Question

2. Find p(0),p(1) and p(2) for each of the following polynomials.
(1) P(x) = x2-x+1
(i) P(x) = 2 + y + 2x2 - y
(ii) P(z)=z 23
(iv) p(t) =(t-1) (t+1)
(v) P(x) = x2 – 3x + 2​

Answer 1

Answer:

(1) P(0)= 1 , P(1)= 2, P(2)= 3

(2)P(0)= 6 , P(1)= 6 , P(2)= 6

(3)P(0)= 0 , P(1)= 23 , P(2)= 46

(4)P(0)= 0 , P(1)= 0 , P(2)= 3

(5)P(0)= 2 , P(1)= 1 , P(2)= 0

Answer 2

= 2 + 2 +8 – 8

= 4

iii. p(z) = z3

Sol : a) If z = 0

p(0) = (0)3

p(0) = 03

= 0

b) If z = 1

p(1) = (1)3

p(1) = 13

= 1

c) If z = 2

p(2) = (2)3

p(2) = 23

= 8

iv. p(t) = (t – 1) (t + 1)

Sol : a) If t = 0

p(0) = (0 – 1) (0 + 1)

p(0) = (– 1) (1)

= – 1

b) If t = 1

p(1) = (1 – 1) (1 + 1)

p(1) = (0) (2)

= 0

c) If t = 2

p(2) = (2 – 1) (2 + 1)

p(0) = (1) (3)

= 3

v. p(x) = x2

– 3x + 2

Sol : a) If x = 0

p(0) = (0)2

– 3(0) + 2

p(0) = 0 – 3(0) + 2

= 0 – 0 + 2

= 0 + 2

= 2

b) If x = 1

p(1) = (1)2

– 3(1) + 2

p(1) = 1 – 3(1) + 2

= 1 – 3 + 2

= 3 – 3

= 0

c) If x = 2

p(2) = (2)2

– 3(2) + 2

= 4 – 6 + 2

= 6 – 6

= 0