# Polynomials

## Exercise 2.2 Part 2

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

(i) p(y) = y^2 - y + 1

Answer: Given; p(y) = y^2 - y + 1

Hence, p(0) = 0^2 - 0 + 1

= 0 – 0 + 1 = 1

When y = 1

Then, p(1) = 1^2 - 1 + 1

= 1 - 1 + 1 = 0 + 1 = 1

When y=2'

Then, p(2) = 2^2 - 2 + 1

= 4 – 2 + 1

= 2 + 1 = 3

(ii) p(t) = 2 + t + 2t^2 - t^3

Answer: Given, p(t) = 2 + t + 2t^2 - t^3

When p(0), i.e. t = 0

Then p(0) = 2 + 0 + 2 x 0^2 - 0^3

= 2 + 0 + 0 – 0 = 2

When, t = 1

Then p(1) = 2 + 1 + 2(1)^2 - 1^3

= 2 + 1 + 2 x 1 – 1

= 3 + 2 – 1

= 5 – 1 = 4

When t = 2

Then p(2) = 2 + 2 + 2 xx 2^2 - 2^3

= 2 + 2 + 2 x 4 – 8

= 4 + 8 – 8 = 4

(iii) p(x) = x^3

Answer: Given, p(x) = x^3

When x = 0

Then p(0) = 0^3

Or, p(0) = 0

When x = 1

Then p(1) = 1^3

Or, p(1) = 1

When x = 2

Then p(2) = 2^3

Or, p(2) = 8

Hence, 0, 1 and 8 are required answers

(iv) p(x) = (x – 1)(x + 1)

Answer: Given, p(x) = (x – 1)(x + 1)

When x = 0

Then p(0) = (0 – 1)(0 + 1)

Or, p(0) = (- 1) x 1 = - 1

When x =1

Then p(1) = (1 – 1)(1 + 1)

Or, p(1) = 0 xx 2 = 0

When x = 2

Then p(2) = (2 – 1)( 2 + 1)

Or, p(2) = 1 xx 3 = 3`

Hence, - 1, 0 and 3 are required answers