The Rational Root Theorem

Intuition

The last two lessons handed you a fast machine. Pick a number $c$ , run synthetic division, and the remainder tells you whether $x - c$ is a factor of your polynomial. What nobody has told you yet is which $c$ to feed the machine. A cubic might have a root at $\tfrac{1}{2}$ , or $-3$ , or $\tfrac{7}{4}$ , and there are infinitely many rational numbers you could try. Testing them one by one is hopeless. The Rational Root Theorem repairs exactly this gap. Reading nothing but the first and last coefficients, it produces a complete, finite list of every rational number that could possibly be a root. The infinite haystack collapses to a handful of candidates, and synthetic division can check each one in seconds. The theorem does not find roots for you; it tells you the only places a rational root could hide.

Learning goals

State that $p$ divides the constant and $q$ the leading coefficient
List every candidate $\pm\tfrac{p}{q}$ , reduced and deduplicated
Test candidates by synthetic division, then peel and finish
Reduce to integer divisors of the constant when monic
Remember the converse is false, since most candidates fail
Prove irrationality from a fully failed candidate list

A pattern hiding in the factored form

Take a factored quadratic whose factors have integer coefficients and expand it:

(2x - 1)(3x + 2) = 6x^2 + x - 2.

Its roots are $x = \tfrac{1}{2}$ and $x = -\tfrac{2}{3}$ . Now look at where each part of those fractions came from. The constant term $-2$ of the expanded polynomial is the product $(-1)\times 2$ of the two constants, so each root’s numerator ( $1$ and $2$ ) divides the constant term. The leading coefficient $6$ is the product $2 \times 3$ of the two $x$ -coefficients, so each root’s denominator ( $2$ and $3$ ) divides the leading coefficient.

That is not a coincidence of this example. When a polynomial with integer coefficients has a rational root, the fraction’s numerator is always locked to the constant term and its denominator to the leading coefficient. The theorem below says exactly that, and the proof shows why the two end coefficients, and only they, control where rational roots can live.

The theorem

The Rational Root Theorem. Let

P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0

be a polynomial whose coefficients $a_n, a_{n-1}, \ldots, a_0$ are all integers, with $a_n \ne 0$ . If the rational number $\tfrac{p}{q}$ , written in lowest terms (so $p$ and $q$ are integers, $q \ge 1$ , and $\gcd(p, q) = 1$ ), is a root of $P(x)$ , then

p \mid a_0 \qquad \text{and} \qquad q \mid a_n,

that is, $p$ divides the constant term and $q$ divides the leading coefficient.

Three remarks before the proof. First, the hypothesis matters: the coefficients must be integers. If your equation has fractional coefficients, multiply through by a common denominator first; that changes no roots and produces an integer-coefficient polynomial the theorem can read. Second, the theorem is a one-way implication. It says every rational root must pass the divisibility test; it does not say that every number passing the test is a root. Most passers fail when you actually substitute them in. Third, if $a_0 = 0$ the conclusion $p \mid 0$ is true for every integer $p$ (since $0 = p \times 0$ ), so the theorem filters out nothing. In that case factor out the largest power of $x$ first: $x = 0$ is a root, and the cofactor has a nonzero constant term the theorem can grip.

Proof of the Rational Root Theorem#

Suppose $P\!\left(\tfrac{p}{q}\right) = 0$ with $\gcd(p, q) = 1$ and $q \ge 1$ . Substituting $x = \tfrac{p}{q}$ into $P(x) = 0$ gives

a_n\left(\frac{p}{q}\right)^{\!n} + a_{n-1}\left(\frac{p}{q}\right)^{\!n-1} + \cdots + a_1\!\left(\frac{p}{q}\right) + a_0 = 0.

Fractions are awkward to reason about, so clear them: multiply both sides by $q^n$ , which is not zero. Each power $\left(\tfrac{p}{q}\right)^{k} = \tfrac{p^k}{q^k}$ picks up exactly the factor $q^{n-k}$ it needs, and the equation becomes

a_n p^n + a_{n-1}p^{n-1}q + \cdots + a_1 p\,q^{n-1} + a_0 q^n = 0.

Every term here is an integer, because the coefficients, $p$ , and $q$ all are. This single equation yields both conclusions, one from each end.

For the first, move the last term across. Every remaining term contains at least one factor of $p$ , so $p$ factors out of the whole bracket:

\begin{aligned} a_0 q^n &= -\left(a_n p^n + a_{n-1}p^{n-1}q + \cdots + a_1 p\,q^{n-1}\right) \\ &= -p\left(a_n p^{n-1} + a_{n-1}p^{n-2}q + \cdots + a_1 q^{n-1}\right). \end{aligned}

The bracket on the right is an integer, so $p$ divides the product $a_0 q^n$ . Now the lowest-terms hypothesis does its work. If $a_0 = 0$ then $p \mid a_0$ holds trivially, so assume $a_0 \ne 0$ and compare prime factorizations. Because $\gcd(p, q) = 1$ , no prime divides both $p$ and $q$ . The prime factors of $q^n$ are exactly the prime factors of $q$ , so by that coprimality no prime factor of $p$ appears in $q^n$ at all. Take any prime $r$ that appears in the factorization of $p$ , say to the exponent $e$ . Since $p$ divides $a_0 q^n$ , the factorization of $a_0 q^n$ contains $r$ at least $e$ times, and none of those copies can come from $q^n$ , which contains no $r$ whatsoever. All of them therefore sit inside $a_0$ . This holds for every prime power in $p$ , so the entire factorization of $p$ is contained in that of $a_0$ , which is precisely the statement $p \mid a_0$ . (Signs never disturb a divisibility claim, since a factor of $-1$ can be absorbed into the cofactor.)

For the second conclusion, run the same argument from the other end. Move the first term across; every remaining term contains at least one factor of $q$ :

\begin{aligned} a_n p^n &= -\left(a_{n-1}p^{n-1}q + \cdots + a_1 p\,q^{n-1} + a_0 q^n\right) \\ &= -q\left(a_{n-1}p^{n-1} + a_{n-2}p^{n-2}q + \cdots + a_1 p\,q^{n-2} + a_0 q^{n-1}\right). \end{aligned}

So $q$ divides $a_n p^n$ . By the identical prime-factor comparison, with the roles of $p$ and $q$ exchanged, no prime factor of $q$ appears in $p^n$ . So, since $q$ divides $a_n p^n$ , every prime power in $q$ must sit inside $a_n$ , and $q \mid a_n$ .

Notice where each hypothesis earned its keep. Integer coefficients made every term of the cleared equation an integer, so divisibility talk made sense at all. Lowest terms separated the primes of $p$ from the primes of $q$ . That separation is what let each divisibility conclusion land on the coefficient rather than on a power of the other variable. And notice, finally, what was proved: a rational root forces the two divisibilities. Nothing in the argument runs backwards, and a fraction that satisfies both divisibilities has no obligation to be a root.

Building the candidate list

The theorem turns root-hunting into a bookkeeping exercise. For a polynomial with integer coefficients and nonzero constant term:

List the positive divisors of the constant term $a_0$ . These are the possible numerators $p$ .
List the positive divisors of the leading coefficient $a_n$ . These are the possible denominators $q$ .
Form every quotient $\tfrac{p}{q}$ , reduce each to lowest terms, discard duplicates, and attach a $\pm$ sign to each survivor.

Take $P(x) = 2x^3 + 3x^2 - 8x + 3$ . The constant term is $3$ , with divisors $1$ and $3$ ; the leading coefficient is $2$ , with divisors $1$ and $2$ . The quotients are $\tfrac{1}{1}, \tfrac{3}{1}, \tfrac{1}{2}, \tfrac{3}{2}$ , so the complete candidate list is

\pm 1, \quad \pm 3, \quad \pm \tfrac{1}{2}, \quad \pm \tfrac{3}{2}.

Eight numbers. Any rational root of $P$ must be one of these eight; every other rational number in existence is ruled out before you compute a single value.

Two small housekeeping points. Duplicates really do occur: for $2x^3 - x^2 - 8x + 4$ the raw quotients include $\tfrac{2}{2}$ and $\tfrac{4}{2}$ . Those two reduce to $1$ and $2$ , which are already on the list, leaving $8$ distinct candidates rather than $12$ . And the signs matter: roots can be negative, so the list always carries both signs even when every coefficient is positive.

Steps 1 and 2 look like two different instructions, but they are one question asked twice: does this whole number divide that coefficient? The figure below is that question with its answer drawn. Set a count of dots and a row width, and the rows either come out even or they do not.

Suppose the constant term is $12$ . Set the count to $12$ and walk the row width from $2$ up to $6$ . Rows of $2$ , $3$ , $4$ and $6$ come out even, and the readout names each width as a factor of $12$ . Rows of $5$ leave two dots hollow, and the readout says plainly that $5$ is not a factor. Two divisors sit outside what the figure can draw, the width $1$ and the width $12$ itself. Those two divisors are free, because every number is divisible by $1$ and by itself. So the numerators are $1, 2, 3, 4, 6, 12$ . Now run the same walk on a leading coefficient of $4$ : set the count to $4$ . Rows of $2$ come out even, while rows of $3$ leave one dot hollow, so the denominators are $1, 2, 4$ . Every candidate the theorem allows is one of those numerators over one of those denominators, and the hollow dots are the whole of what rules a number out.

Dot array

Dots Rows of

12 dots in rows of 6 make 2 full rows with none left over. So 12 = 2 times 6. So 6 is a factor of 12.

A number of dots tested against a row width, with any dot short of completing a row drawn hollow. A width that leaves nothing hollow is a divisor of the count, and that single test is what both halves of the candidate list are built from.

Check your understanding

According to the Rational Root Theorem, which of these numbers CANNOT be a root of $4x^3 - x + 6 = 0$ ?

-\tfrac{3}{2}

\tfrac{2}{3}

\tfrac{1}{4}

-6

The theorem is a filter, not an oracle. For 2x³ + 3x² - 8x + 3 it cuts the infinitely many rationals down to eight candidates; testing each one shows only three are roots.

When the leading coefficient is 1

A polynomial is monic when its leading coefficient is $1$ . Then the theorem’s second condition $q \mid 1$ forces $q = 1$ , and something pleasantly strong follows. Every rational root of a monic integer-coefficient polynomial is an integer, and that integer must divide the constant term. No fractions can occur at all. This special case is sometimes called the integer root theorem.

It bites quickly. Consider $P(x) = x^3 - 4x + 2$ . Any rational root must be an integer dividing $2$ , so the whole candidate list is $\pm 1, \pm 2$ . Substituting gives $P(1) = -1$ , $P(-1) = 5$ , $P(2) = 2$ , and $P(-2) = 2$ . None is zero, so this cubic has no rational roots whatsoever: whatever its roots are, every one of them is irrational or nonreal. Four quick evaluations settled a question about all infinitely many rational numbers.

From candidates to roots

The theorem supplies the shortlist; the tools from earlier in this chapter do the checking. The full workflow for factoring an integer-coefficient polynomial:

Build the candidate list from $a_0$ and $a_n$ .
Test candidates one at a time. Integers before fractions, small before large: synthetic division is quickest, and by the Remainder Theorem the number at the end of the row is exactly $P(c)$ . So one synthetic division tableau both tests the candidate and, on success, hands you the quotient.
When a candidate $c$ gives remainder $0$ , the Factor Theorem converts the discovery into algebra: $x - c$ is a factor, and $P(x) = (x - c)\,Q(x)$ with $Q$ one degree lower. Keep hunting inside $Q$ , and test $c$ itself again, since roots can repeat.
When the quotient drops to a quadratic, stop hunting. Factor it directly or run the quadratic formula, which finds all remaining roots, rational or not.

Worked example 1 Find every root of $2x^3 + 3x^2 - 8x + 3$

The candidate list, built above, is $\pm 1, \pm 3, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}$ . Start with the easiest candidate, $x = 1$ , by direct substitution:

P(1) = 2 + 3 - 8 + 3 = 0.

A root on the first try. By the Factor Theorem $x - 1$ divides $P(x)$ , and synthetic division by $1$ produces the quotient. Bring down the $2$ ; multiply by $1$ and add to $3$ to get $5$ . Then multiply by $1$ and add to $-8$ to get $-3$ ; multiply by $1$ and add to $3$ to get the remainder $0$ :

\begin{array}{c|rrrr} 1 & 2 & 3 & -8 & 3 \\ & & 2 & 5 & -3 \\ \hline & 2 & 5 & -3 & \boxed{0} \end{array}

So $P(x) = (x - 1)\left(2x^2 + 5x - 3\right)$ . The quotient is a quadratic, so the hunt is over: factor it directly,

2x^2 + 5x - 3 = (2x - 1)(x + 3),

which you can confirm by expanding. The complete factorization is $P(x) = (x-1)(2x-1)(x+3)$ , and the roots are

x = 1, \qquad x = \tfrac{1}{2}, \qquad x = -3.

All three sit on the original candidate list, exactly as the theorem promised. Note the other side of that promise: five of the eight candidates turned out not to be roots. The list narrows the search; it never claims every entry succeeds.

Check your understanding

The number $3$ is a root of $P(x) = x^3 - 2x^2 - 2x - 3$ . Divide by $(x - 3)$ using synthetic division. What is the quotient?

x^2 + x + 1

x^2 - x + 1

x^2 + x - 1

x^2 - 5x + 13

Worked example 2 Find every root of $3x^3 - 5x^2 + 5x - 2$

The constant term $-2$ has divisors $1, 2$ and the leading coefficient $3$ has divisors $1, 3$ , so the candidates are $\pm 1, \pm 2, \pm\tfrac{1}{3}, \pm\tfrac{2}{3}$ .

Half the list can be dismissed by a sign observation. For any $x < 0$ , each of the four terms $3x^3$ , $-5x^2$ , $5x$ , $-2$ is negative, so $P(x) < 0$ and no negative number can be a root. Only the four positive candidates need testing:

P(1) = 3 - 5 + 5 - 2 = 1, \qquad P(2) = 24 - 20 + 10 - 2 = 12,

P\!\left(\tfrac{1}{3}\right) = \tfrac{1}{9} - \tfrac{5}{9} + \tfrac{15}{9} - \tfrac{18}{9} = -\tfrac{7}{9}.

Three failures. Test the last candidate, $\tfrac{2}{3}$ , by synthetic division, letting the remainder be the verdict:

\begin{array}{c|rrrr} \tfrac{2}{3} & 3 & -5 & 5 & -2 \\ & & 2 & -2 & 2 \\ \hline & 3 & -3 & 3 & \boxed{0} \end{array}

The remainder is $0$ , so $\tfrac{2}{3}$ is a root and the quotient row gives $P(x) = \left(x - \tfrac{2}{3}\right)\left(3x^2 - 3x + 3\right)$ . Pull the common factor $3$ out of the quadratic and fold it into the linear factor:

P(x) = (3x - 2)\left(x^2 - x + 1\right).

The quadratic ends the hunt. Its discriminant is $(-1)^2 - 4(1)(1) = -3 < 0$ , so by the quadratic formula its roots are the complex conjugate pair from the last chapter:

x = \frac{1 \pm i\sqrt{3}}{2}.

The full set of roots is $x = \tfrac{2}{3}$ together with $x = \tfrac{1 \pm i\sqrt{3}}{2}$ . The rational root theorem could only ever see the first of the three. The conjugate pair lives entirely outside the theorem’s field of view, and only the quadratic formula could catch it.

What the theorem does not say

Two misreadings account for nearly every error with this theorem, and both are worth staring at.

The list contains candidates, not roots. The implication runs from root to divisibility, never backwards. Consider $P(x) = x^2 - 2$ . The number $2$ passes both divisibility checks, since $p = 2$ divides the constant term $-2$ and $q = 1$ divides the leading coefficient. Yet $P(2) = 2 \ne 0$ . A place on the list is an invitation to test, nothing more. In practice most candidates fail, and a polynomial may pass through its entire list without a single success.

The theorem sees only rational roots. Finish the $x^2 - 2$ story: its candidates are $\pm 1, \pm 2$ , and $P(\pm 1) = -1$ , $P(\pm 2) = 2$ , so all four fail. The correct conclusion is that $x^2 - 2$ has no rational roots. It certainly has roots: $x^2 = 2$ holds for $x = \pm\sqrt{2}$ . Those roots are irrational, so no list of fractions could ever contain them. The theorem is structurally blind to those irrational roots, just as it was blind to the nonreal pair in Worked Example 2. An empty verdict from the candidate list never means “no roots”; it means “no rational ones”.

Read positively, that blindness is a weapon. If every candidate fails, you have proved the polynomial has no rational roots, and that is exactly how mathematicians prove numbers irrational.

Worked example 3 Prove that $\sqrt{5}$ is irrational

The number $\sqrt{5}$ is, by definition, a root of the integer-coefficient polynomial

P(x) = x^2 - 5.

Suppose $P$ had a rational root. The polynomial is monic, so any rational root would be an integer dividing $5$ , leaving only four candidates: $\pm 1$ and $\pm 5$ . Test all four; squaring kills the sign, so two computations cover them:

P(\pm 1) = 1 - 5 = -4, \qquad P(\pm 5) = 25 - 5 = 20.

No candidate is a root, so $x^2 - 5$ has no rational roots at all. But $\sqrt{5}$ is a root. A number that is a root of a polynomial with no rational roots cannot be rational, so $\sqrt{5}$ is irrational.

The same four lines generalize. For any positive integer $n$ , a rational root of $x^2 - n$ must be an integer $d$ with $d^2 = n$ . So when that positive integer $n$ is not a perfect square, $x^2 - n$ has no rational roots and $\sqrt{n}$ is irrational. One theorem disposes of $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \ldots$ in a single stroke.

Check your understanding

You build the full candidate list for an integer-coefficient polynomial and test every entry. None is a root. What can you conclude?

A The polynomial has no roots at all B The polynomial has no rational roots C The polynomial has no real roots D The polynomial cannot be factored

Common mistakes

Flipping the fraction. Numerators come from the constant term and denominators from the leading coefficient. For $2x^3 + 3x^2 - 8x + 3$ the candidates are $\pm\tfrac{\text{divisor of } 3}{\text{divisor of } 2}$ , never the other way around. And every candidate carries both signs, even when all coefficients are positive.
Promoting candidates to roots. Passing the divisibility test proves nothing: for $x^2 - 2$ the candidate $2$ passes and still is not a root. Every candidate must be tested by substitution or synthetic division.
Overreading an empty result. When the whole list fails, the polynomial has no rational roots. It may still have irrational or nonreal roots, and usually does.
Applying the theorem to non-integer coefficients. For $x^2 - \tfrac{1}{2}x - \tfrac{1}{2}$ , first multiply by $2$ to get $2x^2 - x - 1$ , which has the same roots and integer coefficients the theorem can read.
Ignoring a zero constant term. If $a_0 = 0$ , the condition $p \mid 0$ holds for every integer, so the theorem filters nothing. Factor out the largest power of $x$ first, then apply the theorem to the cofactor.
Forgetting repeated roots. After dividing out $x - c$ , test $c$ again on the quotient. In $4x^3 - 8x^2 + 5x - 1 = (x - 1)(2x - 1)^2$ the root $\tfrac{1}{2}$ occurs twice.

Key takeaways

The Rational Root Theorem. If a polynomial with integer coefficients has a rational root $\tfrac{p}{q}$ in lowest terms, then $p \mid a_0$ (constant term) and $q \mid a_n$ (leading coefficient).
The candidate list is every $\pm\tfrac{p}{q}$ with $p$ a positive divisor of $a_0$ and $q$ a positive divisor of $a_n$ , reduced and deduplicated. It is finite whenever $a_0 \ne 0$ .
Workflow. List the candidates, test them by synthetic division (the remainder is $P(c)$ ), peel off each discovered factor with the Factor Theorem, and finish the final quadratic with the quadratic formula.
Monic case. If $a_n = 1$ , every rational root is an integer dividing the constant term.
One-way street. Candidates are not roots; most fail. The converse of the theorem is false.
Rational eyes only. Irrational and nonreal roots never appear on the list. A fully failed list proves the polynomial has no rational roots, which is the standard modern proof that numbers like $\sqrt{5}$ are irrational.

The Rational Root Theorem

Learning goals

A pattern hiding in the factored form

The theorem

Building the candidate list

When the leading coefficient is 1

From candidates to roots

What the theorem does not say

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)