if p(y) =4+3y-Y2 +5y3, find p(0), p(2) ,p( -1)
Answers
Answered by
47
Answer:
Step-by-step explanation:
p(y)=4+3y-y²+5y³
=5y³-y²+3y+4
p(0)=5y³-y²+3y+4
=5*(0)³-(0)²+3*0+4
=5*0-0*3+0+4
=0-0+0+4
=4
p(1)=5y³-y²+3y+4
=5*(1)³-(1)²+3*1+4
=5*1-1+3+4
=5-1+7
=4+7
=11
p(2)=5y³-y²+3y+4
=5*(2)³-(2)²+3*2+4
=5*8-4+6+4
=40-4+10
=36+10
=46
so, p(0)=4 , p(1)=11 , p(2)=46
Answered by
11
Step-by-step explanation:
p(y)=4+3y-y²+5y³
=5y³-y²+3y+4
p(0)=5y³-y²+3y+4
=5*(0)³-(0)²+3*0+4
=5*0-0*3+0+4
=0-0+0+4
=4
p(1)=5y³-y²+3y+4
=5*(1)³-(1)²+3*1+4
=5*1-1+3+4
=5-1+7
=4+7
=11
p(2)=5y³-y²+3y+4
=5*(2)³-(2)²+3*2+4
=5*8-4+6+4
=40-4+10
=36+10
=46
so, p(0)=4 , p(1)=11 , p(2)=46
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