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

Firstly, let us put x = 0,
\(\therefore p(0) = 0^2 - 0 +1\)
\(\Rightarrow p(0) = 1 \)

Now, let us put x = 1,
\(\therefore p(1) = 1^2 - 1 +1\)
\(\Rightarrow p(1) = 1 \)

Now, let us put x = 2,
\(\therefore p(2) = 2^2 - 2 +1\)
\(\Rightarrow p(2) = 4 - 2 +1\)
\(\Rightarrow p(2) = 3 \)


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

Firstly, let us put x = 0,
\(\therefore p(0) = 2 + 0 + 2{0}^2 - {0}^3\)
\(\Rightarrow p(0) = 2 \)

Now, let us put x = 1,
\(\therefore p(1) = 2 + 1 + 2{1}^2 - {1}^3 \)
\(\Rightarrow p(1) = 3 + 2 - 1\)
\(\Rightarrow p(1) = 4 \)

Now, let us put x = 2,
\(\therefore p(2) = 2 + 2 + 2{2}^2 - {2}^3\)
\(\Rightarrow p(2) = 4 + 8 - 8\)
\(\Rightarrow p(2) = 4 \)


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

Firstly, let us put x = 0,
\(\therefore, p(0) = {0}^3\)
\(\Rightarrow p(0) = 0 \)

Now, let us put x = 1,
\(\therefore p(1) = {1}^3\)
\(\Rightarrow p(1) = 1 \)

Now, let us put x = 2,
\(\therefore p(2) = {2}^3\)
\(\Rightarrow p(2) = 8\)


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

Firstly, let us put x = 0,
\(\therefore p(0) = (0 - 1)(0 + 1)\)
\(\Rightarrow p(0) = -1 \)

Now, let us put x = 1,
\(\therefore p(1) = (1 - 1)(1 + 1)\)
\(\Rightarrow p(1) = 0 \)

Now, let us put x = 2,
\(\therefore, p(2) = (2 - 1)(2 + 1)\)
\(\Rightarrow p(2) = 1 × 3\)
\(\Rightarrow p(2) = 3 \)