Answer:
The function, h(x) = x³ - 2x² - 4x + 5 has 2 or 0 positive real zeros and exactly 1 negative real zero.
Explanation:
Using Descartes' Rule of Signs
For the positive-root case, the sign on x is not changed:
h(x) = x³ - 2x² - 4x + 5
Counting the number of sign changes while ignoring the actual coefficients; there are two sign changes.
Thus, the function, h(x) has 2 or 0 positive real zeros
For the negative root case, h(-x), that is changing the sign on x
h(-x) = (-x)³ - 2(-x)² - 4(-x) + 5
h(-x) = -x³ - 2x² + 4x + 5
Counting the number of sign changes while ignoring the actual coefficients; there is only one sign change.
Thus, the function, h(x) has exactly 1 negative real zero.