Question

check whether - 1/3 are the zeros of cubic polynomial p(x) =3x^3-5x^2-11x-3​

Answer 1

Answer:    sum of product of zeroes taking two at a time= 3(-1) + (-1)(-1/3) + 3(-1/3) = -11/3 =�(cofficient of x)/cofficient of x3

Step-by-step explanation:    

3, -1, -1/3 will be zero of the cubic polynomial p(x)=3x3-5x2-11x-3 if p(x) = 0 at x= 3, -1, -1/3 so

p(3) = 3 * (3)3 - 5*(3)2 - 11*3 - 3 = 0

p(-1) = 3(-1)3 - 5(-1)2 - 11( -1) - 3 = 0

p(-1/3) = 3(-1/3)3 - 5(-1/3)2 - 11(-1/3) - 3 = 0

So These three values are the zeroes of the polynomial.

Sum of zeroes = 3 + (-1) + (-1/3) = 5/3 = -(cofficient of x2)/cofficient of x3

Product of zeroes = 3(-1)(-1/3) = 1 = -(constant term)/cofficient of x3

sum of product of zeroes taking two at a time= 3(-1) + (-1)(-1/3) + 3(-1/3) = -11/3 =�(cofficient of x)/cofficient of x3