Determine all the possible number of positive and negative real zeros for h(x) = x^3 - 2 x^2 - 4x + 5

The function, h(x) = x³ - 2x² - 4x + 5 has 2 or 0 positive real zeros and exactly 1 negative real zero.

Step-by-step 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.

answer: it is d

step-by-step explanation:

any options?

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