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Question
For the following polynomial, find p(1), p(0) and p(-2).
p(y) = y2 − 2y + 5
Solution
p(y) = y2 − 2y + 5
∴ p(1) = (1)2 − 2 × 1 + 5
= 1 − 2 + 5
= 4
∴ p(0) = (0)2 − 2 × 0 + 5
= 0 − 0 + 5
= 5
∴ p(−2) = (−2)2 − 2 × (−2) + 5
= 4 + 4 + 5
= 13
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