2026-09-29 9 min

Rational Root Theorem Calculator

List all p/q rational root candidates from constant and leading factors, test each in P(x), and collect the true roots. Free online calculator.

Calculator

Theorem Inputs

Integer coefficients (highest degree first)

Integers only, nonzero constant. Results update automatically.

Results update live as you type

Status:P(x) = x^3 - 6x^2 + 11x - 6roots: 1, 2, 3

Rational Root Summary

Rational roots1, 2, 3

Candidates8

p / q sets±{1, 2, 3, 6} / {1}

Per-Candidate Tests

P(-6) = -504

P(-3) = -120

P(-2) = -60

P(-1) = -24

P(1) = 0 ✓ ROOT

P(2) = 0 ✓ ROOT

P(3) = 0 ✓ ROOT

P(6) = 60

Results

Optimal

Rational Roots

1, 2, 3

Candidates with P(x) = 0

Candidates Tested

Unique reduced p/q values × signs

p Factors (const)

±1, ±2, ±3, ±6

Divisors of -6

q Factors (leading)

Divisors of 1

Verdict

3 rational root(s) found

What the candidate sweep proved

Candidates are deduplicated in lowest terms and each is evaluated by exact-feeling Horner arithmetic with a 1e-9 zero tolerance.

Visualizations

|P(candidate)| (log-ish, capped)

100-6

100-3

60-2

24-1

606

Roots vs Rejects

Roots38%

Rejected63%

Sweep Completeness

Done100%

Roots vs Candidates

roots3

Formula

Divisibility forces every reduced rational root into the p/q grid; testing the grid finds them all.

Possible roots: ±(factors of a₀) / (factors of aₙ), reduced

Alternative Form

Root test: P(p/q) = 0 ⟺ (qx − p) is a factor

Variables

Symbol	Name	Unit	Description
a₀	Constant term	integer	Supplies numerators p
aₙ	Leading coefficient	integer	Supplies denominators q
p/q	Candidate	fraction	Reduced, both signs
P(p/q)	Test value	number	Zero ⟺ root

Explanation

Clearing denominators in P(p/q) = 0 forces q|aₙ and p|a₀ — the only place rational roots can hide.

Derivation

aₙ(p/q)ⁿ + … + a₀ = 0, × qⁿ: p divides a₀qⁿ so p|a₀ (coprime to q); symmetric argument gives q|aₙ.

Step-by-Step Calculation

Follow each step to understand how the result is calculated from your inputs.

Step 1

Confirm integer coefficients and read a₀, aₙ.

p | a₀ and q | aₙ

P:x^3 - 6x^2 + 11x - 6

a0:-6

an:1

Result: a₀ = -6, aₙ = 1

Step 2

List factor sets.

all positive divisors

p:1, 2, 3, 6

q:1

Result: p ∈ ±{1, 2, 3, 6}, q ∈ {1}

Step 3

Form reduced p/q candidates with both signs.

reduce by GCD(p, q), dedupe

count:8

Intermediate: -6, -3, -2, -1, 1, 2, 3, 6

Result: 8 candidates

Step 4

Test the first candidates in P(x).

Horner evaluation each

-6:-504

-3:-120

-2:-60

-1:-24

Intermediate: P(-6)=-504; P(-3)=-120; P(-2)=-60; P(-1)=-24; P(1)=0; P(2)=0

Result: See per-candidate table below

Step 5

Collect the survivors.

keep P(candidate) = 0

roots:1, 2, 3

Result: Rational roots: 1, 2, 3

Final Answer

1, 2, 3

Worked Examples

See how the calculator works with different scenarios.

Scenario

Three integer roots.

Inputs

coeffs:1, -6, 11, -6

Calculation

p ∈ ±{1,2,3,6}, q ∈ {1} → test 8 → 1, 2, 3 pass

Result

Roots: 1, 2, 3

Interpretation

Factors as (x−1)(x−2)(x−3).

Understanding Your Result

Your Result: Verified rational roots with full candidate ledger

Typical range: Zero up to n rational roots for degree n (counting multiplicity).

What This Means

The complete set of rational zeros — every fraction that exactly kills the polynomial.

Why It Matters

Each rational root yields a linear factor, stepping cubics and quartics down to solvable quadratics.

Professional Interpretation

Sweep first, factor second: divide out each found root and re-run on the quotient.

Recommendations

•Test ±1 and small fractions first
•Divide out found roots
•Re-run on the quotient
•Clear fractions before starting
•Factor out x when a₀ = 0

Educational Content

Introduction

This calculator turns the rational root theorem into an executed search: factors, candidates, tests, verdict.

Concept Explanation

Divisibility fences rational roots into a finite paddock; evaluation hunts them down one by one.

Applications

▸Factoring cubics/quartics

▸Contest algebra

▸Break-even analysis

▸CAS exact-root stage

▸Homework checking

▸Denominator hunting

Benefits

•Factor lists shown
•Deduped reduced candidates
•Every test evaluated
•Roots collected clearly
•Empty-sweep verdict honest

Limitations

•Integer coefficients only
•Rational roots only
•Long lists for composite constants

Best Practices

•Clear fractions first
•Factor out x when a₀ = 0
•Reduce p/q before testing
•Divide out each hit
•Accept empty sweeps

Common Misconceptions

Myth: Every candidate is a root.

Truth: Candidates are suspects; only P = 0 convictions count.

Myth: Empty sweep means no roots.

Truth: It means no rational roots — irrationals/complex may exist.

Myth: Unreduced fractions add roots.

Truth: Unreduced forms duplicate reduced ones.

Myth: Non-integers work directly.

Truth: Scale to integers first.

Practical Tips

•List p and q factors
•Reduce every p/q
•Test ±1 first
•Synthetic-test each
•Divide out hits
•Re-run on quotient
•Accept empty sweeps

Common Mistakes

Testing unreduced duplicates

Divide p, q by their GCD first.

Forgetting negative candidates

Include both signs of every p/q.

Applying to non-integers

Multiply through to clear fractions.

Ignoring a₀ = 0

Factor out xᵏ before listing.

Stopping after one root

Keep testing — degree n allows n roots.

Assumptions

This calculator makes the following assumptions to provide estimates:

Integer coefficients: All entries are integers.

Nonzero constant: a₀ ≠ 0 (else factor x first).

Univariate: Polynomial in x.

Reduced candidates: p/q tested in lowest terms.

Zero tolerance: |P| below 1e-9 counts as zero.

Limitations

Be aware of these limitations when using this calculator:

Rational only: Irrational/complex roots not listed.

Integer gate: Non-integer input is rejected.

Combinatorial lists: Composite constants give many candidates.

Display rounding: Test values show 6 digits.

Frequently Asked Questions

Answers to common questions about this calculator.

What is the rational root theorem?

For an integer-coefficient polynomial, any rational root p/q (in lowest terms) must have p dividing the constant term and q dividing the leading coefficient. That gives a finite candidate list.

How do I list possible rational roots?

List all factors of the constant (±p) and of the leading coefficient (±q), form every p/q in lowest terms, then test each in P(x). Survivors with P = 0 are the rational roots.

Do I need lowest terms?

Yes for a clean list — 2/4 duplicates 1/2. Reducing each p/q by GCD(p, q) removes duplicates before testing.

What if no candidate works?

Then the polynomial has no rational roots. Its real roots (if any) are irrational and need numeric methods; complex pairs need the quadratic formula or solvers.

Why must coefficients be integers?

The divisibility argument needs integer factors. Clear fractions first by multiplying through, e.g. scale (x/2 terms) before applying the theorem.

What if the constant term is 0?

Then x = 0 is a root — factor out the highest power of x first (P = xᵏ·Q(x)) and run the theorem on Q.

How do I test candidates fast?

Use synthetic division or Horner evaluation per candidate; a zero remainder certifies a root and hands you the reduced quotient for further factoring.

How many candidates can there be?

At most 2·d(a₀)·d(aₙ) before dedup (d = divisor count). Highly composite constants give long lists — test ±1 and small fractions first.

Where is this used?

First step of factoring cubics/quartics by hand, contest problems, and computer algebra systems that try exact roots before going numeric.