Does Having No Roots Make a Degree-2 or Degree-3 Polynomial Irreducible?

A familiar test says that a polynomial of degree 2 or 3 is irreducible if it has no roots. That test is correct when the coefficients lie in a field, such as $\mathbb{Q}$ or $\mathbb{R}$. It is not true for every integral domain, because a factor can have a nonunit leading coefficient, so its root need not belong to the domain.

There is a useful version that does work over any integral domain: a monic polynomial of degree 2 or 3 with no roots in the domain is irreducible. The distinction is about the coefficient ring and the polynomial’s leading coefficient.

Why the test works over a field

Let $F$ be a field and let $f\in F[x]$ have degree 2 or 3. If $f$ is reducible, it factors into two polynomials of positive degree. Since the factor degrees add to 2 or 3, at least one factor has degree 1. A linear factor over $F$ gives a root in $F$.

Conversely, if $f$ has a root $r\in F$, the factor theorem gives

$$f(x)=(x-r)q(x),$$

where $q$ has degree one less than $f$. Both factors have positive degree, so $f$ is reducible. Thus, over a field,

$$f\text{ is irreducible}\quad\Longleftrightarrow\quad f\text{ has no root in }F.$$

This argument applies specifically to degrees 2 and 3: a reducible polynomial of either degree must have a linear factor. For higher degrees, a polynomial can factor into two factors of degree at least 2 without having a root in the field.

Why an integral domain is different

In an integral domain, a linear factor can look like $ax+b$ where $a$ is nonzero but not invertible. Its root would be $-b/a$, which may not belong to the domain. So having no roots in the domain does not rule out such a factor.

For example, in $\mathbb{Z}[x]$ consider

$$f(x)=(2x+1)^2=4x^2+4x+1.$$

This is a reducible degree-2 polynomial: it is a product of two degree-1 polynomials in $\mathbb{Z}[x]$. But it has no integer root, since $2n+1$ is never zero for an integer $n$. Its root, $-1/2$, lies in $\mathbb{Q}$, not in $\mathbb{Z}$.

The same issue occurs in degree 3. For example,

$$g(x)=(2x+1)(x^2+1)=2x^3+x^2+2x+1$$

is reducible in $\mathbb{Z}[x]$ and has no integer root. For an integer $n$, neither $2n+1$ nor $n^2+1$ is zero.

There is another way reducibility can arise in a domain: a nonunit constant can divide every coefficient. For instance, $2x^2+2=2(x^2+1)$ is reducible in $\mathbb{Z}[x]$ and has no integer root. The examples show why “no roots in the domain” alone is not enough.

When the implication does hold over a domain

A useful sufficient condition is that the polynomial be monic, meaning its leading coefficient is 1. More generally, it is enough for the leading coefficient to be a unit (an invertible element) of the domain, since you can multiply the polynomial by the inverse of that coefficient to make it monic.

To see why, suppose a monic degree-2 or degree-3 polynomial factors as $f=gh$. The leading coefficients of $g$ and $h$ multiply to 1, so both are units. Neither factor can be a nonunit constant, and their positive degrees must split as 1 and 1, or 1 and 2. The linear factor has unit leading coefficient, so it can be rescaled to $x-r$ for some $r$ in the domain. Thus $f$ has a root there.

The factor theorem gives the converse: a root in the domain makes a degree-2 or degree-3 polynomial reducible. Therefore, for a monic polynomial $f$ of degree 2 or 3 over an integral domain $R$,

$$f\text{ is irreducible in }R[x]\quad\Longleftrightarrow\quad f\text{ has no root in }R.$$

A practical check in $\mathbb{Z}[x]$

For a monic cubic such as $f(x)=x^3+x+1$, test possible integer roots using the rational root theorem: the only integer candidates are divisors of the constant term, so they are $1$ and $-1$.

$$f(1)=3,\qquad f(-1)=-1.$$

Neither is zero. Since $f$ is monic and has degree 3, the result above shows that it is irreducible in $\mathbb{Z}[x]$. It is also irreducible over $\mathbb{Q}$.

For a nonmonic polynomial in a domain, checking only roots in that domain is not enough. You can instead check roots in the fraction field; for degree 2 or 3, no root there rules out any positive-degree factorization over that field. If the domain is a unique factorization domain and the polynomial is primitive (its coefficients have no common nonunit factor), Gauss’s lemma then lets you conclude irreducibility in the domain’s polynomial ring.

Common mistakes

• Using the field test over every domain: A domain need not contain the quotient that would be a linear factor’s root.
• Assuming primitive means monic: The counterexample $(2x+1)^2$ is primitive in $\mathbb{Z}[x]$, but is still reducible there.
• Ignoring constant factors: A nonunit common divisor of the coefficients can make a polynomial reducible even when it has no positive-degree factorization.
• Applying the degree-2/3 test to higher degrees: A degree-4 polynomial, for example, can split as a product of two quadratics and have no root in the field.