Class 9 Math Exercise 2.2 | Question 2

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)

Explanatory Answer | Exercise 2.2 Question 2

(i) Find p(0), p(1), and p(2) of p(y) = y² – y + 1

p(0) = (0)² – (0) + 1 = 1
p(1) = (1)² – (1) + 1 = 1
p(2) = (2)² – (2) + 1 = 3

(ii) Find p(0), p(1) and p(2) of p(t) = 2 + t + 2t² – t³

p(0) = 2 + 0 + 2(0)² – (0)³ = 2
p(1) = 2 + 1 + 2(1)² – (1)³ = 2 + 1 + 2 – 1 = 4
p(2) = 2 + 2 + 2(2)² – (2)³ = 2 + 2 + 8 – 8 = 4

(iii) Find p(0), p(1) and p(2) of p(x) = x³

p(0) = 0³ = 0
p(1) = 1³ = 1
p(2) = 2³ = 8

(iv) Find p(0), p(1) and p(2) of (x) = (x – 1)(x + 1)

p(0) = (0 – 1)(0 + 1) = - 1 × 1 = -1
p(1) = (1 – 1)(1 + 1) = 0 × 2 = 0
p(2) = (2 – 1)(2 + 1) = 1 × 3 = 3