Problem:
Which of the following polynomials has the greatest real root?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
By Descartes' Rule of Signs, none of these polynomials has a positive root, and each one has exactly one negative root. Because each polynomial is positive at and negative at , it follows that each has exactly one root between -1 and 0 . Note also that each polynomial is increasing throughout the interval . Because for all in the interval , it follows that the polynomial in choice is greater than the polynomial in choice on that interval, which implies that the root of the polynomial in choice is less than the root of the polynomial in choice . Because for all in the interval , it follows that the polynomial in choice is greater than the polynomial in choice on that interval, which implies that the root of the polynomial in choice is less than the root of the polynomial in choice and therefore less than the root of the polynomial in choice . The same reasoning shows that the root of the polynomial in choice is less than the root of the polynomial in choice .
Furthermore, on the interval , so , from which it follows that . Therefore the polynomial in choice is less than on the interval . The polynomial in choice has root . Bernoulli's Inequality shows that for all , which implies that
so the polynomial in choice is negative at the root of the polynomial in choice . This shows that the root of the polynomial in choice is greater than the root in choice .
Because the unique real root of the polynomial in choice is greater than the unique root of the polynomial in each of the other choices, that polynomial has the greatest real root.
The problems on this page are the property of the MAA's American Mathematics Competitions