SOLUTION: I've tried to work this out, I have one answer to the solution set, and I thought I had the others, but according to the answers in the back of the book, I don't! I was wondering i

Question 28800: I've tried to work this out, I have one answer to the solution set, and I thought I had the others, but according to the answers in the back of the book, I don't! I was wondering if you can help.
(Solve each polynomial equation by factoring and then using the zero-product principle) 2x-3=8x(cubed)-12x(squared)
I set it to zero--- 8x(cubed)-12x(squared)-2x+3=0 I then factored by grouping.
(8x(cubed)-12x(squared)) (-2x+3) In the first set, I factored out a -4x(squared) leaving the equation like so.... -4x(squared)(-2x+3) and then the other set, I just set to zero like so....-2x+3=0.
The first equation I set -4x(squared) to zero....-4x(squared)=0, divided by -4 leaving x(squared)=-4 the other set I did the following.... I subtracted 3 from both sides, leaving -2x=-3. I then divided by -2 on bothe sides leaving x=-3/-2, giving me the set of 3/2. According to the book, this is coorect, the other set is where I'm confused. -4x(squared)=0, I divided -4 on both sides leaving x(squared)=0, then I'm not sure what to do! The answers in the back of the book, says the solution set is (-1/2, 1/2, and 3/2)

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
8x^3-12x^2-2x+3=0
4x^2(2x-3)-(2x-3)=0
(2x-3)(4x^2-1)=0
If 2x-3=0 then x=3/2
If 4x^2-1=0 then x^2=1/4 then x=1/2 or x=-1/2
Cheers,
Stan H.
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