Why isn't every reducible cubic polynomial over $F_q[x]$ of the form $(x^2 + ax + b)(x - c)$?

Question

In trying to determine the number of monic irreducible cubic polynomials over $F_q[x]$, where $q$ is prime, I thought that since every reducible cubic must contain a linear factor, each reducible cubic should be able to take the form $$(x+a)(x^2+bx+c)$$ And since there are $q^2$ quadratics over $F_q[x]$, and $q$ linear factors, there must be $q^3$ reducible cubics, but this is obviously a problem because there are then no irreducible cubics!

I've seen ways of determining the number of irreducible cubics here. Why doesn't my reasoning work? What did I miss?

Answer 1

Some reducible cubics look like $$(x+r)(x+s)(x+t)$$ and when $r$, $s$ and $t$ are distinct, your method counts them three times.