Find at least one zero of p(y) if p(y)= y^3+3*y^2-4*y-4. I will give 46 points for the correct answer.
Sammy2712:
Sorry......23 points will be given.
I am extremely sorry for this false claim. I thought all 46 points taken from me will be given to the person who answers the question.
Answers
Answered by
1
1.489 is the zero of this equation
Thanks again, and sorry for the false claim. I thought all the 46 points taken from me will be given to the person who answers it.
Still, you are getting 23 points.
Answered by
1
Step-by-step explanation:
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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