Question

Determine the possible number of positive real zeros and negative real zeros for each polynomial function given by Descartes' Rule of Signs.
P(x) = -3x^3 + 11x^2 + 12x - 8

Answer 1

To apply Descartes' Rule of Signs, we count the number of sign changes in the coefficients of the polynomial.

For P(x) = -3x^3 + 11x^2 + 12x - 8, we have:

Sign changes from -3x^3 to 11x^2: 1 sign change (negative to positive)

Sign changes from 11x^2 to 12x: 0 sign changes

Sign changes from 12x to -8: 1 sign change (positive to negative)

So, there are 1 or 3 positive real zeros for this polynomial.

Next, we consider the sign changes in the coefficients when we substitute -x for x in the polynomial.

For P(-x) = -3(-x)^3 + 11(-x)^2 + 12(-x) - 8:

Simplifying, we have P(-x) = -3x^3 + 11x^2 - 12x - 8

Sign changes from -3x^3 to 11x^2: 1 sign change (negative to positive)

Sign changes from 11x^2 to -12x: 1 sign change (positive to negative)

Sign changes from -12x to -8: 0 sign changes

So, there are 1 or 1 negative real zero for this polynomial.

Therefore, the possible number of positive real zeros is 1 or 3, and the possible number of negative real zeros is 1.