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Rational Root Theorem

Rational Root Theorem
Rational Zero Theorem

A theorem that provides a complete list of possible rational roots of the polynomial equation anxn + an–1xn–1 + ··· + a2x2 + a1x + a0 = 0 where all coefficients are integers.

This list consists of all possible numbers of the form c/d, where c and d are integers. c must divide evenly into the constant term a0. d must divide evenly into the leading coefficient an.

 

Example applying Rational Root Theorem to 6x⁴−2x³+5x²+x−10=0; p=±1,±2,±5,±10; q=±1,±2,±3,±6; lists possible zeros p/q.

 

 

See also

Polynomial facts

Key Formula

Possible rational roots=±pq,where p∣a0 and q∣an\text{Possible rational roots} = \pm\frac{p}{q}, \quad \text{where } p \mid a_0 \text{ and } q \mid a_n
Where:
  • pp = A factor (divisor) of the constant term a₀
  • qq = A factor (divisor) of the leading coefficient aₙ
  • a0a_0 = The constant term of the polynomial (the term with no variable)
  • ana_n = The leading coefficient (the coefficient of the highest-degree term)
  • p∣a0p \mid a_0 = p divides evenly into a₀
  • q∣anq \mid a_n = q divides evenly into aₙ

Worked Example

Problem: Find all rational roots of the polynomial equation 2x³ + 3x² − 8x + 3 = 0.
Step 1: Identify the constant term a₀ and leading coefficient aₙ.
a0=3,an=2a_0 = 3, \quad a_n = 2
Step 2: List all factors of a₀ = 3 (these are the possible values of p) and all factors of aₙ = 2 (these are the possible values of q).
p:±1,±3q:±1,±2p: \pm1, \pm3 \qquad q: \pm1, \pm2
Step 3: Form all possible fractions ±p/q and simplify to remove duplicates.
Possible roots: ±1,±3,±12,±32\text{Possible roots: } \pm1, \pm3, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}
Step 4: Test each candidate by substituting into the polynomial. Start with x = 1.
2(1)3+3(1)2−8(1)+3=2+3−8+3=0  ✓2(1)^3 + 3(1)^2 - 8(1) + 3 = 2 + 3 - 8 + 3 = 0 \; \checkmark
Step 5: Since x = 1 is a root, divide the polynomial by (x − 1) to get a quadratic, then solve.
2x3+3x2−8x+3=(x−1)(2x2+5x−3)=(x−1)(2x−1)(x+3)2x^3 + 3x^2 - 8x + 3 = (x - 1)(2x^2 + 5x - 3) = (x-1)(2x-1)(x+3)
Answer: The rational roots are x = 1, x = 1/2, and x = −3.

Another Example

This example shows two important variations: (1) when the constant term is 0, factor out x first before applying the theorem; (2) when the leading coefficient is 1, all rational roots must be integers.

Problem: Find all rational roots of x⁴ − 6x³ + 11x² − 6x = 0.
Step 1: Factor out the common factor x first, since the constant term is 0. A zero constant term means x = 0 is always a root.
x(x3−6x2+11x−6)=0x(x^3 - 6x^2 + 11x - 6) = 0
Step 2: Apply the Rational Root Theorem to the cubic factor. The constant term is −6 and the leading coefficient is 1.
a0=−6,an=1⇒Possible roots: ±1,±2,±3,±6a_0 = -6, \quad a_n = 1 \quad \Rightarrow \quad \text{Possible roots: } \pm1, \pm2, \pm3, \pm6
Step 3: When the leading coefficient is 1, the only possible rational roots are integer factors of the constant term. Test x = 1.
(1)3−6(1)2+11(1)−6=1−6+11−6=0  ✓(1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 \; \checkmark
Step 4: Divide by (x − 1) and factor the resulting quadratic.
x3−6x2+11x−6=(x−1)(x2−5x+6)=(x−1)(x−2)(x−3)x^3 - 6x^2 + 11x - 6 = (x-1)(x^2 - 5x + 6) = (x-1)(x-2)(x-3)
Answer: The rational roots are x = 0, x = 1, x = 2, and x = 3.

Frequently Asked Questions

What is the difference between the Rational Root Theorem and the Factor Theorem?
The Rational Root Theorem generates a list of candidates that might be roots, based on the constant term and leading coefficient. The Factor Theorem says that if r is a root of a polynomial, then (x − r) is a factor. You typically use the Rational Root Theorem to find candidates, then the Factor Theorem to confirm them and factor the polynomial.
Does the Rational Root Theorem find all roots of a polynomial?
No. It only identifies possible rational roots — fractions and integers. A polynomial can also have irrational roots (like √2) or complex roots (like 3 + 2i), and the theorem will not find those. For example, x² − 2 = 0 has no rational roots at all, even though the theorem suggests ±1 and ±2 as candidates.
Can you use the Rational Root Theorem if the coefficients are not integers?
The theorem requires all coefficients to be integers. However, if your polynomial has fractional coefficients, you can multiply the entire equation by the least common denominator to clear the fractions. This produces an equivalent equation with integer coefficients, and then you can apply the theorem.

Rational Root Theorem vs. Descartes' Rule of Signs

Rational Root TheoremDescartes' Rule of Signs
PurposeLists all possible rational roots of a polynomialCounts the possible number of positive and negative real roots
What it tells youWhich specific values to test as potential rootsHow many positive or negative roots exist (not which ones)
LimitationsOnly finds rational roots; misses irrational and complex rootsGives an upper bound on real roots but not their exact values
When to useWhen you need to find exact rational solutions to a polynomial equationWhen you want to narrow down the sign of potential roots before testing

Why It Matters

The Rational Root Theorem appears throughout Algebra 2 and Precalculus courses, especially when solving higher-degree polynomial equations by hand. It gives you a systematic starting point instead of random guessing, which is essential when factoring cubics or quartics. On standardized tests like the SAT and ACT, knowing this theorem lets you quickly identify which values to test when a problem asks for the zeros of a polynomial.

Common Mistakes

Mistake: Forgetting to include both positive and negative versions of each candidate.
Correction: Every fraction p/q on your list must appear as both +p/q and −p/q. A negative root is just as valid as a positive one.
Mistake: Assuming every candidate on the list is actually a root.
Correction: The theorem only provides possible roots. You must test each candidate by substituting it into the polynomial (or using synthetic division) to confirm whether it is an actual root. Most candidates on the list will not work.

Related Terms