This explanation of Descartes’ Rule of Signs was submitted by Ayush, a HashLearn tutor from IIT, Dhanbad.
Consider a polynomial equation p(x)=0 (descending powers of x)
i) Number of positive roots of the equation p(x)=0 cannot exceed the number of changes of sign in p(x).
ii)Number of negative roots cannot exceed the number of changes of sign of p(-x).
Clearly, there is only one sign change in p(x) i.e., when sign changes from positive to negative between the terms 2×5 and -3x.
Hence p(x) has at most one positive root.
Similarly, in p(-x) there are two sign changes.
Hence p(x) has at most two negative roots.
Clearly, here there are no sign changes in p(x) hence it has no positive roots .
Also , there will be no sign changes in p(-x) as all the powers of x are even, hence it will not even have any negative real root.
Thus the given polynomial has no real roots.
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