Question

For the following polynomial, find p(1), p(0) and p(-2).

p(y) = y² − 2y + 5

Answer 1

p(y) = y² − 2y + 5

∴ p(1) = (1)² − 2 ×  1 + 5

= 1 − 2 + 5

= 4

∴ p(0) = (0)² − 2 × 0 + 5

= 0 − 0 + 5

= 5

∴ p(−2) = (−2)² − 2 × (−2) + 5

= 4 + 4 + 5

= 13