Question

find p(0),p(1),p(2) for the polynomials 1) p(y)=y2-y+1 2) p(t) =2+t+2t2-t3 3) p(x)= x3 4)p(x) =(x-1)(x+1)

Answer 1

   To find  p(0), p(1), or p(2), you have to substitute 0, 1, or 2 in place of the variable.

1) p(y) = y² - y + 1
    p(0) = 0² - 0 +  1 = 1
    p(1) = 1² - 1 + 1 = 1
    p(2) - 2² - 2 + 1 = 3

2) p(t) = 2 + t + 2 t² - t³
      p(0) = 2 
     p(1) = 2 + 1 + 2 1² - 1³ = 4
     p(2) = 2 + 2 + 2 2² - 2³ = 4

3)  p(x) = x³
     p(0) = 0
    p(1) = 1³ = 1
     p(2) = 2³= 8

4) p(x) = (x - 1 ) (x + 1)
        p(0) = -1
        p(1) = 0
        p(2) = 3