Find all linear factors of 'x^3 - 5 x^2 - 12 x + 36' via the rational root theorem. 
Do this by finding rational roots. 
The candidates are 'x = ± p/q' for all 'p' that are divisors of the constant term '36' and for all 'q' that are divisors of the leading coefficient '1'.

The possible rational roots of 'x^3 - 5 x^2 - 12 x + 36' are:

x = ± 1, 
x = ± 2, 
x = ± 3, 
x = ± 4, 
x = ± 6, 
x = ± 9, 
x = ± 12, 
x = ± 18, 
x = ± 36 

Of these, x = 2, x = -3 and x = 6 are roots. This gives 'x - 2', 'x + 3' and 'x - 6' as all factors:

x^3 - 5 x^2 - 12 x + 36 ​= (x - 2) (x + 3) (x - 6)