Key takeaway: The rational root theorem says any rational root p/q of a polynomial with integer coefficients has p dividing the constant term and q dividing the leading coefficient.
Grade 12 – PrecalculusA-APR.B.3Expressions and Equations
Definition of Rational Root Theorem
The rational root theorem turns a hopeless guessing game into a short checklist. Given a polynomial with whole-number coefficients, it limits the fractions that could possibly be roots: the numerator must divide the constant term, and the denominator must divide the leading coefficient.
List every divisor of each number, form all fractions with either sign, and test them with the factor theorem or synthetic division. Passing the test is not guaranteed, since the list only holds candidates. Once one root is found, division lowers the degree and the rest of the roots usually fall out of a quadratic.
Formula and notation
For \(P(x)=a_nx^n+\cdots+a_1x+a_0\) with integer coefficients, every rational root in lowest terms has the form
\[x=\pm\frac{p}{q},\qquad p\mid a_0,\quad q\mid a_n.\]
Here \(p\mid a_0\) means \(p\) divides the constant term, and \(q\) divides the leading coefficient.
Key facts
- With leading coefficient 1, every rational root is a whole number dividing the constant term.
- The theorem finds possible roots only; each candidate still needs a test.
- It cannot find irrational or non-real roots directly, only rational ones.
- The number of candidates is the divisors of a₀ times the divisors of aₙ, with both signs.
Rational Root Theorem example
Find all rational roots of \(P(x)=3x^3+5x^2-26x+8\).
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The constant term 8 has divisors 1, 2, 4 and 8. The leading coefficient 3 has divisors 1 and 3.
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Candidates: the eight whole numbers with both signs, plus one third, two thirds, four thirds and eight thirds, again with both signs. That makes 16 in all.
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Test \(x=2\): \(24+20-52+8=0\), so 2 is a root. Dividing gives \(3x^2+11x-4=(3x-1)(x+4)\).
The roots are 2, \(\frac{1}{3}\) and −4.
Common mistake
Watch out
Ravi builds his candidate fractions as divisors of 3 over divisors of 8, listing \(\frac{3}{8}\) and \(\frac{1}{8}\) among them.
Why it is wrong
The numerator comes from the constant term and the denominator from the leading coefficient: \(\frac{p}{q}\) with \(p\mid 8\) and \(q\mid 3\). Swapping them produces candidates that this rational root theorem rules out, such as \(\frac{1}{8}\).
Related terms and lessons
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