Question

find p(0), p(1), p(2) for each of the following polynomials
(ii) p(y) =2y^2-3y+2​

Answer 1

Answer:

Here is ur ans :

For p(0),

p(y)=2y^2-3y+2

p(0)=2(0)^2-3(0)+2

p(0)= 2

Now for p(1) ,

p(y) =2y^2-3y+2

p(1) =2(1)^2-3(1)+2

p(1) = 1

finally for p(2),

p(2) =2y^2-3y+2

p(2) = 4

Answer 2

Answer:

The given polynomial is

.

$p \: (y) = y - y + 1$

Substitute y =0,

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

Substitutey =1

$p(2) = 2 - 2 + 1 = 3$

There, p(0), p(1), and p(2) are 1,1 and 3 respectively