The Rational Root Theorem: Practice

The theorem's hypothesis is that every coefficient is an integer. The proof clears denominators and then reasons about divisibility, which only makes sense for integers.

$a_n, a_{n-1}, \ldots, a_0 \in \mathbb{Z}$

If an equation has fractional coefficients, multiply through by a common denominator first; the roots do not change, and the new polynomial has integer coefficients the theorem can read.

The leading coefficient is $1$, so the denominator $q$ must divide $1$, forcing $q = 1$: every candidate is an integer. The numerator $p$ must divide the constant term $2$.

$\text{candidates: } \pm 1, \ \pm 2$

Fractions like $\pm\tfrac{1}{2}$ are impossible here, because the leading coefficient $1$ has no divisor other than $1$.

Numerators are divisors of the constant term $-3$, so $p \in \{1, 3\}$. Denominators are divisors of the leading coefficient $2$, so $q \in \{1, 2\}$.

$\pm\tfrac{1}{1}, \ \pm\tfrac{3}{1}, \ \pm\tfrac{1}{2}, \ \pm\tfrac{3}{2}$

That is the eight-number list $\pm 1, \pm 3, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}$.

A rational root $\tfrac{p}{q}$ in lowest terms needs $p \mid 10$, so $p \in \{1, 2, 5, 10\}$, and $q \mid 3$, so $q \in \{1, 3\}$.

$\tfrac{3}{2}\text{ fails both tests: } 3 \nmid 10 \text{ and } 2 \nmid 3$

Each of $\tfrac{5}{3}$, $-10$, and $\tfrac{2}{3}$ passes the divisibility test, so the theorem cannot rule them out (passing still does not make them roots).

Substitute $x = -1$, keeping track of signs: $(-1)^3 = -1$ and $(-1)^2 = 1$.

$P(-1) = -1 + 2 + 5 - 6 = 0$

So $-1$ is a root. (Dividing by $x + 1$ gives $x^2 + x - 6 = (x + 3)(x - 2)$, so the other roots are $-3$ and $2$.)

For a rational root $\tfrac{p}{q}$ in lowest terms, the denominator must divide the leading coefficient. Here that coefficient is $1$.

$q \mid 1 \;\Rightarrow\; q = 1$

So every rational root is an integer $p$, and $p$ divides the constant term. The theorem never guarantees that any rational root exists; it only restricts where one could be.

Factor the quadratic.

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

The roots are $x = -1$ and $x = -2$. Both sit on the candidate list, as the theorem requires, while the positive candidates $1$ and $2$ fail: every coefficient is positive, so no positive number can be a root.

The denominator of a rational root must divide the leading coefficient $4$.

$q \in \{1, 2, 4\}$

The divisors of $9$ are the possible numerators, not denominators; mixing the two roles up is the classic error.

The numerator of a rational root must divide the constant term $-9$, and divisibility ignores the sign.

$p \in \{1, 3, 9\}$

Combining with $q \in \{1, 2, 4\}$ gives the full candidate list $\pm 1, \pm 3, \pm 9, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}, \pm\tfrac{9}{2}, \pm\tfrac{1}{4}, \pm\tfrac{3}{4}, \pm\tfrac{9}{4}$.

The theorem is a one-way implication: a rational root must appear on the list.

$\text{rational root} \;\Rightarrow\; \text{on the list}$

Nothing runs backwards. The polynomial may have three rational roots, one, or none at all; only testing the candidates can tell.

Numerators divide the constant term $-4$, so $p \in \{1, 2, 4\}$; denominators divide the leading coefficient $5$, so $q \in \{1, 5\}$.

$\pm 1, \ \pm 2, \ \pm 4, \ \pm\tfrac{1}{5}, \ \pm\tfrac{2}{5}, \ \pm\tfrac{4}{5}$

The second option is the flipped-fraction trap: it uses divisors of the leading coefficient as numerators.

The Factor Theorem makes the conversion from root to factor.

$P(c) = 0 \;\Longleftrightarrow\; (x - c) \text{ divides } P(x)$

The Rational Root Theorem only supplies the candidates worth testing; the Factor Theorem is what turns a successful test into a factorization step.