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
16x^3-3x^2-64x+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*2^3-3*2^2-64*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
16x^3-3x^2-64x+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 16x^3 needs the term -32x^2 to be able to factor out (x-2). Then we add whatever is necessary to get the original -3x^2 (that is, we add 29x^2) and continue pairing for another factoring of (x-2) etc:
16x^3-3x^2-64x+12 =
= 16x^3-32x^2+29x^2-58x-6x+12 =
= 16x^2(x-2)+29x(x-2)-6(x-2) =
= (x-2)(16x^2+29x-6)
Let's do the same with the quadratic polynomial 16x^2+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:
16x^2+29x-6 = 16x^2+32x-3x-6 =
= 16x(x+2)-3(x+2) = (x+2)(16x-3)
Now we have a complete factorization:
16x^3-3x^2-64x+12 = (x-2)(x+2)(16x-3)
