Question

verify that 3,-1, -1/3 are zeroes of the polynomial p ( x ) = 6x^3 - 10x^2 - 22x - 6

Answer 1

Let zero of the polynomial be 
r(x)= 3
g(x)= -1
f(x)= -1/3
r(x), g(x) and f(x) are zeroes of polynomial p(x) if it divides it completely.
So,by remainder theorem, we get - 


1. p(3) = 6*3^3 - 10*3^2 - 22*3 - 6
            = 6*27 - 10*9 - 66 - 6
            = 162 - 90 -66 - 6
            = 162 -162
            = 0

2. p(-1) = 6*(-1)^3 - 10*(-1)^2 - 22*(-1) - 6
             = -6 - 10 + 22 - 6
             = 22 - 22
             = 0

3. p(-1/3) = 6*(-1/3)^3 - 10*(-1/3)^2 - 22*(-1/3) - 6
                = 6*-1/27 - 10*1/9 - 22*-1/3 -6
                = -2/9 - 10/9 + 22/3 - 6/1
                = -2/9 -10/9 + 66/9 - 54/9  [Taking LCM as 9 we find the values]
                = 66/9 - 66/9
                = 0

Hence, All the values for x give zero for the polynomial p(x).
And Therefore, 3, -1, -1/3 are the zeroes of the polynomial p(x).