Yes, provided that you allow that two of those linear factors may have complex numbers in them.
The fundamental theorem of algebra guarantees that the cubic has 3 roots.
The factor theorem guarantees that if a is a root then (x - a) is a linear factor.
So, while the cubic may not be “factorable over the real numbers,” it does have 3 linear factors, 2 of which may be complex, and all of which may be irrational. The cubic has at least one real root.
