Question

POLYNOMIALS AND
2. Find p(0),p(1) and p(2) for each of the following polynomials.
(1) p(x) = x2-x+1
(i) p(y) = 2 + y + 2y2 -y3

(ii). p(z)=z3
(iv) p(t) = (t-1)(t+1)​

Answer 1

Answer:

1) p(0)=0×2-0+1=1

p(1)=1×2-1+1=2

p(2)=2×2-2+1=1

p(0)=2+0+2×0×2-0×3=2

p(1)=2+1+2×1×2-1×3=4

p(2)=2+2+2×2×2-2×3=6

Answer 2

Sol : a) If x = 0

p(x) = x

2

– x + 1

p(0) = (0)2

– (0) + 1

= 0 – 0 + 1

= 1

b) If x = 1

p(x) = x

2

– x + 1

p(1) = (1)2

– (1) + 1

= 1 – 1 + 1

= 1

c) If x = 2

p(x) = x

2

– x + 1

p(0) = (2)2

– (2) + 1

= 4 – 2 + 1

= 3

ii. p(y) = 2 + y + 2y2

– y

3

Sol : a) If y = 0

p(0) = 2 + 0 + 2(0)2

– (0)3

p(0) = 2 + 0 + 2(0) – (0)

= 2 + 0 + 0 – 0

= 2

b) If y = 1

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

2

– (1)

3

p(1) = 2 + 1 + 2(1) – (1)

= 3 + 2 – 1

= 5 – 1

= 4

c) If y = 2

p(2) = 2 + 2 + 2(2)2

– (2)3

= 2 + 2 + 2(4) – (8)

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