Question

Hi.I need help with math.Please help.Best gets brainiest.

Answer 1

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

Answer 2

$4y^3-4 =y-16y^2$

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! =)