Factorization?

x=2 or x=8 are not the solutions of the given equation.

(2x-4)^2=3x^3-6x=>

4x^2-16x+16=3x^3-6x=>

3x^3-4x^2+10x-16=0=>

The only real solution of the equation is

x=1.493126 approximately.

You may check that if

f(x)=(2x-4)^2-3x^3+6x=0, then

f(2)=/=0

&

f(8)=/=0

 (2x - 4)^2 = 3x^3 - 6x

 4 (x - 2)^2 = 3 x (x^2 - 2)

x = 2 or 8 are not solutions, see below.

(2x – 4)² = 3x³ – 6x

4x² – 16x + 16 = 3x³ – 6x

3x³ – 4x² + 16x – 6x – 16 = 0

3x³ – 4x² + 10x – 16 = 0

now to factor...  which is complicated. A graphical solution is easy...

and it has only one root, x = 1.493 about

(2x – 4)² = 3x³ – 6x

does x = 2 or 8 work?

(2•2 – 4)² = 3•2³ – 6•2

0 = 16 – 12  NO

(2•8 – 4)² = 3•8³ – 6•8

12² = 1536 – 48

144 = 1488  NO

You can try "good guesses".

Or, expand and rewrite as a poly. Then try to find the roots of the poly:

Try RRT, or "the formula"...

Here, the solutions are NOT 2 & 8 btw. 

(2x - 4)² = 3x³ - 6x

I don't see how this has to do anything with factoring, but x = 2 or 8 are not solutions to this equation.  Testing them:

(2 * 2 - 4)² = 3 * 2³ - 6 * 2 and (2 * 8 - 4)² = 3 * 8³ - 6 * 8

(4 - 4)² = 3 * 8 - 12 and (16 - 4)² = 3 * 512 - 48

0² = 24 - 12 and 12² = 1536 - 48

0 = 12 and 144 = 1488

FALSE and FALSE

Neither are solutions.

Looking at a graph, there is only one real (irrational) solution and the other two are complex.

I mean how can I get the answer x=2, and x=8. can someone explain