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.
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
1 answer