P(x)≈[x-1] [x+1] were p(0) , P(1) and P(2)
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Answer:
p(0)= -1, p(1)= 0, p(2)= 3
Verified answer
Answer:
p(x) = [x - 1] [x + 1]
p(0) = [0 - 1 ] [0 + 1]
p(0) = -1
p(1) = [ 1 - 1 ] [1 + 1]
p(1) = 0
p(2) = [ 2 -1 ] [2 +1]
p(2) = 3
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