In this exercise, $4x$ is the greatest common factor. We find $$\begin{aligned} 4x^3+12x^2-12x\amp=4x(\phantom{x^2}+\phantom{3x}-\phantom{3})\\ \amp=4x(x^2+\phantom{3x}-\phantom{3})\\ \amp=4x(x^2+3x-\phantom{3})\\ \amp=4x(x^2+3x-3) \end{aligned}$$ Note that you might fail to recognize that $4x$ is the greatest common factor. At first you might only find that, say, $4$ is a common factor or even $2\text{.}$ This is OK—you can factor out the $4$ first and continue from there: $$\begin{aligned} 4x^3+12x^2-12x\amp=4(\phantom{x^3}+\phantom{3x^2}-\phantom{3})\\ \amp=4(x^3+\phantom{3x^2}-\phantom{3x})\\ \amp=4(x^3+3x^2-\phantom{3x})\\ \amp=4(x^3+3x^2-3x) \end{aligned}$$ Now examine the second factor here and there is still a common factor, $x\text{.}$ Factoring this out too. $$\begin{aligned} \amp=4x(x^2+3x-3) \end{aligned}$$ So there is more than one way to find the answer here.