Question

Find P(0) and P(1) of polynomials P(y)=y square-y+1

Answer 1

Step-by-step explanation:

$p(y) = {y}^{2} - y + 1 \\ \\ p(0) = {0}^{2} - 0 + 1 \\ \\ p(0) = 1 \\ \\ p(1) = {1}^{2} - 1 + 1 \\ \\ p(1) = 1 - 1 + 1 \\ \\ p(1) = 1$

Answer 2

Answer:

Step-by-step explanation:

P(y) = $y^{2} -y + 1$

If p(y) = 0,

p(0) = $0^{2} -0+1$

P(0) = 0-0 +1

p(0) = 1

For p(y) = p(1)

P(1) = $1^{2} - 1 + 1$

P(1) = 1-1+1

P(1) = 1