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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,

\(\therefore p(0) = 0^2 - 0 +1\)

\(\Rightarrow p(0) = 1 \)

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,

\(\therefore, p(2) = (2 - 1)(2 + 1)\)

\(\Rightarrow p(2) = 1 × 3\)

\(\Rightarrow p(2) = 3 \)

- Find the value of the polynomial \(5x - 4x^2 + 3\) at : i) x = 0 ii) x = -1iii) x = 2
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- Find the Zero of the polynomial in each of the following cases. i)\(p(x) = x + 5\)ii)\(p(x) = x - 5\)iii)\(p(x) = 2x + 5\)iv)\(p(x) = 3x - 2\)v)\(p(x) = 3x\)vi)\(p(x) = ax, a \ne 0\)vii)\(p(x) = cx + d ,c \ne 0\), c,d, are real numbers.

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