Respuesta :

apologiabiology
zero product says that if you have
x=0, then the non-zero factors equal zero

so
try to move all to one side to get it to equal zero and factor

4y^3-4=y-16y^2
add 16y^2 to both sides
4y^3+16y^2-4=y
subtract y
4y^3+16y^2-y-4=0
gropu and factor
(4y^3+16y^2)+(-y-4)=0
(4y^2)(y+4)+(-1)(y+4)=0
reverse distribute
ba+ca=a(b+c)
(4y^2-1)(y+4)=0
set each to zero

4y^2-1=0
add 1
4y^2=1
divide by 4
y^2=1/4
square root both sides
y=+/-1/2

y+4=0
minus 4
y=-4



y=-4,-1/2,1/2


answer is A -4,-1/2,1/2
Willchemistry
[tex]4y^3-4 =y-16y^2[/tex]

Subtract [tex]y -16y^2[/tex] from both sides

[tex]4y^3 -4-(y-16y^2) = y -16y^2-(y-16y^2)[/tex]

[tex]4y^3-4-y+16y^2=0[/tex]

Solving by factoring:

Factor [tex]4y^3-4-y + 16 y^2[/tex]

[tex](4y^3+16y^2) + (-y-4)[/tex]

Factor out [tex]4 y^2[/tex] from [tex]4y^3+16y^2 : 4y^2(y+4)[/tex]

Factor out [tex]-1[/tex] from [tex]-y-4 : -y(y+4) = 4y^2(y+4)-(y+4)[/tex]

Factor out common term [tex](y+4)[/tex]

[tex]=(y+4)(4y^2-1)[/tex]

Factor [tex]4y^2-1 = ( 2y+1)(2y-1)[/tex]

[tex]= (y+4)(2y+1)(2y-1) = 0[/tex]

Using the zero factor principle:

Solve [tex]y+4 = 0 : y = -4[/tex]

Solve [tex]2y+1 = 0 : y = - \frac{1}{2} [/tex]

Solve [tex]2y-1 = 0 : y = \frac{1}{2} [/tex]


The final solutions to the equation are :

[tex]y = - 4[/tex]

[tex]y = -\frac{1}{2} [/tex]

[tex]y = \frac{1}{2} [/tex]

Answer A

hope this helps! =)

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