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

(i) p(y) = y2 - y + 1 (ii) p(t) = 2 + t + 2t2 - t3

(iii) p(x) = x3  (iv) p(x) = (x - 1) (x + 1)

Solution:

(i) p(y) = y2 - y + 1

We can calculate,

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

We can calculate,

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

We can calculate,

p(0) = (0)3 = 0

p(1) = (1)3 = 1

p(2) = (2)3 = 8

(iv) p(x) = (x - 1) (x + 1)

We can calculate,

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