Douglas B. answered 05/08/20
Algebra tutor with masters degree in applied math
A very simple example would be f(x) = x3+2. From the rational root theorem, we know that any rational roots of f(x) must be in the set {1,-1,2,-2}. None of these values are zeros of f. Therefore, f has no rational roots and thus it cannot be factored as (x-a)(x-b)(x-c), nor as (x-a)(x^2+bx+c).
Make sense?