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Question

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

(i) p(y) = y2y + 1 (ii) p(t) = 2 + t + 2t2t3

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

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Solution

(i) p(y) = y2y + 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 + 2t2t3

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


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