How this calculation works
Remove zero roots, enumerate reduced rational candidates from the remaining integer polynomial and evaluate them exactly. Repeated division finds multiplicities. The residual factor is retained so irrational or complex zeros are not falsely reported as absent.
Rational candidates = ±(factor of constant)/(factor of leading coefficient)A worked example
x³ − 6x² + 11x − 6 has rational zeros 1, 2 and 3, each once. Dividing out their linear factors leaves the constant factor 1.
Good to know
Integer coefficients within ±10,000, degree at most 12, and at most 10,000 candidate fractions. Exact rational arithmetic has an intermediate-size bound. Enumerates rational zeros only; remaining factors may have irrational or complex zeros. The zero polynomial vanishes at every real number.
Reference: OpenStax: rational-zero and factor theorems (opens in a new tab) ↗
Candidate fractions are possibilities, not confirmed zeros
For an integer polynomial with nonzero constant, the rational-zero theorem bounds candidate roots by ±p/q: p divides the constant and q divides the leading coefficient. Most candidates may still fail when substituted.
Every candidate is evaluated exactly here. A zero result confirms a linear factor, and repeated division finds how many times it occurs. For x³−6x²+11x−6, candidates 1, 2 and 3 each succeed once.
A zero constant requires separate treatment because zero is then a root. Each factor x is removed first; the remaining nonzero-constant polynomial can then use the theorem. Leading zero entries are discarded before determining the actual degree.
Retain the factor that rational search does not resolve
The polynomial x²−2 has no rational zeros, but its real zeros are ±√2. The remaining polynomial factor therefore carries important information. None in the rational-zero result must not be read as no real or complex zeros.
A quadratic equation solver can address a residual quadratic under its own real-root convention. This finder does not approximate general irrational or complex roots, and it does not claim full symbolic factorization over those fields.
The bounded search accepts integer coefficients within ±10,000 and degree at most 12. It stops rather than silently truncating if the candidate set would exceed the supported limit. A nonzero constant has no zeros; the zero polynomial vanishes for every real input.
A few questions, answered
Can I enter rational coefficients?
This search requires integer coefficients after parsing. Clear denominators explicitly if that is appropriate for your polynomial, keeping the resulting integers within the limits.
What does multiplicity mean?
It is the number of repeated factors (x−r) associated with that zero.
Is the residual factor always a constant?
No. It can contain irrational or complex zeros that the rational search does not resolve.