Let's assume that factors contain integer numbers and will look like (x−a). In this case the value of a, if substituted as x would evaluate this expression to 0, that is a would be a solution of an equation
16x3−3x2−64x+12=0
Integer solutions to such an equation with integer coefficients must be among the divisors of a free member 12, that is they are supposed to be equal to ±2 and ±3.
Let's check if any of these four candidates is a solution.
Starting with x=2, we get
16⋅23−3⋅22−64⋅2+12=
=16⋅8−3⋅4−64⋅2+12=128−12−128+12=0
Since x=2 is solution, our expression can be represented as
16x3−3x2−64x+12=(x−2)⋅A
where A we can determine very simply by the following procedure.
To have (x−2) as a factor, we transform the original expression in such a way that (x−2) will be a factor for every pair of terms. The term 16x3 needs the term −32x2 to be able to factor out (x−2). Then we add whatever is necessary to get the original −3x2 (that is, we add 29x2) and continue pairing for another factoring of (x−2) etc:
16x3−3x2−64x+12=
=16x3−32x2+29x2−58x−6x+12=
=16x2(x−2)+29x(x−2)−6(x−2)=
=(x−2)(16x2+29x−6)
Let's do the same with the quadratic polynomial 16x2+29x−6.
Again, looking for integer divisors of the free member 6 to find the value that evaluates this polynomial to 0. These are ±2 and ±3. Obviously, 2 is not good, but −2 fits the bill:
16(−2)2+29(−2)−6=
=16⋅4−29⋅2−6=64−58−6=0
Therefore, we can extract (x+2) from the quadratic polynomial. Let's use the same technique as above:
16x2+29x−6=16x2+32x−3x−6=
=16x(x+2)−3(x+2)=(x+2)(16x−3)
Now we have a complete factorization:
16x3−3x2−64x+12=(x−2)(x+2)(16x−3)
