Question

Findp(0) p(1) p(2) for the given polynomial p(y) = y^2 -y+ 1

Answer 1

p(0) = 1

p(1) = 1

p(2) = 3

Step-by-step explanation:

Given -

p(y) = y² - y + 1

to find -

the value of p(0), p(1), p(2)

Solution -

p(y) = y² - y + 1

p(0) = 0² - 0 + 1

=> 0 - 0 + 1

=> 1

p(0) = 1

p(1) = 1² - 1 + 1

=> 1 - 1 + 1

=> 0 + 1

=> 1

p(1) = 1

p(2) = 2² - 2 + 1

=> 4 - 2 + 1

=> 2 + 1

=> 3

p(2) = 3

Hence p(0) = 1, p(1) = 1 and p(2) = 3 for p(y).

hope it helps.