What if my polynomial has fractions or decimals as coefficients?

The Rational Root Theorem requires all coefficients to be integers. Multiply the entire polynomial by an appropriate constant to clear fractions and decimals without changing the roots, then apply the theorem.

Can the Rational Root Theorem miss a rational root?

No. If \frac{p}{q} is a rational root (in lowest terms), the theorem guarantees that p divides a_0 and q divides a_n. However, not all candidates are roots—you must verify each by substitution.

How is the Rational Root Theorem different from the Factor Theorem?

The Rational Root Theorem predicts which rational numbers might be roots of a polynomial. The Factor Theorem states that if r is a root of P(x), then (x-r) is a factor. Use the theorem to generate candidates, then the Factor Theorem to confirm and factor.

Do I have to test every candidate if I've already found enough roots?

While a cubic has at most three roots (so you could stop early), testing all candidates systematically is best practice to ensure you have found every rational root and to avoid overlooking solutions.

Rational Root Theorem

The Rational Root Theorem tells you which rational numbers could possibly be roots of a polynomial with integer coefficients.

x = \pm\frac{p}{q}, \quad p \mid a_0, ; q \mid a_n

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What each symbol means

What Rational Root Theorem takes

Symbol	Meaning
$p$	The numerator of a potential rational root written as $p/q$ in lowest terms; must be an integer divisor of $a_0$ .
$q$	The denominator of a potential rational root written as $p/q$ in lowest terms; must be an integer divisor of $a_n$ .
$a_0$	The constant term of the polynomial—the term with no variable $x$ —which the numerator $p$ must divide.
$a_n$	The leading coefficient, which is the coefficient of the highest-power term in the polynomial; the denominator $q$ must divide it.
$x$	The unknown variable in the polynomial equation; you are solving for which rational values of $x$ make the polynomial equal zero.

When to use it

You reach for the Rational Root Theorem when you need to find rational roots of a polynomial equation with integer coefficients.

Level

Usually taught in: Algebra II

Worked examples

1. Find all rational roots of a cubic with integer roots

Problem

Find all rational roots of

x^3 - 6x^2 + 11x - 6 = 0

$a_0 = -6, \quad a_n = 1$
Identify the constant term $a_0$ (the term with no $x$ ) and the leading coefficient $a_n$ (coefficient of $x^3$ ).
$\text{Divisors of } -6: \pm 1, \pm 2, \pm 3, \pm 6$
Find all integer divisors of the constant term; any potential numerator $p$ must be on this list.
$\text{Divisors of } 1: \pm 1$
Find all integer divisors of the leading coefficient; any potential denominator $q$ must be on this list.
$x = \pm 1, \pm 2, \pm 3, \pm 6$
These are all possible rational roots; since $q \in \{\pm 1\}$ , the candidates are just integers.
$(1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$
Test $x = 1$ by substituting into the polynomial; it equals zero, so $x = 1$ is a root.
$(2)^3 - 6(2)^2 + 11(2) - 6 = 8 - 24 + 22 - 6 = 0$
Test $x = 2$ ; substitution yields zero, confirming $x = 2$ is a root.
$(3)^3 - 6(3)^2 + 11(3) - 6 = 27 - 54 + 33 - 6 = 0$
Test $x = 3$ ; substitution yields zero, so $x = 3$ is a root.

Answer: $x = 1, 2, \text{ or } 3$

The Rational Root Theorem generated four candidates ( $\pm 1, \pm 2, \pm 3, \pm 6$ ), but testing revealed only three are actual roots. Since $a_n = 1$ , we knew any rational root had to be an integer, which reduced the search to just eight candidates instead of sixteen.

2. Find rational roots when the leading coefficient is not 1

Problem

Find all rational roots of

2x^3 + 3x^2 - 8x - 12 = 0

$a_0 = -12, \quad a_n = 2$
The constant term is $-12$ and the leading coefficient of $x^3$ is $2$ .
$\text{Divisors of } -12: \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$
Find all divisors of $a_0 = -12$ for possible numerators.
$\text{Divisors of } 2: \pm 1, \pm 2$
Find all divisors of $a_n = 2$ for possible denominators.
$x = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{7}{2}$
List all combinations of divisors as $\frac{p}{q}$ ; there are now fractional candidates because $a_n \neq 1$ .
$2(2)^3 + 3(2)^2 - 8(2) - 12 = 16 + 12 - 16 - 12 = 0$
Test $x = 2$ ; it equals zero, so $x = 2$ is a root.
$2(-2)^3 + 3(-2)^2 - 8(-2) - 12 = -16 + 12 + 16 - 12 = 0$
Test $x = -2$ ; it also satisfies the equation, so $x = -2$ is a root.
$2 \left( -\frac{3}{2} \right)^3 + 3 \left( -\frac{3}{2} \right)^2 - 8 \left( -\frac{3}{2} \right) - 12 = -\frac{27}{4} + \frac{27}{4} + 12 - 12 = 0$
Test the fractional candidate $x = -\frac{3}{2}$ ; it satisfies the equation, showing that fractions are possible roots when $a_n \neq 1$ .

Answer: $x = 2, -2, \text{ or } -\frac{3}{2}$

When the leading coefficient is not $1$ , fractional candidates become possible, and the candidate list grew significantly. However, testing still revealed only three roots among all the candidates. The theorem provides the search space, but verification remains essential.

3. Apply the theorem to a word problem about basketball jersey sales

Problem

During a basketball season, the profit (in thousands of dollars) from selling team jerseys is modeled by

P(x) = x^3 - 4x^2 - 7x + 10

, where

x

is the number of thousands of jerseys sold. Find the rational values of

x

where profit is zero.

$a_0 = 10, \quad a_n = 1$
From the profit function, the constant term is $10$ and the leading coefficient is $1$ .
$\text{Divisors of } 10: \pm 1, \pm 2, \pm 5, \pm 10$
List all divisors of the constant term.
$\text{Divisors of } 1: \pm 1$
The leading coefficient is $1$ , so rational roots must be integers.
$x = \pm 1, \pm 2, \pm 5, \pm 10$
These eight integer candidates are the only possible rational roots.
$(1)^3 - 4(1)^2 - 7(1) + 10 = 1 - 4 - 7 + 10 = 0$
Test $x = 1$ thousand jerseys; profit is zero at this sales level.
$(-2)^3 - 4(-2)^2 - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0$
Test $x = -2$ ; while negative (not meaningful for quantity), it is a valid mathematical root.
$(5)^3 - 4(5)^2 - 7(5) + 10 = 125 - 100 - 35 + 10 = 0$
Test $x = 5$ thousand jerseys; profit is zero at this higher sales level.

Answer: $x = 1, -2, \text{ or } 5 \text{ (only } x = 1 \text{ and } x = 5 \text{ are meaningful for jersey quantity)}$

The Rational Root Theorem narrowed the search to eight integer candidates. In context, negative quantities have no meaning, so only $x = 1$ and $x = 5$ represent practical break-even points—but mathematically, all three are legitimate roots of the profit polynomial.

Common mistakes

Where Rational Root Theorem usually goes wrong

The mistake	Why it is wrong	The fix
Students assume all divisors of the constant term are guaranteed roots, writing " $\pm 1, \pm 2, \pm 3, \pm 6$ are all roots of $x^3 - 6x^2 + 11x - 6 = 0$ because they divide 6."	The Rational Root Theorem only provides a list of candidates; the actual roots must be verified by substituting each candidate into the polynomial.	Test each candidate by substitution: keep those that yield zero (true roots) and discard the rest.
Students include $x = \sqrt{2}$ as a candidate for $x^3 - 3x + 1 = 0$ because $2$ divides the constant term.	The Rational Root Theorem applies only to rational roots of the form $\frac{p}{q}$ where both $p$ and $q$ are integers; $\sqrt{2}$ is irrational and lies outside the theorem's scope.	Only test values where the numerator $p$ and denominator $q$ are both integers that divide the respective polynomial coefficients.
For $2x^3 - 5x + 4 = 0$ , a student uses $a_0 = -5$ (the coefficient of $x$ ) instead of $a_0 = 4$ (the constant term).	$a_0$ must be the constant term, which is the term that contains no variable $x$ at all; $-5$ is the coefficient of the $x$ term, not the constant.	Identify $a_0 = 4$ and $a_n = 2$ correctly; the numerator $p$ divides $4$ and the denominator $q$ divides $2$ .

Tips and when to use something else

The Rational Root Theorem gives you a list of candidates—you must test every one by substituting into the polynomial to determine which are actual roots.
If the leading coefficient is $1$ , any rational root must be an integer; you can eliminate all fractional candidates immediately.
To speed up the testing process, use synthetic division instead of direct substitution, or apply numerical methods like Newton-Raphson if no rational candidates work.
Factor out any common monomial factors from the polynomial first; this simplifies the coefficients and reduces the number of candidates to test.

Related formulas

Factor TheoremThe Factor Theorem tells you when a linear binomial divides a polynomial: (x - c) is a factor of P(x) exactly when P(c) = 0.Remainder TheoremThe Remainder Theorem lets you find the remainder when a polynomial is divided by (x - c) by simply evaluating the polynomial at c.Polynomial Long DivisionPolynomial long division breaks a fraction into quotient and remainder, letting you simplify rational expressions or test polynomial divisibility.Synthetic DivisionA fast method for dividing a polynomial by a binomial of the form (x - c), giving quotient and remainder efficiently without long division.Complex NumbersSolve equations with no real solutions and multiply combinations of real and imaginary numbers using the complex number system.

Frequently asked questions

What if my polynomial has fractions or decimals as coefficients?: The Rational Root Theorem requires all coefficients to be integers. Multiply the entire polynomial by an appropriate constant to clear fractions and decimals without changing the roots, then apply the theorem.
Can the Rational Root Theorem miss a rational root?: No. If $\frac{p}{q}$ is a rational root (in lowest terms), the theorem guarantees that $p$ divides $a_0$ and $q$ divides $a_n$ . However, not all candidates are roots—you must verify each by substitution.
How is the Rational Root Theorem different from the Factor Theorem?: The Rational Root Theorem predicts which rational numbers might be roots of a polynomial. The Factor Theorem states that if $r$ is a root of $P(x)$ , then $(x-r)$ is a factor. Use the theorem to generate candidates, then the Factor Theorem to confirm and factor.
Do I have to test every candidate if I've already found enough roots?: While a cubic has at most three roots (so you could stop early), testing all candidates systematically is best practice to ensure you have found every rational root and to avoid overlooking solutions.

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Reviewed 2026-09-18