The Rational Root Theorem: Finding Zeros of Polynomials

Quick answer

The rational root theorem narrows the search for zeros. If p/q, in lowest terms, is a rational zero of a polynomial with integer coefficients, then p divides the constant term and q divides the leading coefficient. List those candidates, test them with synthetic division, and each zero r gives a factor x − r. Dividing it out leaves a smaller polynomial; repeat, and finish with the quadratic formula once the quotient is quadratic. That finds every zero, rational, irrational or complex.

What you'll learn

List the possible rational zeros of a polynomial
Test candidates with synthetic division
Factor a polynomial completely once one zero is found
Explain why a rational zero must have that form

Where to look for zeros

The factor theorem says that $x - r$ is a factor exactly when $r$ is a zero. But a cubic like $2x^3 - 3x^2 - 11x + 6$ does not announce its zeros, and guessing at random is hopeless. The rational root theorem turns the guessing into a short list.

The rational root theorem

Suppose $a_nx^n + \cdots + a_1x + a_0$ has integer coefficients. If $\tfrac{p}{q}$ , in lowest terms, is a zero, then $p$ divides the constant term $a_0$ and $q$ divides the leading coefficient $a_n$ .

For $2x^3 - 3x^2 - 11x + 6$ : $p$ divides $6$ , so $p$ is $1, 2, 3$ or $6$ , and $q$ divides $2$ , so $q$ is $1$ or $2$ . The candidates are

\pm 1,\ \pm 2,\ \pm 3,\ \pm 6,\ \pm\tfrac{1}{2},\ \pm\tfrac{3}{2}

Twelve candidates instead of infinitely many.

Why the zeros leave fingerprints

Put $x = \tfrac{p}{q}$ into the polynomial, set it equal to $0$ , and multiply through by $q^n$ to clear the fractions:

a_np^n + a_{n-1}p^{n-1}q + \cdots + a_1pq^{n-1} + a_0q^n = 0

Every term except the last contains the factor $p$ , so $p$ divides $a_0q^n$ too. Since $p$ and $q$ share no factors, $p$ must divide $a_0$ . The same argument with every term except the first shows that $q$ divides $a_n$ . A rational zero leaves its fingerprints on the first and last coefficients.

The strategy

List the candidates $\pm\tfrac{p}{q}$ .
Test them with synthetic division. A remainder of $0$ means a zero.
The bottom row of the division is the remaining factor, one degree lower.
Repeat on that factor. Once it is quadratic, factor it or use the quadratic formula.

A graph can save time: a candidate far from any $x$ -intercept need not be tested.

Example 1 — Three rational zeros

Find all zeros of $2x^3 - 3x^2 - 11x + 6$ .

y = 2x³ − 3x² − 11x + 6

y = 2x³ − 3x² − 11x + 6

The graph suggests $3$ . Test it with synthetic division:

	$2$	$-3$	$-11$	$6$
add		$6$	$9$	$-6$
result	$2$	$3$	$-2$	$0$

The remainder is $0$ , so $x - 3$ is a factor and the rest is $2x^2 + 3x - 2 = (2x - 1)(x + 2)$ . The zeros are $3$ , $\tfrac{1}{2}$ and $-2$ , all on the list.

Worked examples

Common mistakes

Practice problems

List the possible rational zeros of $3x^3 + 2x - 8$ .

Answer

$\pm 1, \pm 2, \pm 4, \pm 8, \pm\tfrac{1}{3}, \pm\tfrac{2}{3}, \pm\tfrac{4}{3}, \pm\tfrac{8}{3}$

Full solution

$p$ divides $8$ : $1, 2, 4, 8$ . $q$ divides $3$ : $1, 3$ . Form every $\pm\tfrac{p}{q}$ .
Find all zeros of $x^3 - 7x + 6$ .

Answer

$1$ , $2$ and $-3$

Full solution

$1 - 7 + 6 = 0$ , so $1$ is a zero. Dividing by $x - 1$ leaves $x^2 + x - 6 = (x + 3)(x - 2)$ .
Find all zeros of $x^3 + 2x^2 - 5x - 6$ .

Answer

$-1$ , $2$ and $-3$

Full solution

$-1 + 2 + 5 - 6 = 0$ , so $-1$ is a zero. Dividing by $x + 1$ leaves $x^2 + x - 6 = (x + 3)(x - 2)$ .
Find all zeros of $2x^3 + x^2 - 13x + 6$ .

Answer

$2$ , $\tfrac{1}{2}$ and $-3$

Full solution

$16 + 4 - 26 + 6 = 0$ , so $2$ is a zero. Dividing by $x - 2$ leaves $2x^2 + 5x - 3 = (2x - 1)(x + 3)$ .
Find all zeros of $x^3 - 3x^2 + x + 1$ .

Answer

$1$ and $1 \pm \sqrt{2}$

Full solution

$1 - 3 + 1 + 1 = 0$ . Dividing by $x - 1$ leaves $x^2 - 2x - 1$ , and the quadratic formula gives $1 \pm \sqrt{2}$ .
Find all zeros of $x^4 - x^3 - 7x^2 + x + 6$ .

Answer

$1$ , $-1$ , $3$ and $-2$

Full solution

$1$ and $-1$ both give $0$ . Dividing by $(x - 1)(x + 1) = x^2 - 1$ leaves $x^2 - x - 6 = (x - 3)(x + 2)$ .
Show that $x^3 + x + 1$ has no rational zeros.

Answer

The only candidates, $\pm 1$ , give $3$ and $-1$ .

Full solution

The constant and leading coefficients are both $1$ , so the only candidates are $\pm 1$ . Neither gives $0$ .
Solve $3x^3 - 4x^2 - 5x + 2 = 0$ .

Answer

$x = 2$ , $\tfrac{1}{3}$ or $-1$

Full solution

$24 - 16 - 10 + 2 = 0$ , so $2$ is a zero. Dividing by $x - 2$ leaves $3x^2 + 2x - 1 = (3x - 1)(x + 1)$ .
A polynomial with integer coefficients has leading coefficient $1$ and constant term $5$ . Can $\tfrac{1}{5}$ be a zero?

Answer

No

Full solution

The denominator of a rational zero must divide the leading coefficient, $1$ . So every rational zero is an integer: $\pm 1$ or $\pm 5$ .
A student lists the possible rational zeros of $2x^3 + 5x^2 + 3x - 3$ as $\pm 1, \pm 2, \pm\tfrac{1}{3}, \pm\tfrac{2}{3}$ . What went wrong?

Hint

Which coefficient gives the numerators?

Answer

The student swapped the roles: the list is $\pm 1, \pm 3, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}$ .

Full solution

Numerators divide the constant term, $-3$ ; denominators divide the leading coefficient, $2$ . Indeed $\tfrac{1}{2}$ is a zero: $2 \cdot \tfrac{1}{8} + 5 \cdot \tfrac{1}{4} + 3 \cdot \tfrac{1}{2} - 3 = 0$ , and it is not on the student’s list.

Frequently asked questions

What does the rational root theorem say?

If a polynomial has integer coefficients and p/q in lowest terms is a zero, then p divides the constant term and q divides the leading coefficient.

How do I list the possible rational zeros?

Write every factor p of the constant term and every factor q of the leading coefficient, and form all fractions ±p/q.

How do I test a possible zero?

Substitute it, or use synthetic division. A remainder of 0 means it is a zero, and the bottom row gives the remaining factor.

What if none of the candidates works?

Then the polynomial has no rational zeros. Its real zeros, if any, are irrational, and a graph or numerical method can estimate them.

Does the theorem find irrational zeros?

Not directly. But once the rational zeros are divided out, a remaining quadratic can be solved with the quadratic formula, which gives the irrational and complex zeros.

What to learn next

The Fundamental Theorem of AlgebraA polynomial of degree n has exactly n complex zeros, counted with multiplicity. Use that count and conjugate pairs to find every zero and build polynomials.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSA.APR.B.3Arithmetic with Polynomials and Rational ExpressionsIdentify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
CCSS.MATH.CONTENT.HSA.APR.B.2Arithmetic with Polynomials and Rational ExpressionsKnow and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x - a is p(a), so p(a) = 0 if and only if (x - a) is a factor of p(x).

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Where to look for zeros

Why the zeros leave fingerprints

The strategy

Worked examples

Common mistakes

Practice problems

Frequently asked questions

What does the rational root theorem say?

How do I list the possible rational zeros?

How do I test a possible zero?

What if none of the candidates works?

Does the theorem find irrational zeros?

What to learn next

Standards alignment

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