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{{short description|Relationship between the rational roots of a polynomial and its extreme coefficients}}
In [[algebra]], the '''rational root theorem''' (or '''rational root test''', '''rational zero theorem''', '''rational zero test''' or '''{{math|''p''/''q''}} theorem''') states a constraint on [[rational number|rational]] [[Equation solving|solutions]] of a [[polynomial equation]]
<math display="block">a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0</math>
with [[integer]] coefficients <math>a_i\in\mathbb{Z}</math> and <math>a_0,a_n \neq 0</math>.  Solutions of the equation are also called [[root of a polynomial|roots]] or zeros of the [[polynomial]] on the left side.

The theorem states that each [[rational number|rational]] solution {{math|1=''x''&nbsp;=&nbsp;''<sup>p</sup>⁄<sub>q</sub>''}}, written in lowest terms so that {{math|''p''}} and {{math|''q''}} are [[relatively prime]], satisfies:
* {{math|''p''}} is an integer [[divisor|factor]] of the [[constant term]] {{math|''a''<sub>0</sub>}}, and
* {{math|''q''}} is an integer factor of the leading [[coefficient]] {{math|''a<sub>n</sub>''}}.

The rational root theorem is a special case (for a single linear factor) of [[Gauss's lemma (polynomial)|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&nbsp;{{math|1=''a<sub>n</sub>''&nbsp;=&nbsp;1}}.

==Application==

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 {{math|1=''x'' = ''r''}} is found, a linear polynomial {{math|(''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.

===Cubic equation===

The general [[cubic equation]]
<math display="block">ax^3 + bx^2 + cx + d = 0</math>
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 [[algebraic expression|algebraically]] uses [[Cubic function|cube roots]]. But if the test finds a rational solution {{math|''r''}}, then factoring out {{math|(''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.

==Proofs==

===Elementary proof===

Let <math>P(x) \ =\ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0</math> with <math>a_0, \ldots, a_n \in \mathbb{Z}.</math> 

Suppose {{math|1=''P''(''p''/''q'') = 0}} for some [[coprime]] {{math|''p'', ''q'' ∈ '''ℤ'''}}:
<math display="block">P\left(\tfrac{p}{q}\right) = a_n\left(\tfrac{p}{q}\right)^n + a_{n-1}\left(\tfrac{p}{q}\right)^{n-1} + \cdots + a_1 \left(\tfrac{p}{q}\right) + a_0 = 0.</math>

To clear denominators, multiply both sides by {{math|''q''<sup>''n''</sup>}}: 
<math display="block">a_n p^n + a_{n-1} p^{n-1}q + \cdots + a_1 p q^{n-1} + a_0 q^n = 0.</math>

Shifting the {{math|''a''<sub>0</sub>}} term to the right side and factoring out {{mvar|p}} on the left side produces:
<math display="block">p \left (a_np^{n-1} + a_{n-1}qp^{n-2} + \cdots + a_1q^{n-1} \right ) = -a_0q^n.</math>

Thus, {{mvar|p}} divides {{math|''a''<sub>0</sub>''q<sup>n</sup>''}}. But {{mvar|p}} is coprime to {{mvar|q}} and therefore to {{math|''q<sup>n</sup>''}}, so by [[Euclid's lemma]] {{mvar|p}} must divide the remaining factor {{math|''a''<sub>0</sub>}}.

On the other hand, shifting the {{math|''a''<sub>''n''</sub>}} term to the right side and factoring out {{mvar|q}} on the left side produces:
<math display="block">q \left (a_{n-1}p^{n-1} + a_{n-2}qp^{n-2} + \cdots + a_0q^{n-1} \right ) = -a_np^n.</math>

Reasoning as before, it follows that {{mvar|q}} divides {{math|''a<sub>n</sub>''}}.<ref>{{cite book |first1=D. |last1=Arnold |first2=G. |last2=Arnold |title=Four unit mathematics |publisher=Edward Arnold |year=1993 |isbn=0-340-54335-3 |pages=120–121 }}</ref>

=== Proof using Gauss's lemma ===

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 (polynomial)|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 {{math|'''Q'''[''X'']}}, then it also factors in {{math|'''Z'''[''X'']}} as a product of primitive polynomials. Now any rational root {{math|''p''/''q''}} corresponds to a factor of degree 1 in {{math|'''Q'''[''X'']}} of the polynomial, and its primitive representative is then {{math|''qx'' − ''p''}}, assuming that {{math|''p''}} and {{math|''q''}} are coprime. But any multiple in {{math|'''Z'''[''X'']}} of {{math|''qx'' − ''p''}} has leading term divisible by {{math|''q''}} and constant term divisible by {{math|''p''}}, which proves the statement. This argument shows that more generally, any irreducible factor of {{math|''P''}} can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of&nbsp;{{math|''P''}}.

==Examples==

===First===

In the polynomial
<math display="block">2x^3+x-1,</math>
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.

===Second===
In the polynomial
<math display="block">x^3-7x+6</math>
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).

===Third===

Every rational root of the polynomial
<math display="block">P=3x^3 - 5x^2 + 5x - 2 </math>
must be one of the 8 numbers
<math display="block">\pm 1, \pm2, \pm\tfrac{1}{3}, \pm \tfrac{2}{3} .</math>
These 8 possible values for {{mvar|x}} can be tested by evaluating the polynomial. It turns out there is exactly one rational root, which is <math display=inline>x=2/3.</math> 

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

Firstly, if <math>x<0,</math> all terms of {{mvar|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 <math display=inline>1, 2, \tfrac{1}{3},  \tfrac{2}{3} .</math>

One has <math>P(1)=3-5+5-2=1.</math> So, {{math|1}} is not a root. Moreover, if one sets {{math|1=''x'' = 1 + ''t''}}, one gets without computation that <math>Q(t)=P(t+1)</math> is a polynomial in {{mvar|t}} with the same first coefficient {{math|3}} and constant term {{math|1}}.<ref>{{cite journal |last=King |first=Jeremy D. |title=Integer roots of polynomials |journal=Mathematical Gazette |volume=90 |date= November 2006 |pages=455–456 |doi=10.1017/S0025557200180295 |doi-access=free }}</ref> The rational root theorem implies thus that a rational root of {{mvar|Q}} must belong to <math display=inline>\{\pm1, \pm\frac 13 \},</math> and thus that the rational roots of {{mvar|P}} satisfy <math display=inline>x = 1+t \in \{2, 0, \tfrac{4}{3}, \tfrac{2}{3}\}.</math> This shows again that any rational root of {{mvar|P}} is positive, and the only remaining candidates are {{math|2}} and {{math|2\3}}.

To show that {{math|2}} is not a root, it suffices to remark that if <math>x=2,</math> then <math>3x^2</math> and <math>5x-2</math> are multiples of {{math|8}}, while <math>-rx^2</math> is not. So, their sum cannot be zero.

Finally, only <math>P(2/3)</math> needs to be computed to verify that it is a root of the polynomial.

==See also==
{{Portal|Mathematics}}
*[[Fundamental theorem of algebra]]
*[[Integrally closed domain]]
*[[Descartes' rule of signs]]
*[[Gauss–Lucas theorem]]
*[[Properties of polynomial roots]]
*[[Content (algebra)]]
*[[Eisenstein's criterion]]

==Notes==
{{Reflist}}

==References==
{{refbegin}}
* {{ cite book | first1 = Charles D. | last1 = Miller | first2 = Margaret L. | last2 = Lial | first3 = David I. | last3 = Schneider | title = Fundamentals of College Algebra | publisher = Scott & Foresman/Little & Brown Higher Education | edition = 3rd | year = 1990 | isbn = 0-673-38638-4 | pp = 216–221 }}
* {{ cite book | first1 = Phillip S. | last1 = Jones | first2 = Jack D. | last2 = Bedient | title = The historical roots of elementary mathematics | publisher = Dover Courier Publications | year = 1998 | isbn = 0-486-25563-8 | pp = 116–117 | url = {{Google books|7xArILpcndYC|page=116|plainurl=yes}} }}
* {{ cite book | first = Ron | last = Larson | title = Calculus: An Applied Approach | publisher = Cengage Learning | year = 2007 | isbn = 978-0-618-95825-2 | pp = 23–24 | url = {{Google books|bDG7V0OV34C|page=23|plainurl=yes}} }}
{{refend}}

==External links==
*{{MathWorld|urlname=RationalZeroTheorem|title=Rational Zero Theorem}}
*[http://planetmath.org/encyclopedia/RationalRootTheorem.html ''RationalRootTheorem''] at [[PlanetMath]]
* [http://www.cut-the-knot.org/Generalization/RationalRootTheorem.shtml Another proof that n<sup>th</sup> roots of integers are irrational, except for perfect nth powers] by Scott E. Brodie
*[http://www.purplemath.com/modules/rtnlroot.htm ''The Rational Roots Test''] at purplemath.com

[[Category:Theorems about polynomials]]
[[Category:Polynomial factorization algorithms]]