Notice that 27, x^3 and 1/8 are all cubes. If you were allowed complex numbers as coefficients, then this could be completely factored as (3x - 1/2)(3 omega x - 1/2)(3 omega ^2x - 1/2), where omega is the complex cube root of 1.
As it is, you are only interested in real numbers, not complex ones. So the only linear factor is (3x - 1/2). The other factor is (9x^2+(3/2)x+1/4).
Here are a couple of ways of finding that quadratic factor:
(1) Using complex arithmetic:
(3 omega x - 1/2)(3 omega ^2x - 1/2) = 9 omega^3 x^2 - (3/2)(omega + omega^2)x + 1/4 = 9x^2 + (3/2)x + 1/4
(since omega^3 = 1 and omega + omega^2 = -1).
(2) Using real arithmetic, solve:
27x^3 - 1/8 = (3x - 1/2)(ax^2 + bx + c)
= 3ax^3 + (3b - (1/2)a)x^2 + (3c - (1/2)b)x - (1/2)c
Comparing coefficients, we get:
3a = 27
(3b - (1/2)a) = 0
(3c - (1/2)b) = 0
(1/2)c = 1/8
yielding a = 9, b = 3/2 and c = 1/4.