Verify that 2, 3 and 1/2 are the zeroes of the polynomial p(x)= 2x³ – 11x² + 17x – 6

p(x) = 2x³ – 11x² + 17x – 6

‘a’ is said to be zero of p(x) if p(a) = 0

which means that we substitute x by a in p(x) and if we get zero then we say that ‘a’ is zero of p(x)

consider a = 2

⇒ p(a) = 2(2)³ – 11(2)² + 17(2) – 6

⇒ p(a) = 16 – 44 + 34 – 6

⇒ p(a) = 16 – 6 – 44 + 34

⇒ p(a) = 10 – 10

⇒ p(a) = 0

Consider a = 3

⇒ p(a) = 2(3)³ – 11(3)² + 17(3) – 6

⇒ p(a) = 2(27) – 99 + 51 – 6

⇒ p(a) = 54 – 99 + 51 – 6

⇒ p(a) = 54 + 51 – 99 – 6

⇒ p(a) = 105 – 105

⇒ p(a) = 0

Consider a = 1/2

⇒ p(a) = 2(1/2)³ – 11(1/2)² + 17(1/2) – 6

⇒ p(a) = 2(1/2)³ – 11(1/2)² + 17(1/2) – 6

⇒ p(a) = 6

Hence 2 and 3 are zeroes of polynomial p(x) = 2x³ – 11x² + 17x – 6 and 1/2 is not a zero of p(x)