Answer :
Sure, let's evaluate the polynomials step-by-step for [tex]\(x = 0, 1, 2\)[/tex].
### (i) [tex]\( p(x) = x^2 - x + 1 \)[/tex]
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ p(0) = 0^2 - 0 + 1 = 1 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = 1^2 - 1 + 1 = 1 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ p(2) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3 \][/tex]
Thus, the values are [tex]\( p(0) = 1 \)[/tex], [tex]\( p(1) = 1 \)[/tex], and [tex]\( p(2) = 3 \)[/tex].
### (ii) [tex]\( p(y) = 2 + y + 2y^2 - y^3 \)[/tex]
1. For [tex]\( y = 0 \)[/tex]:
[tex]\[ p(0) = 2 + 0 + 0 - 0 = 2 \][/tex]
2. For [tex]\( y = 1 \)[/tex]:
[tex]\[ p(1) = 2 + 1 + 2 \cdot 1^2 - 1^3 = 2 + 1 + 2 - 1 = 4 \][/tex]
3. For [tex]\( y = 2 \)[/tex]:
[tex]\[ p(2) = 2 + 2 + 2 \cdot 2^2 - 2^3 = 2 + 2 + 8 - 8 = 4 \][/tex]
Thus, the values are [tex]\( p(0) = 2 \)[/tex], [tex]\( p(1) = 4 \)[/tex], and [tex]\( p(2) = 4 \)[/tex].
### (iii) [tex]\( p(z) = z^3 \)[/tex]
1. For [tex]\( z = 0 \)[/tex]:
[tex]\[ p(0) = 0^3 = 0 \][/tex]
2. For [tex]\( z = 1 \)[/tex]:
[tex]\[ p(1) = 1^3 = 1 \][/tex]
3. For [tex]\( z = 2 \)[/tex]:
[tex]\[ p(2) = 2^3 = 8 \][/tex]
Thus, the values are [tex]\( p(0) = 0 \)[/tex], [tex]\( p(1) = 1 \)[/tex], and [tex]\( p(2) = 8 \)[/tex].
### (iv) [tex]\( p(t) = (t-1)(t+1) \)[/tex]
1. For [tex]\( t = 0 \)[/tex]:
[tex]\[ p(0) = (0-1)(0+1) = -1 \cdot 1 = -1 \][/tex]
2. For [tex]\( t = 1 \)[/tex]:
[tex]\[ p(1) = (1-1)(1+1) = 0 \cdot 2 = 0 \][/tex]
3. For [tex]\( t = 2 \)[/tex]:
[tex]\[ p(2) = (2-1)(2+1) = 1 \cdot 3 = 3 \][/tex]
Thus, the values are [tex]\( p(0) = -1 \)[/tex], [tex]\( p(1) = 0 \)[/tex], and [tex]\( p(2) = 3 \)[/tex].
### (v) [tex]\( p(x) = x^2 - 3x + 2 \)[/tex]
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ p(0) = 0^2 - 3 \cdot 0 + 2 = 2 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = 1^2 - 3 \cdot 1 + 2 = 1 - 3 + 2 = 0 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ p(2) = 2^2 - 3 \cdot 2 + 2 = 4 - 6 + 2 = 0 \][/tex]
Thus, the values are [tex]\( p(0) = 2 \)[/tex], [tex]\( p(1) = 0 \)[/tex], and [tex]\( p(2) = 0 \)[/tex].
In summary, here are the results for each polynomial:
1. [tex]\( p(x) = x^2 - x + 1 \)[/tex]:
[tex]\[ p(0) = 1, \quad p(1) = 1, \quad p(2) = 3 \][/tex]
2. [tex]\( p(y) = 2 + y + 2y^2 - y^3 \)[/tex]:
[tex]\[ p(0) = 2, \quad p(1) = 4, \quad p(2) = 4 \][/tex]
3. [tex]\( p(z) = z^3 \)[/tex]:
[tex]\[ p(0) = 0, \quad p(1) = 1, \quad p(2) = 8 \][/tex]
4. [tex]\( p(t) = (t-1)(t+1) \)[/tex]:
[tex]\[ p(0) = -1, \quad p(1) = 0, \quad p(2) = 3 \][/tex]
5. [tex]\( p(x) = x^2 - 3x + 2 \)[/tex]:
[tex]\[ p(0) = 2, \quad p(1) = 0, \quad p(2) = 0 \][/tex]
### (i) [tex]\( p(x) = x^2 - x + 1 \)[/tex]
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ p(0) = 0^2 - 0 + 1 = 1 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = 1^2 - 1 + 1 = 1 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ p(2) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3 \][/tex]
Thus, the values are [tex]\( p(0) = 1 \)[/tex], [tex]\( p(1) = 1 \)[/tex], and [tex]\( p(2) = 3 \)[/tex].
### (ii) [tex]\( p(y) = 2 + y + 2y^2 - y^3 \)[/tex]
1. For [tex]\( y = 0 \)[/tex]:
[tex]\[ p(0) = 2 + 0 + 0 - 0 = 2 \][/tex]
2. For [tex]\( y = 1 \)[/tex]:
[tex]\[ p(1) = 2 + 1 + 2 \cdot 1^2 - 1^3 = 2 + 1 + 2 - 1 = 4 \][/tex]
3. For [tex]\( y = 2 \)[/tex]:
[tex]\[ p(2) = 2 + 2 + 2 \cdot 2^2 - 2^3 = 2 + 2 + 8 - 8 = 4 \][/tex]
Thus, the values are [tex]\( p(0) = 2 \)[/tex], [tex]\( p(1) = 4 \)[/tex], and [tex]\( p(2) = 4 \)[/tex].
### (iii) [tex]\( p(z) = z^3 \)[/tex]
1. For [tex]\( z = 0 \)[/tex]:
[tex]\[ p(0) = 0^3 = 0 \][/tex]
2. For [tex]\( z = 1 \)[/tex]:
[tex]\[ p(1) = 1^3 = 1 \][/tex]
3. For [tex]\( z = 2 \)[/tex]:
[tex]\[ p(2) = 2^3 = 8 \][/tex]
Thus, the values are [tex]\( p(0) = 0 \)[/tex], [tex]\( p(1) = 1 \)[/tex], and [tex]\( p(2) = 8 \)[/tex].
### (iv) [tex]\( p(t) = (t-1)(t+1) \)[/tex]
1. For [tex]\( t = 0 \)[/tex]:
[tex]\[ p(0) = (0-1)(0+1) = -1 \cdot 1 = -1 \][/tex]
2. For [tex]\( t = 1 \)[/tex]:
[tex]\[ p(1) = (1-1)(1+1) = 0 \cdot 2 = 0 \][/tex]
3. For [tex]\( t = 2 \)[/tex]:
[tex]\[ p(2) = (2-1)(2+1) = 1 \cdot 3 = 3 \][/tex]
Thus, the values are [tex]\( p(0) = -1 \)[/tex], [tex]\( p(1) = 0 \)[/tex], and [tex]\( p(2) = 3 \)[/tex].
### (v) [tex]\( p(x) = x^2 - 3x + 2 \)[/tex]
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ p(0) = 0^2 - 3 \cdot 0 + 2 = 2 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = 1^2 - 3 \cdot 1 + 2 = 1 - 3 + 2 = 0 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ p(2) = 2^2 - 3 \cdot 2 + 2 = 4 - 6 + 2 = 0 \][/tex]
Thus, the values are [tex]\( p(0) = 2 \)[/tex], [tex]\( p(1) = 0 \)[/tex], and [tex]\( p(2) = 0 \)[/tex].
In summary, here are the results for each polynomial:
1. [tex]\( p(x) = x^2 - x + 1 \)[/tex]:
[tex]\[ p(0) = 1, \quad p(1) = 1, \quad p(2) = 3 \][/tex]
2. [tex]\( p(y) = 2 + y + 2y^2 - y^3 \)[/tex]:
[tex]\[ p(0) = 2, \quad p(1) = 4, \quad p(2) = 4 \][/tex]
3. [tex]\( p(z) = z^3 \)[/tex]:
[tex]\[ p(0) = 0, \quad p(1) = 1, \quad p(2) = 8 \][/tex]
4. [tex]\( p(t) = (t-1)(t+1) \)[/tex]:
[tex]\[ p(0) = -1, \quad p(1) = 0, \quad p(2) = 3 \][/tex]
5. [tex]\( p(x) = x^2 - 3x + 2 \)[/tex]:
[tex]\[ p(0) = 2, \quad p(1) = 0, \quad p(2) = 0 \][/tex]
