p(0), p(1) and p(2) for each of the following polynomials:
(1.) p(t)= 2 +t + 2t² - t³

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Solution

p(t) = 2 + t + 2t² – t³
Plug t = 0 we get
=> p(0) = 2 + 0 +2(0)² – (0)³
=> p(0) = 2 + 0 +0 – 0
=> p(0) = 2

p(t) = 2 + t + 2t² – t³
Plug t = 1 we get
=> p(1) = 2 + 1 +2(1)² – (1)³
=> p(1) = 2 + 1 + 2 – 1
=> p(1) = 4

p(t) = 2 + t + 2t² – t³
Plug t = 2 we get
=> p(2) = 2 + 2 +2(2)² – (2)³
=> p(2) = 2 + 2 + 8 – 8
=> p(2) = 4

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