Sign changes give a set of possibilities, not a number — here they all are, with the true one marked.
Count the sign variations
Every combination the rule allows — and which one is true.
Any degree from 1 to 12. Missing powers are fine — they are skipped, not counted.
x³ − 6x² + 11x − 6 — degree 3
2 combinations are possible
The rule narrows it to these; solving picks one out.
Variations in f(x)
3
so 3 or 1 positive real roots
Variations in f(−x)
0
so 0 negative real roots
Roots at zero
0
counted as neither positive nor negative
Complex pairs
0
always an even number of complex roots
f(x), term by term
x³→−6x²→11x→−6
Signs: + − + − — 3 variations, so 3 or 1 roots.
f(−x) = − x³ − 6x² − 11x − 6
−x³→−6x²→−11x→−6
Signs: − − − − — 0 variations, so 0 roots.
Every combination the rule permits
Possible counts of positive, negative, zero and complex roots
Positive
Negative
At zero
Complex pairs
Total
3
0
0
0
3 — this one
1
0
0
1
3
The roots themselves
1, 2, 3
f(x) has 3 sign variations, so there are 3 or 1 positive real roots. f(−x) has 0 sign variations, so there is exactly 0 negative real roots. 2 combinations are consistent with the rule. Solving shows the true one is 3 positive and 0 negative, with 0 complex pairs.
Every row above sums to the degree, and the true split is one of them.
The counts drop by two at a time because non-real roots come in conjugate pairs.
What this tool shows
Every combination of positive, negative and complex roots the rule allows — and, beside them, the one that is actually true.
Sign variations in f(x) and in f(−x), term by term
Every (positive, negative, complex) combination the rule permits
Roots at the origin factored out and counted separately
The true split, found by solving, marked among the possibilities
Every possibility the rule allows Zero coefficients skipped, not counted The true split shown beside them Free, no signup
Free, no signup — every row shown sums to the degree.
Updated 6 September 2026 · Works in any browser, no installation
Descartes’ rule gives the number of sign variations, or that number reduced by an even amount — a set of possibilities rather than a count. x³ − 6x² + 11x − 6 has 3 variations in f(x) and 0 in f(−x), which leaves 2 combinations; solving shows the true one is 3 positive and 0 negative.
At a glance
Formula shown
The number of positive real roots equals the number of sign variations in f(x), or that number reduced by a positive even integer. The same rule on f(\u2212x) gives the negative roots.
Scenario support
Any polynomial with real coefficients, degree 1 to 12.
Educational estimate
Planning support from the values you enter — not professional advice.
Why the rule gives possibilities rather than a count
The honest statement is: the number of positive real roots is the number of sign variations, or that number less an even amount. Not “equals”. A page that reports one number has quietly dropped the second half of the theorem.
x³ − 6x² + 11x − 6 has 3 variations, so it has 3 or 1 positive real roots — the rule cannot distinguish those two on its own. Combined with the 0 variations in f(−x), 2 whole combinations survive, and the rule stops there.
That is still useful. It bounds the search, it often rules out cases entirely, and occasionally — as in x³ + x + 1 — only one combination survives and the rule has settled the question completely without any solving at all.
Zero coefficients are skipped, not counted
x⁵ − 1 has signs (+, −) and therefore 1 variation, not four. The missing x⁴, x³, x² and x terms are not zeros in the sign sequence — they are absent from it.
This trips people up because writing the polynomial out in full, with zero coefficients in place, makes it look like there should be more transitions. There are not. A term with coefficient zero contributes nothing to the sign pattern: it cannot start a new sign and it cannot end one.
The tool marks skipped powers explicitly under each row for that reason. x⁵ − 1’s single variation gives exactly 1 positive real root, which happens to be the whole answer for that polynomial.
Why the step down is two and not one
Because non-real roots of a real polynomial come in conjugate pairs. If a + bi is a root then so is a − bi — you cannot have one without the other, since the coefficients would stop being real.
So every pair of real roots the polynomial “loses” relative to the variation count is replaced by a pair of complex ones, two at a time. That is why the possible counts step down by two, and why they end at 0 or 1 depending on the parity of the variation count rather than at some other number.
It also fixes the arithmetic that has to close: positive + negative + roots at zero + 2×(complex pairs) = the degree. Every row in the table above satisfies that, and any combination that does not is not a possibility the rule ever allowed.
The f(−x) half, and what actually changes
Substituting −x flips the sign of every odd-power term and leaves the even ones alone. That is the entire operation — there is no need to re-expand anything.
The variations in f(−x) count the positive roots of f(−x), and a positive root of f(−x) is a negative root of f(x). That is the whole argument, and it is why the same rule does double duty.
x⁴ − 5x² + 4 is even, so f(−x) is identical to f(x) and both halves give 2 variations — which is what you would expect from a polynomial symmetric about the y-axis, whose roots come in ± pairs. The true split is 2 positive and 2 negative, exactly balanced.
A root at zero is in neither list
x = 0 is neither positive nor negative, and the rule counts it in neither place. If the constant term is zero the arithmetic will not close until that root is accounted for separately.
x³ − x has a root at the origin. Factoring x out leaves x² − 1, and the rule is applied to that: 1 variation each way, giving 1 positive and 1 negative, plus the 1 at zero. Three roots, correctly accounted for.
Skip that step and the totals come out one short, which usually gets blamed on the rule rather than on the missing root. The tool factors x out automatically and says so.
When the rule settles it on its own
Sometimes only one combination survives, and then no solving is needed at all. x³ + x + 1 is one: f(x) has 0 variations and f(−x) has 1, which leaves exactly one row — 0 positive, 1 negative, 1 complex pair.
x⁴ + 1 is the strongest version of this. Neither f(x) nor f(−x) has a single sign change, so there are no positive and no negative real roots — and since there is no root at zero either, all 4 roots are complex. The rule has proved that without any root-finding whatsoever.
That is worth knowing before reaching for a numerical solver. A method that searches for real roots will find none here, and it is better to know in advance that none exists than to conclude the search failed.
Sources and methodology
The rule is a classical result; the reference below carries its canonical statement alongside the related bounds on polynomial zeros.
Method. Every figure on this page comes from src/lib/descartes-rule-of-signs.ts. The sign counting is exact over the rational coefficients. The true root split shown beside the possibilities comes from the Aberth–Ehrlich method, which refines all roots simultaneously in the complex plane rather than by successive deflation — so it does not miss a conjugate pair the way deflation can. That engine is verified on every change against 52 hand-written assertions, including that the true split is always one of the rows the rule permits, and that every row sums to the degree. The count and the per-case breakdown are published on the formula verification page.
Related calculators
Where this goes next:
Rational ZerosList every rational root candidate from the rational root theorem, test each one, divide out the hits by synthetic division, and use the bound rules to strike the rest off without testing them.
Cubic EquationSolve any cubic exactly when it has a rational root — deflate and finish with the quadratic formula — and by the trigonometric form when it does not, with the discriminant saying which case you are in.
DiscriminantEvaluate b² − 4ac, the cubic and quartic discriminants, or D = f_xx·f_yy − (f_xy)² for the second-derivative test, in exact arithmetic so the sign is never a rounding artefact.
Quadratic FormulaSolve any quadratic with exact roots — surds stay surds and a negative discriminant gives the complex pair — plus the vertex, the factored form and every step of the working.
Polynomial DivisionDivide polynomials by long division or synthetic division, with the quotient, the remainder, the remainder theorem check and every stage of the working.
Factoring TrinomialsFactor a trinomial by the AC method with every factor pair of ac listed, the common factor taken out first, and the answer multiplied back out to check it.
This calculator applies Descartes' Rule of Signs to polynomials with real coefficients, degree 1 to 12. The rule gives possible counts rather than definite ones, and the page presents them that way. The true split shown alongside comes from finding the roots numerically by the Aberth–Ehrlich method; those root values are accurate to roughly the last few digits of a double, and are there to demonstrate the rule rather than to be used as high-precision root values.
Published the Descartes' rule of signs page: sign variations in f(x) and f(−x), with zero coefficients skipped rather than counted.
Lists every combination of positive, negative and complex roots the rule permits — because the rule gives possibilities, not a count — and then says which one is true.
The true split is found numerically by the Aberth–Ehrlich method, so the rule is shown doing its job rather than asserted.
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