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Question
Find p(0), p(1) and p(2) for the following polynomial:-
p(y) = y2 – y + 1
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Solution
p(y) = y2 − 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
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