Question

2. Find p(0),p(1) and p(2) for each of the following polynomials:
1. p (y)= y2-y+1
2.p(t) = 2 +t +2t square - t cube​

Answer 1

Answer:

firstly , p(0)

put y =0

(0)square-(0)+1

0-0+1

=1

now put y=1

(1)square-(1)+1

1-1+1

=1

also put y =2

(2)square-(2)+1

8-2+1

6+1

=7

hence @ p(0)=1,p(1)=1,p(2)=7

(b)p(t)=2+t+2t square- t cube

firstly,put t=0

2+(0)+2(0)square-(0)cube

2+0+2-0

=4

now put t=1

2+(1)+2(1)square-(1)cube

2+1+2-1

=4

also put t=2

2+(2)+2(2)square-(2)cube

2+2+2(4)_(8)

2+2+8-8

=4