Question

if p(y)=4+3 y minus y square+ 5 y cube find p0 P2 P - 1​

Answer 1

Answer:

Step-by-step explanation:

p(y)=4-3y+y^2-5y^3

put y=0

∴P(0)=4

put y=2

∴p(2)=4-(3*2)+(5*2*2*2)+(4)

        =4-6+40+4=42

put p=-1

∴P(-1)=4+(3*1)+(-5*1)+1

        =4+3-5+1=3