Question

2. 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

Answer:

Hope it helps you buddie ....,

Answer 2

Answer:

Step-by-step explanation:

(i) p(y) = y 2 − y + 1

p(0) = (0)2 − (0) + 1 = 0 + 0 + 1 = 1

p(1) = (1)2 − (1) + 1 = 1 − 1 + 1 = 1

p(2) = (2)2 − (2) + 1 = 4 − 2 + 1 = 3

(ii) p(t) = 2 + t + 2t 2 − t 3

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

p(1) = 2 + (1) + 2(2)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) = x 3

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