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

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?

## 1 Answer

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.