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04/07/2021 1:35 pm
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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
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04/07/2021 1:37 pm
(i) p(y) = y2–y+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 + 2t2 − t3
∴ 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