Question

p(y)=2+y+2(y)^2-y^3 y=0,1,-1,2​

Answer 1

Answer:

0 I sthe correct answer of quest

Answer 2

Step-by-step explanation:

p(y)=2+y+2(y)^2-y^3 (y=0)

p(0)=2+0+2(0)^2-(0)^3

p=(0) =2+0+0-0

p(0)=2

p(y)=2+y+2(y)^2-y^3 (y=1)

p(1)=2+1+2+2-1

p(1)=4

p(y)=2+y+2(y)^2-y^3 (y=-1)

p(-1)=2+(-1)+2(-1)^2-(-1)^3

p(-1)=2-1+2+3

p(-1) =6

p(y)=2+y+2(y)^2-y^3 (y=2)

p(2)=2+2+2(2)^2-(2)^3

p(2)=2+2+8-8

p(2)=4

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