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)

(i) p(y) = y² - y + 1

Putting y = 0 we get,

p(0) = (0)² - 0 + 1

= 0 - 0 + 1

= 1

∴ p(0) = 1

Putting y = 1 we get,

p(1) = (1)² - 1 + 1

= 1 - 1 + 1

= 2 - 1

= 1

∴ p(1) = 1

Putting y = 2 we get,

p(2) = (2)² - 2 + 1

= 4 - 2 + 1

= 5 - 2

= 3

∴ p(2) = 1

Hence, for p(y) = y² - y + 1, p(0) = 1, p(1) = 1 and p(2) = 1

(ii) p(t) = 2 + t + 2t² - t³

Putting t = 0 we get,

p(0) = 2 + 0 + 2(0²) - 0³

= 2 + 0 + 0 - 0

= 2

∴ p(0) = 2

Putting t = 1 we get,

p(1) = 2 + 1 + 2(1²) - 1³

= 2 + 1 + 2 - 1

= 4

∴ p(1) = 4

Putting t = 2 we get,

p(2) = 2 + 2 + 2(2²) - 2³

= 4 + 2(4) - 8

= 4 + 8 - 8

= 4

∴ p(2) = 4

Hence, for p(t) = 2 + t + 2t² - t³, p(0) = 2, p(1) = 4 and p(2) = 4

(iii) p(x) = x³

Putting x = 0 we get,

p(0) = (0)³

= 0

∴ p(0) = 0

Putting x = 1 we get,

p(1) = (1)³

= 1

∴ p(1) = 1

Putting x = 2 we get,

p(2) = (2)³

= 8

∴ p(2) = 8

Hence, for p(x) = x³, p(0) = 0, p(1) = 1 and p(2) = 8

(iv) p(x) = (x - 1)(x + 1)

Putting x = 0 we get,

p(0) = (0 - 1)(0 + 1)

= (-1) x 1

= (-1)

∴ p(0) = -1

Putting x = 1 we get,

p(1) = (1 - 1)(1 + 1)

= 0 x 2

= 0

∴ p(1) = 0

Putting x = 2 we get

p(2) = (2 - 1) (2 + 1)

= 1 x 3

= 3

∴ p(2) = 3

Hence, for p(x) = (x - 1) (x + 1), p(0) = -1, p(1) = 0 and p(2) = 3