Is there a quick technique to determine if a polynomial (Degree 3 or greater) is factorable?

Question

Say for example the following cubic, ${2x^3-x^2-2x+1}$. This cubic is factorable to $(x-1)(x+1)(2x-1)$

Are there any techniques or tricks to quickly determine if a polynomial like this is factorable? What about quartics?

Answer 1

The answer strictly depends on what field you have the polynomial defined on: in complex numbers $\mathbb C$, the only irreducible polynomials are the linear ones, for $\mathbb C$ is algebraic closed; in $\mathbb R$, every polynomial of degree $\geq 3$ is factoriable; nevertheless in $\mathbb Q$ (or equivalently in $\mathbb Z$) every degree is admissible to have decomposable polynomials.

In case of $\mathbb Q$, you have Gauss’ Theorem which specifies prime and irreducible polynomials. Even the “reduction modulo p theorem” and Eisentstein Criterion are useful tools.

Answer 2

One of my favorite techniques for factoring cubic polynomials is "factoring by grouping."

The main idea is to look at the two terms of highest degree somewhat independently of the other two terms, and to see what can be factored out of each pair of terms. If, after factoring something out of each pair of terms, what remains in the same, then you've discovered a factorization.

For example: $$\begin{align} 2x^3-x^2-2x+1&=(2x^3-x^2)+(-2x+1)\\ &=\color{red}{x^2}(2x-1)\color{red}{-1}(2x-1)\\ &=(\color{red}{x^2-1})(2x-1)\\ &=(x-1)(x+1)(2x-1) \end{align}$$

I don't know any similar techniques for quartics, though I'm fairly sure they exist.

Answer 3

A cubic polynomial is factorable iff it has a linear factor. This is always the case for factoring over the real numbers (by the Intermediate Value Theorem) and complex numbers (buy the Fundamental Theorem of Algebra).

A polynomial has a linear factor over the rational numbers, say, $x - a$, iff it has a root, respectively, $a$, and the Rational Root Theorem says that any rational root of a polynomial $p(x) := a_n x^n + \cdots + a_1 x + a_0$ has the form $$\pm \frac{r}{s} \qquad \textrm{for some } r \mid a_0, s \mid a_n.$$

In the given example, $p(x) = 2 x^3 - x^2 - 2 x + 1$, the theorem gives that the only possible roots are $\pm \frac{1}{2}, \pm 1$, and these can be checked manually.

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For quartic polynomials, factorability of a polynomial does not imply that it has a linear factor (consider $(x^2 + 1)^2$ over the real numbers), but it does imply that it has a factorization of one of the forms $$p(x) = (\textrm{linear}) \cdot (\textrm{cubic}), \qquad p(x) = (\textrm{quadratic}) \cdot (\textrm{quadratic}) . $$ We can test for the first possibility using the Rational Room Theorem, and we can test the latter by writing $$p(x) = (a x^2 + b x + c) (d x^2 + e x + f) .$$ A polynomial with integer coefficients is factorable over the rationals iff it is factorable over the integers, which leaves finitely many possibilities for the pairs $(a, d), (c, f)$.

For example, it's easy to see that $q(x) := x^4 + x^2 + 1$ has no real roots. Checking for a quadratic factorization like the above, we have $a d = 1$, so we have $a = d = \pm 1$, and by moving a factor over, we may as well assume $a = d = 1$. Likewise $c = f = \pm 1$. The case $q(x) = (x^2 + b x - 1)(x^2 + e x - 1)$ doesn't have any solutions $b, e$ (in fact, both of these quadratic factors have real roots, which in particular would be real roots of $q$, a contradiction), but expanding $q(x) = (x^2 + b x + 1) (x^2 + e x + 1)$ quickly gives the solutions $b = -e = 1$.

For larger degrees we can generalize our technique from the quartic case, though often it's less work to establish existence/nonexistence of a factorization through some other means, like those mentioned in Blumer's answer.