Question

find p(0),p(1)and p(2) for the polynomial:
p(t)=2+1+2t/2-t​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

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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