Verified Answer :

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

\(4y^3-4 =y-16y^2\)

Subtract \(y -16y^2\) from both sides

\(4y^3 -4-(y-16y^2) = y -16y^2-(y-16y^2)\)

\(4y^3-4-y+16y^2=0\)

Solving by factoring:

Factor \(4y^3-4-y + 16 y^2\)

\((4y^3+16y^2) + (-y-4)\)

Factor out \(4 y^2\) from \(4y^3+16y^2 : 4y^2(y+4)\)

Factor out \(-1\) from \(-y-4 : -y(y+4) = 4y^2(y+4)-(y+4)\)

Factor out common term \((y+4)\)

\(=(y+4)(4y^2-1)\)

Factor \(4y^2-1 = ( 2y+1)(2y-1)\)

\(= (y+4)(2y+1)(2y-1) = 0\)

Using the zero factor principle:

Solve \(y+4 = 0 : y = -4\)

Solve \(2y+1 = 0 : y = - \frac{1}{2}\)

Solve \(2y-1 = 0 : y = \frac{1}{2}\)


The final solutions to the equation are :

\(y = - 4\)

\(y = -\frac{1}{2}\)

\(y = \frac{1}{2}\)

Answer A

hope this helps! =)

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