Using Descartes' Rule of Signs, we can tell that the polynomial P(x) = x5 − 2x4 + 5x3 − x2 + 4x − 4 has, from smallest to largest, 1 , 2 , or 3 positive real zeros and 4 negative real zeros.

Accepted Solution

Answer:In the question,According to Descartes' Rule of signs change we can say that the number of sign changes of the co-efficient of the polynomial is the number of positive zeroes of the polynomial.And,On putting x = -x,The number of sign changes a polynomial obtained has, is equal to the number of negative zeroes.So,In the polynomial,[tex]P(x)=x^{5}-2x^{4}+5x^{3}-x^{2}+4x-4[/tex]So, we can see that the number of sign changes are from 1 to -2, -2 to 5, 5 to -1, -1 to 4, 4 to -4.So, there are 5 number of co-efficient sign changes taking place.Therefore, there are 5 positive zeroes.Now, at x = -x,[tex]P(-x)=(-x)^{5}-2(-x)^{4}+5(-x)^{3}-(-x)^{2}+4(-x)-4\\P(-x)=-x^{5}-2x^{4}-5x^{3}-x^{2}-4x-4[/tex]Here, we can see that there are no sign changes in the polynomial.Therefore, there are 0 negative zeroes.