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2.Find p(0), p(1)and p(2) for each of the following polynomials :
i)\(p(y) = y^2 -y + 1\)
ii)\(p(t) = 2 + t + 2t^2 - t^3\)
iii)\(p(x) = x^3\)
iv)\(p(x) = (x - 1)(x + 1)\)
Answer :

i)\(p(y) = y^2 - y + 1\)
Firstly, let us put x = 0,
Therefore, \(p(0) = 0^2 - 0 +1\)
Hence, p(0) = 1
Now, let us put x = 1,
Therefore, \(p(1) = 1^2 - 1 +1\)
Hence, p(1) = 1
Now, let us put x = 2,
Therefore, \(p(2) = 2^2 - 2 +1\)
i.e. \(p(2) = 4 - 2 +1\)
Hence, p(2) = 3

ii)\(p(t) = 2 + t + 2t^2 - t^3\)
Firstly, let us put x = 0,
Therefore, \(p(0) = 2 + 0 + 2{0}^2 - {0}^3\)
Hence, p(0) = 2
Now, let us put x = 1,
Therefore, \(p(1) = 2 + 1 + 2{1}^2 - {1}^3\)
\(p(1) = 3 + 2 - 1\)
Hence, p(1) = 4
Now, let us put x = 2,
Therefore, \(p(2) = 2 + 2 + 2{2}^2 - {2}^3\)
i.e. \(p(2) = 4 + 8 - 8\)
Hence, p(2) = 4

iii)\(p(x) = x^3\)
Firstly, let us put x = 0,
Therefore, \(p(0) = {0}^3\)
Hence, p(0) = 0
Now, let us put x = 1,
Therefore, \(p(1) = {1}^3\)
Hence, p(1) = 1
Now, let us put x = 2,
Therefore, \(p(2) = {2}^3\)
Hence, p(2) = 8

iv)\(p(x) = (x - 1)(x + 1)\)
Firstly, let us put x = 0,
Therefore, \(p(0) = (0 - 1)(0 + 1)\)
Hence, p(0) = -1
Now, let us put x = 1,
Therefore, \(p(1) = (1 - 1)(1 + 1)\)
Hence, p(1) = 0
Now, let us put x = 2,
Therefore, \(p(2) = (2 - 1)(2 + 1)\)
i.e. \(p(2) = 1 × 3\)
Hence, p(2) = 3