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)#
