Question

P(y)=4y3+4y2-y-1 is a multiple of (2y+1)

Answer 1

hi friend
here's your answer looking for

p(y)= 4y^3+4y^2-y-1

so, if 2y-1 is a factor of p(y)

then,

zero of p(y) is
y = 1/2

so,
p(1/2)= 4(1/2)^3+4(1/2)^2-1/2-1

=4(1/8)+4(1/4)-1/2-1

=1/2+1-1/2-1

= 0

so, 2y+1 is multiple of p(y)

hope it's helpful to you
From brainly star Amaytripathi19

Answer 2

P(y) =4y^3+4y^2-y-1
P(-1/2)=4*(-1/2)^3+4*(-1/2)^2-(-1/2)-1
=4*(-1/8)+4*1/4+1/2-1
=-1/2+1+1/2-1
=0
Hope it helps u