P(x) = x^4 + x^3 + x^2 + x + 11

Question

Use Descartes' Rule of Signs to determine how many positive and how many negative real zeros the polynomial can have. Then determine the possible total number of real zeros. (Enter your answers as comma-separated lists.)

number of positive zeros possible	0
number of negative zeros possible	(i got this wrong)
number of real zeros possible (i got this wrong too)

Answer 1

P(-x) = (-x)⁴ + (-x)³ + (-x)² + (-x) + 11 = x⁴ - x³ + x² - x + 11

P(-x) has 4 sign changes

So, by DesCartes Rule of Signs the number of possible negative zeros is 4 or 2 or none.

Answer 2

P(-x) = x⁴ - x³ + x² - x + 11, possilbe negative reals?