ALGEBRA QUESTION- factoring?

Question

How do you solve for x for this equation by factoring?
I know the answer is x=0, x=1/2, or x= -1/2

Here is the equation:
8x^3 - 2x=0

Answer 1

8x^3-2x=0
2x(4x^2-1)=0
2x{(2x)^2-(1)^2}=0
2x(2x+1)(2x-1)=0
either 2x=0 or 2x+1=0 or 2x-1=0
If 2x=0,then x=0
If 2x+1=0,then 2x= -1 or x= -1/2
If2x-1=0,then 2x=1 or x=1/2
x=0 or -1/2 or 1/2

Answer 2

2x is the common factor in the equation. hence we take out the common factor and use it to divide the two terms
2x(4x^2 - 1) = 0
Here we a product of two numbers being equal to zero. To find x equate each factor to zero.
2x = 0 and 4x^2 - 1 = 0
x = 0/2 4x^2 = 1
x = 0 x^2 = 1/4
x = sq. rt 1/4
x = 1/2

Answer 3

Factor out 2x and set equal to 0 which =0
Factor (4x^2-1) = 0
(2x+1)(2x-1) = 0
Set equal to 0
2x+1 = 0
2x - 1 = 0
Solve for x

Answer 4

There are 2x in each term, so rewrite the equation as
2x(4x^2 -1) = 0
either 2x=0 or 4x^2-1=0
x=0       x^2=1/4
x=0       x=+/- 1/2

Answer 5

X is a factor pretty obviously since you can divide it out of both terms so you have:
x(8X^2-2)

So if you divide both sides by X you get:

8X^2-2=0

You can subtract 2 on both sides and get:

8X^2=2

Divide both sides by 8 and you get:

X^2=1/4.

Just remember that when you take the square root of both sides, you have two answers. Both (-1/2*-1/2) and (1/2*1/2) both give 1/4.

Answer 6

2x (4x² - 1) = 0
2x = 0
x = 0

4x² - 1 = 0
x² = 1 / 4
x = ± 1 / 2

x = 0 , x = ± 1 / 2

Answer 7

8x³ - 2x=0
2x(4x² - 1) = 0
Factoring: 2x(2x - 1)(2x + 1) = 0
>#<