Question

Find p(0), p(1) and p(2) for each of the following
polynomials.
(i) p(y) = y3 – 3y + 1
(ii) p(x) = x2 – x + 2

Answer 1

$(i) p(y) = {y}^{3} - 3y + 1 \\ \\ = > p(0) = {0}^{3} - 3(0) + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1 \\ \\ = > p(1) = {1}^{3} - 3(1) + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =1 - 3 + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = - 1 \\ \\ = > p(2) = {2}^{3} - 3(2) + 1\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =8 - 6 + 1\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =3$

Do simmiler in (ii)

Hope you got it