One possibility is to note the $8x^3$ and $3$ at each end and then, thinking of the rational root theorem, see if any of $x=\pm3,\pm1,\pm\frac32,\pm\frac12,\pm\frac34,\pm\frac14,\pm\frac38,\pm\frac18$ make your expression zero.
Three of them do in this case, namely $x=-\frac32,+\frac12 \text{ and } -\frac12$, which gives you the complete factorisation in this particular example of $(2x+3)(2x-1)(2x+1)$, but even if you only noticed one then you could make a start.
This only works when there is a rational factor.