number of zeros cannot be more than n and hence the number of positive zeros cannot be more than n . . . (b) Bounds for the number of Imaginary (Nonreal Complex)roots Using the Descartes rule, we can compute a lower bound for the number of imaginary roots.
Let m denote the number of sign changes in coefficients of P x ( )of degree n ; let k denote the number of sign changes in coefficients of P . Then there are at least n m ) imaginary roots for the polynomial P x ( ). Using the other conclusion of the rule, namely, the difference between the number of roots and the corresponding sign changes is even, we can sharpen the bounds in particular cases. Example .
Show that the polynomial has at least six imaginary roots. Clearly there are sign changes for the given polynomial P x ( ) and hence number of positive roots of P x ( ) cannot be more than two. Further, as P there is one sign change for P and hence the number of negative roots cannot be more than one. Clearly is not a root.
So maximum number of real roots is and hence there are atleast six imaginary roots. Remark From the above discussion we note that the Descartes rule gives only upper bounds for the number of positive roots and number of negative roots; the Descartes rule neither gives the exact number of positive roots nor the exact number of negative roots. But we can find the exact number of positive, negative and nonreal roots in certain cases. Also, it does not give any method to find the roots.
- - Theory of Equations Example . Discuss the nature of the roots of the following polynomial equations: (i) x (ii) x Let P x ( ) be the polynomial under consideration. (i) The number of