Rational Zeros Calculator
Apply the rational root theorem to any polynomial — see the candidate list, which ones are real zeros, and what is left over.
Enter the coefficients, highest power first
Choose the degree, then fill in every coefficient from the leading term down to the constant. A missing power is a coefficient of 0, and fractions such as 1/2 are cleared automatically before the theorem is applied.
How the rational zeros calculator works
A rational zeros calculator finds the possible rational zeros of a polynomial using the Rational Root Theorem. Possible rational zeros have the form ±p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. Testing these values identifies which candidates are actual zeros.
Each genuine zero is reported with its multiplicity. The theorem does not promise that a rational zero exists — what it guarantees is that none can hide outside the list, so if every candidate fails, that is a proof rather than a dead end, and the remaining zeros must be irrational or complex.
Take p from the constant term and q from the leading coefficient.
Form every ±p/q, reduce, and drop the duplicates.
Each hit gives a factor, and dividing it out shrinks the problem.
The theorem in symbols
The statement
p ∣ a₀ and q ∣ aₙ ⟹ x = p/q
For aₙxⁿ + … + a₀ with integer coefficients, any rational zero in lowest terms has this shape.
The factor it gives
x = p/q ⟺ (qx − p) is a factor
Writing the factor this way keeps every coefficient a whole number, which Gauss's lemma guarantees.
Descartes' rule of signs
positives ≤ sign changes in p(x)
Counting sign changes in p(x) and p(−x) bounds how many positive and negative real zeros can exist, each dropping by 2.
Clearing fractions
p(x) = 0 ⟺ k·p(x) = 0
Multiplying by the lowest common denominator makes the coefficients integers without moving any zero.
Why it works: substitute x = p/q into the polynomial and multiply through by qⁿ. Every term except a₀qⁿ carries a factor of p, and every term except aₙpⁿ carries a factor of q. Since p and q share no factor, p must divide a₀ and q must divide aₙ.
Method reference: OpenStax states the rational zero theorem, the factor theorem and Descartes' rule together, with worked examples, in Zeros of Polynomial Functions.
Reading the outcome
What the candidate list tells you
A short candidate list is good news and a long one is only tedious, not harder. What matters is how many of them turn out to be zeros, because that decides how much of the polynomial you can factor by hand.
| Outcome | What it means | What is left | Next step |
|---|---|---|---|
| Zeros account for the whole degree | The polynomial factors completely over the rationals | A constant | Nothing — write the factored form and stop. |
| Some candidates work | Partial factorisation, degree reduced by each hit | A smaller polynomial | Finish a leftover quadratic with the quadratic formula. |
| No candidate works | Proof that no rational zero exists | The original polynomial | Use the cubic formula, or a numerical method for higher degrees. |
| A candidate works more than once | A repeated zero, with multiplicity above 1 | Degree drops by the multiplicity | Keep testing the same value until the division stops working. |
Test the small ones first: ±1 costs nothing to check — just add the coefficients, and add them with alternating signs. Textbook polynomials are built around small zeros far more often than chance would suggest.
Worked examples
Rational zero examples
Each row is one of the presets above. Load it to see the full candidate list and which entries survive the test.
| Polynomial | Candidates | Rational zeros | Factored form |
|---|---|---|---|
| 2x³ − 3x² − 8x + 12 | 16 | −2, 3/2, 2 | (x + 2)(2x − 3)(x − 2) |
| x⁴ − 2x³ − 3x² + 8x − 4 | 6 | −2, 1 (twice), 2 | (x + 2)(x − 1)²(x − 2) |
| x³ + 3x − 5 | 4 | None | Irreducible over the rationals |
| 2x⁴ − 2x³ − 4x² | 4 | −1, 0 (twice), 2 | 2x²(x + 1)(x − 2) |
Faster by hand
How to shrink the candidate list
A list of sixteen candidates is rarely sixteen tests. Three cheap observations usually cut it to two or three worth trying.
1. Use Descartes' rule first
If there can be no negative real zeros, half the candidate list disappears before you test anything.
2. Check ±1 in your head
p(1) is the sum of the coefficients, and p(−1) is that sum with alternating signs. Both take seconds.
3. Divide as soon as you hit
Every hit lowers the degree, and the shorter quotient makes every remaining test cheaper.
Common factors matter: in 2x⁴ − 2x³ − 4x², pull out 2x² first. The theorem then applies to the much smaller x² − x − 2, and the zero at x = 0 was already found by the factoring rather than by any test.
Practice reference: OpenStax works through candidate lists, synthetic division and repeated zeros in Zeros of Polynomial Functions.
Reading the result
What a failed search actually proves
An empty result is one of the strongest outcomes here. The candidate list is exhaustive, so testing all of it settles the question completely rather than leaving it open.
Every candidate fails
No rational zero exists. The polynomial cannot be factored into linear pieces over the rationals.
Some succeed
Each one peels off a factor. What survives has no rational zeros left, by the same exhaustive argument.
A cubic still has a real zero
Odd degree guarantees at least one real zero. If none is rational, it is irrational — not missing.
Interesting fact
The theorem that proves √2 is irrational
Apply the rational root theorem to x² − 2. The constant term is 2 and the leading coefficient is 1, so the only candidates are ±1 and ±2, and none of them is a zero. But √2 certainly is a zero of x² − 2 — therefore √2 cannot be rational. The same one-line argument disposes of √3, ∛5, and the whole family: a modest bookkeeping rule about divisors turns out to settle a question that reportedly unsettled the Pythagoreans, who had built a worldview on the assumption that every magnitude was a ratio of whole numbers.
Source: MacTutor History of Mathematics, Real numbers: the discovery of irrationals.
Showing your work
What to record alongside the zeros
The candidate list is part of the answer, not scratch work. It is what turns "I found these zeros" into "these are all of them", and it is usually where the marks are.
Minimum working to show
- The divisors of the constant term and of the leading coefficient
- The full ±p/q list, reduced and deduplicated
- The division that follows each successful test
- The quotient at the end, and how you finished it
When zeros need interpreting
Polynomials from real models — break-even quantities, box volumes, equilibrium concentrations — often admit only one meaningful zero. A negative count or a length beyond the material available is algebraically valid and physically impossible, so name the zero you keep and say why the others are discarded.
Example line: 2x³ − 3x² − 8x + 12 · p ∈ {1, 2, 3, 4, 6, 12}, q ∈ {1, 2} · 16 candidates · x = 2 works → divide by (x − 2) → 2x² + x − 6 · factor → (2x − 3)(x + 2) · zeros 2, 3/2, −2.
Common mistakes to avoid
Putting p and q the wrong way round. The numerator divides the constant term; the denominator divides the leading coefficient. Swapping them produces a list that misses the real zeros.
Forgetting the negatives. Every candidate comes in a ± pair. Testing only the positives is the fastest way to miss half the answer.
Stopping after one hit. A zero can repeat, and there may be others further down the list. Keep dividing and keep testing.
Treating an empty list as failure. No rational zeros is a complete, provable answer — the remaining zeros are simply irrational or complex.
Frequently Asked Questions
What is the rational zeros theorem?
The rational zeros theorem, also called the rational root theorem, says that if a polynomial with integer coefficients has a rational zero p/q in lowest terms, then the numerator p must divide the constant term and the denominator q must divide the leading coefficient. It does not promise that any rational zero exists — it only narrows an infinite search down to a finite list of possible zeros worth testing. Every rational zero of the polynomial is somewhere on that list, so a candidate list that all fail proves no rational zero exists: one of the few results in algebra that rules an answer out as firmly as it rules one in.
How do you list the possible rational zeros of a polynomial?
Write down every divisor of the constant term, call them p, and every divisor of the leading coefficient, call them q. Form all the fractions ±p/q, reduce them to lowest terms, and remove duplicates. For 2x³ − 3x² − 8x + 12 the divisors of 12 are 1, 2, 3, 4, 6 and 12, and the divisors of 2 are 1 and 2, giving sixteen distinct candidate zeros once repeats are dropped. Each candidate that survives testing is also an x-intercept of the matching polynomial function, so a glance at the graph often shows which ones are worth trying first.
Does every polynomial have a rational zero?
No. A polynomial such as x³ − 2 has candidates ±1 and ±2, and none of them work, so the only real root of the equation x³ − 2 = 0 is the irrational cube root of 2. The theorem is a filter rather than a guarantee: when all the candidates fail, you have proved that no rational zero exists and that the remaining solutions are irrational or complex. Numerical methods or the cubic and quadratic formulas take over at that point.
What do you do after finding one rational zero?
Divide the polynomial by the corresponding factor, which lowers the degree by one and leaves a smaller polynomial with the same remaining zeros. The factor theorem is what licenses this: a genuine zero leaves a remainder of 0, so the division comes out exact. For a zero p/q in lowest terms the factor is (qx − p), so a zero of 3/2 gives the factor (2x − 3), and synthetic division handles it far faster than long division because the divisor is linear. Repeat on the quotient, testing the same candidate again in case the zero is repeated, and stop when the quotient is a quadratic you can finish with the quadratic formula.
Do the coefficients have to be integers?
The theorem is stated for integer coefficients, but fractional ones are no obstacle. Multiplying the whole polynomial equation by the lowest common denominator clears the fractions without moving a single zero, because scaling something that equals zero changes nothing. This rational zeros calculator does that automatically, so you can type 1/2 or 0.25 and still get a valid candidate list. What the theorem genuinely requires is a polynomial in the first place — a whole-number exponent on every power of the variable, with no roots and no negative powers.
How does the rational root theorem relate to factoring?
They are the same task seen from two sides. Each rational zero p/q corresponds to a linear factor (qx − p), so finding the roots is factoring, and factoring reveals the roots. A polynomial whose candidates all fail cannot be factored into linear pieces with rational coefficients, which is exactly why a quadratic like x² + 1 resists factoring over the rationals while x² − 1 does not. Read the other way round, a complete factorisation hands you every solution at once, with no further testing needed.
Before you submit the answer
- Coefficients are integers, or fractions were cleared first
- p came from the constant, q from the leading coefficient
- Both signs of every candidate were tested
- Each hit was retested for a repeated zero
- Every zero was substituted back into the original
- The leftover factor was finished or named as irreducible
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Disclaimer
This calculator handles single-variable polynomials of degree 2 to 6 with rational coefficients. Radicals, variables in an exponent, several variables at once, and non-polynomial terms are outside its scope.
Candidates are generated and tested in exact integer arithmetic, so a zero reported here is exact rather than approximate, and an empty result is a proof rather than a failed search. Very large coefficients produce very long candidate lists and are capped for practicality.
The theorem finds rational zeros only. Irrational and complex zeros are real answers that this method cannot reach — the leftover factor is reported so they can be pursued with a quadratic, cubic, or numerical method.
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