Is there an easy way to tell whether a cubic or quartic polynomial is factorable over the integers?

Question

For a quadratic, it is easy to tell whether it is factorable. If the discriminant is a perfect square, the quadratic is factorable. Otherwise, the quadratic is not factorable. Is there anything similar for cubic and quartic polynomials? For a cubic, the only method that I can think of is to use the rational root theorem. If every possible root fails, the cubic is not factorable since it must contain at least one linear factor. However, this can take a while to verify that the cubic is indeed non-factorable, especially if both the constant and leading coefficient has a lot of factors. Is there any easy value to compute, such as the quadratic discriminant, that will make it obvious whether the cubic is factorable over the integers? As far as I can tell, the cubic discriminant does not give any information on whether the cubic is factorable or not. For quartics, the situation is even harder. Even if the rational root theorem fails, it may still be factorable into two quadratics. Is there any easy value to compute for the quartic to determine whether it is factorable over the integers?

Answer 1

There isn't a one-size-fits-all solution. There are strategies that one could try, for example reduction modulo a prime or a prime power. There are some well known criteria for detecting whether a polynomial is not factorable over the integers/rational numbers, for example:

• Eisenstein's criterion: if $f=a_nx^n+\dots+a_1x+a_0$ is a polynomial with integer coefficients and $p$ a prime that divides all coefficients except $a_n$, $p$ does not divide $a_n$ and $p^2$ does not divide $a_0$, then $f$ is irreducible over the rational numbers.
• Reduction modulo a prime: let $f$ be a polynomial with integer coefficients and $\overline f$ obtained from $f$ by reducing the coefficinets of $f$ modulo a prime number $p$. If $\deg{\overline{f}}=\deg{f}$ and $\overline{f}$ is irreducible over the finite field $\mathbb Z_p$, then $f$ is irreducible over the rationals.

An example that defeats both criteria is $f = x^4 + 1$. This polynomial is clearly irreducible over the rationals, however Eisenstein's criterion can't be applied. Furthermore, for any prime $p$ the reduction $\overline{f}=f \operatorname{mod} p$ is reducible in $\mathbb Z_p$ but $f$ is irreducible.

There are algorithmic solutions like Berlekamp's algorithm, or the Cantor–Zassenhaus algorithm.

Out of all of these, the rational root test is often going to be simplest approach, unless the number of candidates to be tested is obscenely large.