What is the possible number of negative real roots of the function f(x) = x⁵ - 2x³ + 7x² + 2x - 2?

Solution:

The Descartes’ Rule of of Signs states that the number of positive roots is at the most equal to the number of sign changes.

It could also be less than that by 2, 4 … i.e. If there are k sign changes in f(x) the number of positive roots could be k, k - 2, k - 4 ..

This rule can be extended to negative roots as follows.

The number of negative roots of f(x) is equal to the number of positive roots of g(x) = f(-x) = 0.

Therefore the negative real roots of the given equation are two or zero.

To find the negative real roots of the function can be obtained by first replacing x by -x as shown below :

f(-x) = (-x)⁵ - 2(-x)³ + 7(-x)² + 2(-x) - 2

f(-x) = -x⁵ + 2x³ + 7x² - 2x - 2 --- (1)

To determine the real roots of equation (1) one has to determine the number of sign changes which occur in the equation.

The number of times the sign changes is 2.

Hence the possible number of negative real roots are maximum 2.

There is a possibility of even having no negative real roots.

Therefore the number of negative real roots is either 2 or zero.

---

What is the possible number of negative real roots of the function f(x) = x⁵ - 2x³ + 7x² + 2x - 2?

Summary:

The possible number of negative roots of the given function x⁵ - 2x³ + 7x² + 2x - 2 is either 2 or zero.