Answer :
Sure! Let's evaluate the given polynomials at specified points:
### (i) [tex]\( p(y) = y^2 - y + 1 \)[/tex]
1. For [tex]\( y = 0 \)[/tex]:
[tex]\[ p(0) = 0^2 - 0 + 1 = 1 \][/tex]
2. For [tex]\( y = 1 \)[/tex]:
[tex]\[ p(1) = 1^2 - 1 + 1 = 1 \][/tex]
3. For [tex]\( y = 2 \)[/tex]:
[tex]\[ p(2) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3 \][/tex]
So, the values are:
[tex]\[ p(0) = 1, \, p(1) = 1, \, p(2) = 3 \][/tex]
### (ii) [tex]\( p(t) = 2 + t + 2t^2 - t^3 \)[/tex]
1. For [tex]\( t = 0 \)[/tex]:
[tex]\[ p(0) = 2 + 0 + 2 \cdot 0^2 - 0^3 = 2 \][/tex]
2. For [tex]\( t = 1 \)[/tex]:
[tex]\[ p(1) = 2 + 1 + 2 \cdot 1^2 - 1^3 = 2 + 1 + 2 - 1 = 4 \][/tex]
3. For [tex]\( t = 2 \)[/tex]:
[tex]\[ p(2) = 2 + 2 + 2 \cdot 2^2 - 2^3 = 2 + 2 + 2 \cdot 4 - 8 = 2 + 2 + 8 - 8 = 4 \][/tex]
So, the values are:
[tex]\[ p(0) = 2, \, p(1) = 4, \, p(2) = 4 \][/tex]
### (iii) [tex]\( p(x) = x^3 \)[/tex]
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ p(0) = 0^3 = 0 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = 1^3 = 1 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ p(2) = 2^3 = 8 \][/tex]
So, the values are:
[tex]\[ p(0) = 0, \, p(1) = 1, \, p(2) = 8 \][/tex]
### (iv) [tex]\( p(x) = (x - 1)(x + 1) \)[/tex]
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ p(0) = (0 - 1)(0 + 1) = (-1)(1) = -1 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = (1 - 1)(1 + 1) = (0)(2) = 0 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ p(2) = (2 - 1)(2 + 1) = (1)(3) = 3 \][/tex]
So, the values are:
[tex]\[ p(0) = -1, \, p(1) = 0, \, p(2) = 3 \][/tex]
Now summarizing all the evaluated results:
(i) [tex]\( p(y) = y^2 - y + 1 \)[/tex]:
[tex]\[ p(0) = 1, \, p(1) = 1, \, p(2) = 3 \][/tex]
(ii) [tex]\( p(t) = 2 + t + 2t^2 - t^3 \)[/tex]:
[tex]\[ p(0) = 2, \, p(1) = 4, \, p(2) = 4 \][/tex]
(iii) [tex]\( p(x) = x^3 \)[/tex]:
[tex]\[ p(0) = 0, \, p(1) = 1, \, p(2) = 8 \][/tex]
(iv) [tex]\( p(x) = (x - 1)(x + 1) \)[/tex]:
[tex]\[ p(0) = -1, \, p(1) = 0, \, p(2) = 3 \][/tex]
These are the desired function values for each polynomial at [tex]\( 0 \)[/tex], [tex]\( 1 \)[/tex], and [tex]\( 2 \)[/tex].
### (i) [tex]\( p(y) = y^2 - y + 1 \)[/tex]
1. For [tex]\( y = 0 \)[/tex]:
[tex]\[ p(0) = 0^2 - 0 + 1 = 1 \][/tex]
2. For [tex]\( y = 1 \)[/tex]:
[tex]\[ p(1) = 1^2 - 1 + 1 = 1 \][/tex]
3. For [tex]\( y = 2 \)[/tex]:
[tex]\[ p(2) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3 \][/tex]
So, the values are:
[tex]\[ p(0) = 1, \, p(1) = 1, \, p(2) = 3 \][/tex]
### (ii) [tex]\( p(t) = 2 + t + 2t^2 - t^3 \)[/tex]
1. For [tex]\( t = 0 \)[/tex]:
[tex]\[ p(0) = 2 + 0 + 2 \cdot 0^2 - 0^3 = 2 \][/tex]
2. For [tex]\( t = 1 \)[/tex]:
[tex]\[ p(1) = 2 + 1 + 2 \cdot 1^2 - 1^3 = 2 + 1 + 2 - 1 = 4 \][/tex]
3. For [tex]\( t = 2 \)[/tex]:
[tex]\[ p(2) = 2 + 2 + 2 \cdot 2^2 - 2^3 = 2 + 2 + 2 \cdot 4 - 8 = 2 + 2 + 8 - 8 = 4 \][/tex]
So, the values are:
[tex]\[ p(0) = 2, \, p(1) = 4, \, p(2) = 4 \][/tex]
### (iii) [tex]\( p(x) = x^3 \)[/tex]
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ p(0) = 0^3 = 0 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = 1^3 = 1 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ p(2) = 2^3 = 8 \][/tex]
So, the values are:
[tex]\[ p(0) = 0, \, p(1) = 1, \, p(2) = 8 \][/tex]
### (iv) [tex]\( p(x) = (x - 1)(x + 1) \)[/tex]
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ p(0) = (0 - 1)(0 + 1) = (-1)(1) = -1 \][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[ p(1) = (1 - 1)(1 + 1) = (0)(2) = 0 \][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ p(2) = (2 - 1)(2 + 1) = (1)(3) = 3 \][/tex]
So, the values are:
[tex]\[ p(0) = -1, \, p(1) = 0, \, p(2) = 3 \][/tex]
Now summarizing all the evaluated results:
(i) [tex]\( p(y) = y^2 - y + 1 \)[/tex]:
[tex]\[ p(0) = 1, \, p(1) = 1, \, p(2) = 3 \][/tex]
(ii) [tex]\( p(t) = 2 + t + 2t^2 - t^3 \)[/tex]:
[tex]\[ p(0) = 2, \, p(1) = 4, \, p(2) = 4 \][/tex]
(iii) [tex]\( p(x) = x^3 \)[/tex]:
[tex]\[ p(0) = 0, \, p(1) = 1, \, p(2) = 8 \][/tex]
(iv) [tex]\( p(x) = (x - 1)(x + 1) \)[/tex]:
[tex]\[ p(0) = -1, \, p(1) = 0, \, p(2) = 3 \][/tex]
These are the desired function values for each polynomial at [tex]\( 0 \)[/tex], [tex]\( 1 \)[/tex], and [tex]\( 2 \)[/tex].