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)\(p(x) = (x - 1)(x + 1)\)

i)\(p(y) = y^2 -y + 1\)

ii)\(p(t) = 2 + t + 2t^2 - t^3\)

iii)\(p(x) = x^3\)

iv)\(p(x) = (x - 1)(x + 1)\)

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