Rational Root Theorem
In short: The rational root theorem narrows the hunt for fraction zeros: a candidate p/q must have a numerator that divides the last coefficient and a denominator that divides the first.
PrecalculusAlgebraA-APR.B.3
What is Rational Root Theorem?
The rational root theorem turns a blind search into a short list. Rather than guessing zeros of a polynomial at random, you build every fraction that could possibly work and test only those.
Suppose \(f(x) = a_nx^n + \cdots + a_1x + a_0\) has whole-number coefficients. Any fraction zero, in lowest terms, looks like \(\frac{p}{q}\) with \(p\) a factor of \(a_0\) and \(q\) a factor of \(a_n\). Passing the screen does not guarantee a candidate works; failing it guarantees the candidate does not.
How to spot it
- A polynomial of degree 3 or more has integer coefficients and must be factored; the rational root theorem gives a starting list.
- No obvious zero appears, and the quadratic formula does not apply directly.
- A question asks for all possible rational zeros before testing any of them.
- A leading coefficient other than 1 suggests candidates that are fractions.
The method
- Under the rational root theorem, list all factors \(p\) of the constant term and all factors \(q\) of the leading coefficient.
- Form every ratio \(\pm\frac{p}{q}\) and remove duplicates.
- Test candidates with direct substitution or synthetic division until one gives remainder zero.
- Divide it out, then repeat on the lower-degree quotient until it factors.
Rational Root Theorem: solved example
Problem
List the possible rational zeros of \(f(x) = 2x^3 + 3x^2 – 8x + 3\), then find all of its zeros.
Solution
Factors of the constant term 3: \(\pm 1, \pm 3\). Factors of the leading coefficient 2: \(\pm 1, \pm 2\). Candidates: \(\pm 1, \pm 3, \pm\frac{1}{2}, \pm\frac{3}{2}\).
Test 1: \(2 + 3 – 8 + 3 = 0\), so \(x – 1\) is a factor, with quotient \(2x^2 + 5x – 3 = (2x – 1)(x + 3)\). The zeros are \(1, \frac{1}{2}, -3\).
Try it
Question 1
List the possible rational zeros of \(x^3 – 4x^2 + x + 6\).
Show the answer
The leading coefficient is 1, so the candidates are the factors of 6: \(\pm 1, \pm 2, \pm 3, \pm 6\).
Question 2
Could \(\frac{3}{4}\) be a rational zero of \(6x^3 – 5x + 8\)?
Show the answer
No. The numerator 3 would need to divide the constant term 8, and it does not.
Rational Root Theorem vs. Factor Theorem
The list-making rule only suggests candidates and never confirms a zero. The factor theorem is the test: if a candidate \(c\) gives \(f(c) = 0\), then \(x – c\) is a factor. In a typical problem the first creates the shortlist and the second checks each name on it.
Keep going
Lessons
Worksheets and quizzes
- Polynomial Division and the Factor Theorem – Worksheet
- Polynomial Division and the Factor Theorem – Quiz
- Zeros and Graphs of Polynomial Functions – Worksheet
- Zeros and Graphs of Polynomial Functions – Quiz
Related words
- Factor Theorem
- Remainder Theorem
- Synthetic Division
- Fundamental Theorem of Algebra
- Zeros of a Function
Whole dictionary: A–Z · Precalculus · Algebra