To solve this type of problem, We have to find out at least one root by hit and trial using the factor theorem.

Here is a trick:
If a polynomial function has integer coefficients, then every rational zero will have the form qp​, where p is a factor of the constant and q is a factor of the leading coefficient.
Here
p=±1,±2,±3,±4,±6,±12 and
q=±1
Find every combination of ±qp​.
These are the possible roots of the polynomial function.
±1,±2,±3,±4,±6,±12

Substituting x=1 in f(x), we get
f(−1)​=(−1)3−13(−1)12=−1+13−12=13+13=0​
∴(x+1) is a factor of f(x)

Now, dividing f(x) by (x+1), we get
x3−13x−12=(x+1)(x2−x−12)
=(x+1)(x2−4x+3x−12)
=(x+1){x(x−4))+3(x−4)}
=(x+1)(x+3)(x−4)