Question

verify whether 2,3,1/2 are tje zeros of the polynomial 2x³-11x²+17x-6 and verify the relationship between zeros and coefficients​

Answer 1

Answer:

p(x) = 2x³-11x²+17x-6

Substituting the value of x = 2

p(2) = 2(2)³ - 11(2)² + 17(2) - 6

       = 16 - 44 + 34 - 6 = 50 - 50 = 0

Since p(2) = 0

Therefore , x = 2 is the zero of p(x)

Substituting the value of x = 3

p(3) = 2(3)³ - 11(3)² + 17(3) - 6

       = 54 - 99 + 51 - 6 = 105 - 105 = 0

Since p(3) = 0

Therefore , x = 3 is the zero of p(x)

Substituting the value of x = 1/2

p(1/2) = 2(1/2)³ - 11(1/2)² + 17(1/2) - 6

          = 1/4 - 11/4 + 17/2 - 6

          = -10/4 + 34/4 - 24/4

          = -34/4 + 34/4 = 0

Since p(1/2) = 0

Therefore , x = 1/2 is the zero of p(x)