Question

Find p(0), p(1) and p(2) for each of the following polynomials: 

(i) p(y) = y 2 – y + 1 

(ii) p(t) = 2 + t + 2t 2 – t 3 

(iii) p(x) = x 3 

(iv) p(x) = (x – 1) (x + 1)

Answer 1

(i) p(y) = ^y2 − y + 1 p(0) = 

(0)² − (0) + 1 = 1 p(1) = (1)² − (1) + 1 

= 1 p(2) = (2)² − (2) + 1 = 3

(ii) p(t) = 2 + t + 2t² − t3 p(0) 

= 2 + 0 + 2 (0)2 − (0)3 = 2 p(1) = 2 + (1) + 2(1)² − (1)³ 

= 2 + 1 + 2 − 1 = 4 p(2) 

= 2 + 2 + 2(2)² − (2)³ 

= 2 + 2 + 8 − 8 = 4

(iii) p(x) = x³ p(0) = (0)³ = 0 p(1) 

= (1)³ = 1 p(2) = (2)³ = 8

(iv) p(x) = (x − 1) (x + 1) p(0) = (0 − 1) (0 + 1) = (− 1) (1) = − 1 p(1) = (1 − 1) (1 + 1) = 0 (2) 

= 0 p(2) = (2 − 1 ) (2 + 1) = 1(3) = 3