Suppose \(p\) is a nonconstant real polynomial such that for some real \(a\), \(p(a)\ne0\) and \(p'(a)=p''(a)=0\). Prove that \(p\) has at least one nonreal zero.
Solution
Observe that if \(q ( z )\) is a real-rooted polynomial with distinct roots, then by Rolle’s theorem \(q ^ { \prime } ( z )\) is also real-rooted (since it has degree one less than the degree of \(q )\) and has the property that between every two roots of \(q ^ { \prime }\) there is a root of q. Since polynomials with distinct roots are dense in the set of real-rooted polynomials, this implies that if \(q\) is any real-rooted polynomial and \(q ^ { \prime } ( z )\) has a double root at z then \(q ( z ) = 0\)
For the given polynomial, \(p ^ { \prime } ( z )\) has a double root at \(a\), but \(p ( a ) \neq 0\) , so \(p\) cannot be real-rooted.