Find p(0), p(1) and p(2) for each of the following polynomials

Question 2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² – y + 1    (ii) p(t) = 2 + t + 2t² – t3            (iii) p(x) = x3   (iv) p(x) = (x – 1) (x + 1)
Solution: (i)p(y) = y² – y + 1    
Plug y = 0 we get
=>p(0) = (0)² – 0 + 1
=>p(0) = 0 – 0  + 1
=> 1

Plug y = 1 we get
=>p(1) = (1)² – 1 + 1
=>p(1) = 1 – 1  + 1
=> 1

Plug y = 2 we get
=>p(2) = (2)² – 2 + 1
=>p(2) = 4 – 2  + 1
=> 3

(ii) p(t) = 2 + t + 2t² – t³           
Plug t = 0 we get
=> p(t) = 2 + t + 2t² – t³
=> p(0) = 2 + 0 +2(0)² – (0)³
=> p(0) = 2 + 0 +0 – 0
=> p(0) = 2

p(t) = 2 + t + 2t² – t³   
Plug t = 1 we get
=> p(t) = 2 + t + 2t² – t³
=> p(1) = 2 + 1 +2(1)² – (1)³
=> p(1) = 2 +  1 + 2 – 1
=> p(1) = 4

p(t) = 2 + t + 2t² – t³   
Plug t = 2 we get
=> p(t) = 2 + t + 2t² – t³
=> p(2) = 2 + 2 +2(2)² – (2)³
=> p(2) = 2 + 2 + 8 – 8
=> p(2) =  4

(iii) p(x) = x³
=>plug x = 0 we get            
=>p(x) = x³
=>p(0) = (0)³
=> p(0) = 0

p(x) = x3
=>plug x = 1 we get            
=>p(x) = x³
=>p(1) = (1)³
=> p(1) = 1

p(x) = x³
=>plug x = 2 we get            
=>p(x) = x³
=>p(2) = (2)³
=> p(2) = 8

(iv) p(x) = (x – 1) (x + 1)
Plug x = 0 we get
=>p(x) = (x – 1) (x + 1)
=> p(0) = (0 – 1) (0 + 1)
=> p(0) = (- 1)(1)
=> p(0)  = - 1

Plug x = 1 we get
=>p(x) = (x – 1) (x + 1)
=> p(1) = (1 – 1) (1 + 1)
=> p(1) = (0)(2)
=> p(1)  = 0

Plug x = 2 we get
=>p(x) = (x – 1) (x + 1)
=> p(2) = (2 – 1) (2 + 1)
=> p(2) = (1)(3)
=> p(2)  = 3