Question

p(y)=y²-2y+5 For the given polynomial, find p(1), p(0) and p(- 2).

Answer 1

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Given :
p(y) = y^2 - 2y + 5.

For p(1), p (0) and p (-2)...
p (1) = 1 - 2×1 + 5 = 6 - 2 = 4.

p (0) = 0 - 2×0 + 5 = 5.

p (-2) = (-2)^2 - 2 (-2) + 5 = 4 + 4 + 5 = 13.


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Answer 2

Given p(y) = y² - 2y + 5 ,

i ) p(1) = 1² - 2×1 + 5

= 1 - 2 + 5

= 4

p(1) = 4

ii ) p(0) = 0² - 2(0) + 5

= 0 - 0 + 5

p(1) = 5

iii ) p(-2) = (-2)² - 2(-2) + 5

= 4 + 4 + 5

= 13

p(-2) = 13

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