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4x^3 - 12x = 0
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4x is a common factor so factor that out first \[\implies 4x(x^2 - 3) = 0\] now equate the factors to zero \[\implies 4x = 0 \quad ; \quad x^2 - 3 = 0\] now solve for x \[\implies \frac{4x}4 = \frac 04 \quad ; \quad x^2 = 3\] \[\implies x = 0 \quad ; \quad x = \pm \sqrt 3\] does that help?
The first part I get like 4x=0 but for x^2 - 3 = 0, isn't that supposed to be like (x+3)(x-3)=0? Then x=+/- 3?
no. (x+3)(x-3) is x^2 - 9
x^2 - 3 is \((x + \sqrt 3)(x - \sqrt 3)\) if you want to write it like that
Ohhh! Okay, okay! I understand it now! Thank You!
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