3x3-18x+12=3(x3-6x+4).
Consider x3-6x+4 and what the rational zeroes might be. Their product has to be have a magnitude of 4, so that gives us factors 1, 2, 4 only. Let x=1, then x3-6x+4=1-6+4=-1 and when x=-1 it's -1+6+4=9m so ±1 is not a factor. Now try x=2: 8-12+4=0, so x=2 is a factor. We can use synthetic division to find the quadratic:
2 | 1 0 -6 4
1 2 4 | -4
1 2 -2 | 0 = x2+2x-2.
To find the factors of x2+2x-2:
x2+2x-2=0,
x2+2x=2,
x2+2x+1=3,
(x+1)2=3,
x+1=±√3,
x=√3-1 or -(√3+1).
So complete factorisation is 3(x-2)(x+1-√3)(x+1+√3).
