f(x) = x³ + 3x² + 36x + 108

We can take x² out of the first two terms:

f(x) = x²(x + 3) + 36x + 108

We can take 36 out of the last two terms:

f(x) = x²(x + 3) + 36(x + 3)

We can take (x + 3) out of each of those two terms:

f(x) = (x + 3)(x² + 36)

We were told in an earlier algebra course that we
can factor the difference of two squares, but we
cannot factor the sum of two squares.  But that
was before we had studied about imaginary numbers.
But now we have learned that i² = -1 and so -i² = 1.
So we can multiply the 36 by -i² without changing 
its value, so we get

f(x) = (x + 3)[x² + 36·(-i²)]

And now we have:

f(x) = (x + 3)(x² - 36i²)

And we do know how to factor the difference of two
squares, so we get:

f(x) = (x + 3)(x - 6i)(x + 6i)

That's it.  The trick is to change the sum of two
squares into the difference of two squares by multiplying
the second square by -i², since that is just the same as 
multiplying by 1.

Edwin