Question

2
find P(0), P (1), P (2)
if p(x)= (x-1) (x+1)​

Answer 1

Answer:

$p(x) = (x - 1)(x + 1) \\ \\ = {x}^{2} + x - x - 1 \\ \\ = {x}^{2} - 1 \\ \\ p(0) = - 1 \\ \\ p(1) = 0 \\ \\ p(2) = {2}^{2} - 1 = 4 - 1 = 3$