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| In [[algebra]], the '''rational root theorem''' (or '''rational root test''') states a constraint on rational solutions (or [[root of a function|roots]]) of the [[polynomial]] equation
| | 55 years old Electrician (Special Class ) Valentin Jury from Didsbury, likes theatre, 10 hatha yoga positions and woodworking. Plans to give up work and take the family to numerous great heritage listed destinations in the world like Chongoni Rock-Art Area. |
| :<math>a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\!</math>
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| with [[integer]] coefficients.
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| If ''a''<sub>0</sub> and ''a''<sub>''n''</sub> are nonzero,
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| then each [[rational number|rational]] solution ''x'',
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| when written as a fraction ''x'' = ''p''/''q'' in lowest terms (i.e., the [[greatest common divisor]] of ''p'' and ''q'' is 1), satisfies
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| * ''p'' is an integer [[divisor|factor]] of the [[constant term]] ''a''<sub>0</sub>, and
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| * ''q'' is an integer factor of the leading [[coefficient]] ''a''<sub>''n''</sub>.
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| 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 a special case of the rational root theorem if the leading coefficient ''a''<sub>''n''</sub> = 1.
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| == Proofs ==
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| ===A proof===
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| Let ''P''(''x'') = ''a''<sub>''n''</sup>''x''<sup>''n''</sup> + ''a''<sub>''n''−1</sup>''x''<sup>''n''−1</sup> + ... + ''a''<sub>1</sup>''x'' + ''a''<sub>0</sup> for some ''a''<sub>0</sub>, ..., ''a''<sub>''n''</sub> ∈ '''Z''', and suppose ''P''(''p''/''q'') = 0 for some [[coprime]] ''p'', ''q'' ∈ '''Z''':
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| :<math>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>
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| If we shift the constant term to the right hand side, factor a ''p'' and multiply by ''q''<sup>''n''</sup>, we get
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| :<math>\qquad p(a_np^{n-1} + a_{n-1}qp^{n-2} + \cdots + a_1q^{n-1}) = -a_0q^n.</math>
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| We see that ''p'' times the integer quantity in parentheses equals −''a''<sub>0</sub>''q''<sup>''n''</sup>, so ''p'' divides ''a''<sub>0</sub>''q''<sup>''n''</sup>. But ''p'' is coprime to ''q'' and therefore to ''q''<sup>''n''</sup>, so by (the generalized form of) [[Euclid's lemma]] it must divide the remaining factor ''a''<sub>0</sub> of the product.
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| If we instead shift the leading term to the right hand side and multiply by ''q''<sup>''n''</sup>, we get
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| :<math>\qquad q(a_{n-1}p^{n-1} + a_{n-2}qp^{n-2} + \cdots + a_0q^{n-1}) = -a_np^n.</math>
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| And for similar reasons, we can conclude that ''q'' divides ''a''<sub>''n''</sub>.<ref>{{cite book|author=D. Arnold, G. Arnold|title=Four unit mathematics|publisher=Edward Arnold|year=1993|isbn=0-340-54335-3|pages=120–121}}</ref>
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| === Proof using Gauss's lemma ===
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| 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 ''p'' and ''q'' are coprime. But any multiple in {{math|ℤ[''X'']}} of {{math|''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''.
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| == Example ==
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| For example, every rational solution of the equation
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| :<math>3x^3 - 5x^2 + 5x - 2 = 0\,\!</math>
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| must be among the numbers symbolically indicated by
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| :± <math>\tfrac{1,2}{1,3}\,,</math>
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| which gives the list of 8 possible answers:
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| :<math>1, -1, 2, -2, \frac{1}{3}, -\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}\,.</math>
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| These root candidates can be tested using the [[Horner's method]] (for instance). In this particular case there is exactly one rational root. If a root candidate does not satisfy the equation, it can be used to shorten the list of remaining candidates. For example, ''x'' = 1 does not satisfy the equation as the left hand side equals 1. This means that substituting ''x'' = 1 + ''t'' yields a polynomial in ''t'' with constant term 1, while the coefficient of ''t''<sup>3</sup> remains the same as the coefficient of ''x''<sup>3</sup>. Applying the rational root theorem thus yields the following possible roots for ''t'':
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| :<math>t=\pm\tfrac{1}{1,3}</math>
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| Therefore,
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| :<math>x = 1+t = 2, 0, \frac{4}{3}, \frac{2}{3}</math>
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| Root candidates that do not occur on both lists are ruled out. The list of rational root candidates has thus shrunk to just ''x'' = 2 and ''x'' = 2/3.
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| If a root ''r''<sub>1</sub>
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| is found, Horner's method will also yield a polynomial of degree ''n'' − 1 whose roots, together with ''r''<sub>1</sub>, are exactly the roots of the original polynomial. It may also be the case that none of the candidates is a solution; in this case the equation has no rational solution. If the equation lacks a constant term ''a''<sub>0</sub>, then 0 is one of the rational roots of the equation.
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| ==See also==
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| *[[Descartes' rule of signs]]
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| *[[Gauss–Lucas theorem]]
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| *[[Properties of polynomial roots]]
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| *[[Content (algebra)]]
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| == Notes ==
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| <references/>
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| == References ==
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| *Charles D. Miller, Margaret L. Lial, David I. Schneider: ''Fundamentals of College Algebra''. Scott & Foresman/Little & Brown Higher Education, 3rd edition 1990, ISBN 0-673-38638-4, pp. 216–221
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| *Phillip S. Jones, Jack D. Bedient: ''The historical roots of elementary mathematics''. Dover Courier Publications 1998, ISBN 0-486-25563-8, pp. 116–117 ({{Google books|7xArILpcndYC|online copy|page=116}})
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| *Ron Larson: ''Calculus: An Applied Approach''. Cengage Learning 2007, ISBN 978-0-618-95825-2, pp. 23–24 ({{Google books|bDG7V0OV34C|online copy|page=23}})
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| ==External links==
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| *{{MathWorld|urlname=RationalZeroTheorem|title=Rational Zero Theorem}}
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| *[http://planetmath.org/encyclopedia/RationalRootTheorem.html ''RationalRootTheorem''] at [[PlanetMath]]
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| * [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
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| *[http://www.purplemath.com/modules/rtnlroot.htm ''The Rational Roots Test''] at purplemath.com
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| [[Category:Polynomials]]
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| [[Category:Theorems in algebra]]
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| [[Category:Root-finding algorithms]]
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55 years old Electrician (Special Class ) Valentin Jury from Didsbury, likes theatre, 10 hatha yoga positions and woodworking. Plans to give up work and take the family to numerous great heritage listed destinations in the world like Chongoni Rock-Art Area.