Question

Stuck on a factoring question!!
it's 8y^3=2y
(^ is exponent.)
From there I got,
8y^3-2y=0
2y(4y^2-1)=0
I don't know where to go from there!

Answer 1

first do exponents

Answer 2

a^2 - b^2 = (a+b)(a-b)

Answer 3

then i think it should be easy to do it

Answer 4

We use difference of squares to get 2y(2y-1)(2y+1).

Answer 5

once you have factors equal to 0
split up the factors and solve each separately

--> 2y = 0 --> y=0

--> 4y^2 -1 = 0
4y^2 = 1
y^2 = 1/4
y = +-sqrt(1/4) = +-1/2

Answer 6

Are you asked to solve or just factor?

Answer 7

both

Answer 8

Okay. 2y(2y-1)(2y+1) = 0
y = 0, y = 1/2, y = -1/2

Answer 9

Alternatively, you could also do that...but be sure your factoring is not shaky