The rational root calculator above applies the rational root theorem properly rather than half of it. It builds the complete candidate list from the factors of the constant term and the leading coefficient, evaluates the polynomial at every one of them using exact integer arithmetic so no candidate is missed to rounding, reports which are genuine roots and with what multiplicity, divides them out, and shows what is left. It also runs Descartes' rule of signs, which tells you in advance how many positive and negative real roots you can possibly have.
Arb Digital publishes this page because the theorem is routinely taught as "list the candidates" and then abandoned. Listing them is the easy half. The useful half is systematically testing them, recognising a repeated root when one appears, and knowing what the leftover factor means when the list runs out. This page does all of that and shows the working, so it can be used to check a hand solution rather than to replace it.
What This Rational Root Calculator Does
It accepts a polynomial of degree up to eight with integer coefficients and returns every rational root it has, expressed as an exact fraction rather than a decimal. It reports how many candidates the theorem produced, how many turned out to be roots, and the factored form of the polynomial with those roots pulled out. The evaluation is done with exact integer arithmetic, which matters: testing a candidate like 1/3 in floating point can produce a result of 1e−16 rather than 0, and a naive tool will report that as "not a root".
This is a broader job than our quadratic equation solver or the cubic equation calculator, both of which are locked to a fixed degree and return all roots including irrational and complex ones. This page works at any degree but only finds rational roots — that is the whole point of the theorem, and the limitation is a feature, because a rational root is the one kind you can factor out cleanly and by hand.
How to Use It
- Write the polynomial in descending order. Highest power first, down to the constant, with nothing skipped.
- Enter a 0 for every missing term. x⁴ − 1 is 1, 0, 0, 0, -1. Leaving the zeros out shifts every coefficient into the wrong power.
- Clear any fractions first. Multiply the whole polynomial by the least common denominator. The roots are unchanged and the theorem then applies.
- Read the candidate list. Every candidate is shown with a tick or a cross, so you can check your own list against it.
- Look at what is left over. The remaining factor holds all the irrational and complex roots, and if it is a quadratic you can finish it with the quadratic formula.
The Theorem and How It's Calculated
The rational root theorem says that if a polynomial with integer coefficients has a rational root p/q written in lowest terms, then p must divide the constant term and q must divide the leading coefficient. That is a strong restriction. It does not say such a root exists; it says that if one does, it is on a finite list you can write down in a minute.
Take the default, 2x³ − 3x² − 23x + 12. The constant is 12, whose positive factors are 1, 2, 3, 4, 6 and 12. The leading coefficient is 2, whose factors are 1 and 2. So p/q ranges over ±1, ±2, ±3, ±4, ±6, ±12, ±1/2 and ±3/2 — sixteen candidates once duplicates like 2/2 are removed. Testing them finds three roots: x = 4, because 128 − 48 − 92 + 12 = 0; x = −3, because −54 − 27 + 69 + 12 = 0; and x = 1/2, because 0.25 − 0.75 − 11.5 + 12 = 0. Dividing each out in turn leaves the constant 2, so the polynomial factors completely as 2(x − 4)(x + 3)(x − 1/2), which is the same thing as (x − 4)(x + 3)(2x − 1).
The calculator does the testing with synthetic division on exact fractions. When a candidate is a root, the remainder is exactly zero and the quotient is one degree lower, so the next candidate is tested against a smaller polynomial. That is also why a repeated root is caught: the calculator retries the same value against the reduced polynomial and counts how many times it divides. Paul's Online Notes covers the same process in Finding Zeroes of Polynomials.
Why the Candidate List Gets Long, and How to Shorten It
The number of candidates is roughly twice the number of divisors of the constant times the number of divisors of the leading coefficient. That grows quickly. A polynomial ending in 60 with a leading coefficient of 12 produces 12 divisors times 6 divisors times two signs — around a hundred candidates before duplicates are removed. Testing all of them by hand is not realistic, so it is worth knowing the standard ways to cut the list down.
Descartes' rule of signs is the first. Count the sign changes in the coefficients as written: that is the maximum number of positive real roots, and the actual number is that figure or less by an even amount. Substitute −x and count again for the negatives. If the rule says there are no negative roots, half the candidate list can be discarded without testing anything. For the default polynomial the signs run plus, minus, minus, plus, giving two changes, so there are either two positive roots or none; and P(−x) gives one change, so there is exactly one negative root. Both predictions match what the calculator finds.
The second trick is bounds. If synthetic division by a positive candidate produces a bottom row with no negative entries, no larger value can be a root either, so every bigger candidate can be dropped. There is a matching rule for negative candidates, where the bottom row alternates in sign. The third is simply to test small integers first — most textbook problems are built around roots like 1, −1, 2 and 3, and finding one immediately reduces the degree and shrinks every subsequent step. Our divisors calculator will generate the factor lists, and the prime factorization calculator is quicker still for a large constant term.
When There Are No Rational Roots
This is the case students find hardest to accept, and it is extremely common. Consider x² − 2. The constant is 2 and the leading coefficient is 1, so the candidates are ±1 and ±2. None of them work, and the theorem has therefore proved something genuinely useful: x² − 2 has no rational roots at all. Its actual roots are ±√2, which the theorem could never have found because it only ever looks at fractions.
The same reasoning proves x² + 1 has no rational roots, and in that case no real roots either. So an empty result is not a failure of the method. It is a proof of absence, and it tells you that whatever comes next — the quadratic formula, a numerical method, or factoring over a larger number system — will not be integer factoring. Any polynomial of degree three or higher with no rational roots cannot be factored into linear pieces with integer coefficients, which is exactly the information you need before you waste an hour trying.
Multiplicity, and Reading the Leftover Factor
A root can divide a polynomial more than once. In x³ − 3x + 2 the value 1 is a root, and after dividing out (x − 1) the quotient x² + x − 2 still has 1 as a root, so the factor is (x − 1)² and the full factorisation is (x − 1)²(x + 2). The calculator reports that multiplicity explicitly, because it changes the graph: the curve touches the axis and turns back at a double root rather than crossing it, and a triple root crosses with a flattening. The polynomial graph plotter shows that difference clearly.
Whatever remains after every rational root is removed is where the rest of the roots live. If it is degree two, finish it with the quadratic formula and you have the complete solution — that is the standard route through a cubic or quartic in a first algebra course. If it is degree three or more with no rational roots, there is no elementary way to finish, and numerical methods take over. A useful check throughout is that the degrees add up: the number of rational roots counted with multiplicity, plus the degree of the leftover factor, always equals the degree you started with. Our polynomial division calculator will perform any of those divisions step by step.
Edge Cases Worth Knowing
If the constant term is zero, x itself is a factor and zero is a root. The theorem does not apply directly in that form, because p would have to divide zero, which every integer does. The fix is to factor out the highest power of x first and apply the theorem to what remains, which is what this calculator does automatically before it builds any candidate list.
If the coefficients share a common factor, dividing it out first shortens the candidate list without changing a single root — 4x² − 12x + 8 has the same roots as x² − 3x + 2 but a longer candidate list. And if the leading coefficient is 1, every denominator q must be 1, so the candidates are simply the plus-and-minus factors of the constant term. That special case is the one worth memorising, because monic polynomials are common and the list is short. MathWorld's entry on the rational zero theorem states the general result compactly.
Arb Digital designs and writes pages that show their working — the kind that hold attention and earn links instead of a quick bounce.
Browse All Free Tools Talk To Our TeamCommon Mistakes to Avoid
- Skipping missing powers — entering x⁴ − 1 as 1, -1 describes x − 1 instead. Every absent term needs an explicit zero.
- Putting the leading coefficient's factors on top — p, the numerator, comes from the constant term; q, the denominator, comes from the leading coefficient. Reversing them produces the wrong list.
- Assuming a candidate is a root because it is close — a floating-point evaluation of 1e−15 is not zero. Exact fraction arithmetic is the only reliable test, which is what this page uses.
- Stopping after the first root — always divide it out and keep testing the reduced polynomial, since the same value may be a repeated root and other candidates may still work.
- Applying the theorem with fractional coefficients — it only holds for integer coefficients. Multiply through by the common denominator before you begin.
Related Free Tools From Arb Digital
Factor a quadratic directly with the factoring trinomials calculator, add and multiply expressions using the polynomial arithmetic calculator, solve the leftover quadratic with the quadratic equation solver, or find where the polynomial is positive and negative with the quadratic inequality calculator. The full free online tools hub lists every algebra calculator we publish.
Frequently Asked Questions
It states that any rational root of a polynomial with integer coefficients, written in lowest terms as p over q, must have p dividing the constant term and q dividing the leading coefficient.
No. It only limits where a rational root could be. Many polynomials, such as x squared minus 2, have candidates but no rational roots at all, and that is a valid and useful conclusion.
Multiply the whole polynomial by the least common denominator first. That leaves the roots unchanged and gives you the integer coefficients the theorem requires.
Synthetic division is faster than substituting. If the remainder is zero the candidate is a root, and the bottom row already gives you the quotient polynomial to carry on with.
Counting the sign changes in the coefficients gives the maximum number of positive real roots, with the true number lower by an even amount. Repeating on the polynomial with x replaced by minus x does the same for negative roots.
That the same factor divides the polynomial more than once. On a graph the curve touches the axis and turns back at a root of even multiplicity, rather than crossing it.
Because every integer divides zero, so the candidate list would be meaningless. Factor out the highest power of x first, then apply the theorem to what remains.
This page explains an algebraic method for educational purposes only, and results should be checked against your own working before being used in assessed coursework.
