1.

For each of the following polynomial, find p(1), p(0) and p(- 2).p(y) = y2 – 2y + 5

Answer»

p(y) = y2 – 2y + 5 

∴ p(1) = 12 – 2(1) + 5 

= 1 – 2 + 5

∴ P(1) = 4

 p(y) = y2 – 2y + 5 

∴ p(0) = 02 – 2(0) + 5

 = 0 – 0 + 5 

∴ p(0) = 5

p(y) = y2 – 2y + 5 

∴ p(- 2) = (- 2)2 – 2(- 2) + 5 

= 4 + 4 + 5 

∴ p(-2) = 13


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