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In algebra, the **rational root theorem** (or **rational root test**, **rational zero theorem**, **rational zero test** or ** p/q theorem**) states a constraint on rational solutions of a polynomial equation
with integer coefficients and . Solutions of the equation are also called roots or zeros of the polynomial on the left side.

The theorem states that each rational solution *x* = * ^{p}⁄_{q}*, written in lowest terms so that

*p*is an integer factor of the constant term*a*_{0}, and*q*is an integer factor of the leading coefficient*a*._{n}

The rational root theorem is a special case (for a single linear factor) of Gauss's lemma on the factorization of polynomials. The **integral root theorem** is the special case of the rational root theorem when the leading coefficient is *a _{n}* = 1.

The theorem is used to find all rational roots of a polynomial, if any. It gives a finite number of possible fractions which can be checked to see if they are roots. If a rational root *x* = *r* is found, a linear polynomial (*x* – *r*) can be factored out of the polynomial using polynomial long division, resulting in a polynomial of lower degree whose roots are also roots of the original polynomial.

The general cubic equation
with integer coefficients has three solutions in the complex plane. If the rational root test finds no rational solutions, then the only way to express the solutions algebraically uses cube roots. But if the test finds a rational solution *r*, then factoring out (*x* – *r*) leaves a quadratic polynomial whose two roots, found with the quadratic formula, are the remaining two roots of the cubic, avoiding cube roots.

Let with

Suppose *P*(*p*/*q*) = 0 for some coprime *p*, *q* ∈ **ℤ**:

To clear denominators, multiply both sides by *q*^{n}:

Shifting the *a*_{0} term to the right side and factoring out p on the left side produces:

Thus, p divides *a*_{0}*q ^{n}*. But p is coprime to q and therefore to

On the other hand, shifting the *a*_{n} term to the right side and factoring out q on the left side produces:

Reasoning as before, it follows that q divides *a _{n}*.

Should there be a nontrivial factor dividing all the coefficients of the polynomial, then one can divide by the greatest common divisor of the coefficients so as to obtain a primitive polynomial in the sense of Gauss's lemma; this does not alter the set of rational roots and only strengthens the divisibility conditions. That lemma says that if the polynomial factors in **Q**[*X*], then it also factors in **Z**[*X*] as a product of primitive polynomials. Now any rational root *p*/*q* corresponds to a factor of degree 1 in **Q**[*X*] of the polynomial, and its primitive representative is then *qx* − *p*, assuming that *p* and *q* are coprime. But any multiple in **Z**[*X*] of *qx* − *p* has leading term divisible by *q* and constant term divisible by *p*, which proves the statement. This argument shows that more generally, any irreducible factor of *P* can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of *P*.

In the polynomial any rational root fully reduced should have a numerator that divides 1 and a denominator that divides 2. Hence the only possible rational roots are ±1/2 and ±1; since neither of these equates the polynomial to zero, it has no rational roots.

In the polynomial the only possible rational roots would have a numerator that divides 6 and a denominator that divides 1, limiting the possibilities to ±1, ±2, ±3, and ±6. Of these, 1, 2, and –3 equate the polynomial to zero, and hence are its rational roots (in fact these are its only roots since a cubic polynomial has only three roots).

Every rational root of the polynomial must be one of the 8 numbers These 8 possible values for x can be tested by evaluating the polynomial. It turns out there is exactly one rational root, which is

However, these eight computations may be rather tedious, and some tricks allow to avoid some of them.

Firstly, if all terms of P become negative, and their sum cannot be 0; so, every root is positive, and a rational root must be one of the four values

One has So, 1 is not a root. Moreover, if one sets *x* = 1 + *t*, one gets without computation that is a polynomial in t with the same first coefficient 3 and constant term 1.^{[2]} The rational root theorem implies thus that a rational root of Q must belong to and thus that the rational roots of P satisfy This shows again that any rational root of P is positive, and the only remaining candidates are 2 and 2\3.

To show thet 2 is not a root, is suffices to remark that is then and are nultiples of 8, while is not. So, their sum cannot be zero.

Finally, only needs to be computed to verify that it is a root of the polynomial.

- Miller, Charles D.; Lial, Margaret L.; Schneider, David I. (1990).
*Fundamentals of College Algebra*(3rd ed.). Scott & Foresman/Little & Brown Higher Education. pp. 216–221. ISBN 0-673-38638-4. - Jones, Phillip S.; Bedient, Jack D. (1998).
*The historical roots of elementary mathematics*. Dover Courier Publications. pp. 116–117. ISBN 0-486-25563-8. - Larson, Ron (2007).
*Calculus: An Applied Approach*. Cengage Learning. pp. 23–24. ISBN 978-0-618-95825-2.

- Weisstein, Eric W. "Rational Zero Theorem".
*MathWorld*. *RationalRootTheorem*at PlanetMath- Another proof that n
^{th}roots of integers are irrational, except for perfect nth powers by Scott E. Brodie *The Rational Roots Test*at purplemath.com