Question

2. Find p(0), p(1) and p(2) for each of the following polynomials:
(1) ply) = y2 – y + 1 (i) p(t) = 2+t+ 2t2 –
(iii) p(x) = x
(iv) p(x) = (x - 1)(x + 1)​

Answer 1

Step-by-step explanation:

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

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

=0-0+1 =1-1+1 =4-2+1

=1 =1 =3

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

p(0)=2+0+2×0^2-0^3

=2+0-0

=2

p(1)=2+1+2×1^2-1^3

=3+2-1

=4

p(2)=2+2+2×2^2-2^3

=4+2×4-8

=4+8-8

=4

iii)p(x)=x^3

p(0)=0^3

=0

p(1)=1^3

=1

p(2)=2^3

=8

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

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

=0^2-1^2

=0-1

=0

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

=1^2-1^2

=1-1

=0

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

=2^2-1^2

=4-1

=3