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Question:

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
Substitute t = 0, 1, 2 in p(y) to get the required values.
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
Similar way,
(ii) p(t) = 2 + t + 2t² - t³
p(0) = 2 + (0) + 2(0)² - (0)³ = 2 + 0 + 0 - 0 = 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