Rational root theorem

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

a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}=0

with integer coefficients $a_{i}\in \mathbb {Z}$ and $a_{0},a_{n}\neq 0$ . 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 $p$ and $q$ are relatively prime, satisfies:

$p$ 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 $a n = 1$ .

Application edit

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.

Cubic equation edit

The general cubic equation

ax^{3}+bx^{2}+cx+d=0

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.

Proofs edit

Elementary proof edit

Let $P(x)\ =\ a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}$ with $a_{0},\ldots ,a_{n}\in \mathbb {Z} .$

Suppose $P (p / q) = 0$ for some coprime $p, q \in ℤ$ :

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.

To clear denominators, multiply both sides by $q n$ :

a_{n}p^{n}+a_{n-1}p^{n-1}q+\cdots +a_{1}pq^{n-1}+a_{0}q^{n}=0.

Shifting the $a 0$ term to the right side and factoring out $p$ on the left side produces:

p\left(a_{n}p^{n-1}+a_{n-1}qp^{n-2}+\cdots +a_{1}q^{n-1}\right)=-a_{0}q^{n}.

Thus, $p$ divides $a 0 q n$ . But $p$ is coprime to $q$ and therefore to $q n$ , so by Euclid's lemma $p$ must divide the remaining factor $a 0$ .

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

q\left(a_{n-1}p^{n-1}+a_{n-2}qp^{n-2}+\cdots +a_{0}q^{n-1}\right)=-a_{n}p^{n}.

Reasoning as before, it follows that $q$ divides $a n$ .^[1]

Proof using Gauss's lemma edit

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 $P$ .

Proof using Vieta's formulas edit

Given a polynomial of degree n:

P(x)={a_{n}x^{n}}+{a_{n-1}x^{n-1}}+\cdots +{a_{1}x}+{a_{0}}

,

{a_{i}}\in \mathbb {Z}

Let $r_{1}$ , ${r_{2}}$ , $\cdots$ , ${r_{n}}$ be roots of $P(x)$ , where ${r_{i}}\in {\mathbb {C} }$ . By Vieta's formulas:

{r_{1}}{r_{2}}\cdots {r_{n}}={(-1)^{n}}{a_{0} \over a_{n}}

Suppose there exist some coprime $p,q\in$ $\mathbb {Z}$ such that ${\textstyle P({p \over q})=0}$ . Then $p \over q$ must be a factor of ${a_{0} \over a_{n}}$ because $p \over q$ is a root of $P(x)$ . Which implies $p|a_{0}$ and $q|a_{n}$ because $p$ and $q$ are coprime.

Examples edit

First edit

In the polynomial

2x^{3}+x-1,

any rational root fully reduced would have to have a numerator that divides evenly into 1 and a denominator that divides evenly into 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 edit

In the polynomial

x^{3}-7x+6

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 edit

Every rational root of the polynomial

3x^{3}-5x^{2}+5x-2

must be among the numbers

\pm {\tfrac {1,2}{1,3}}=\pm \left\{1,2,{\tfrac {1}{3}},{\tfrac {2}{3}}\right\}.

These 8 root candidates $x = r$ can be tested by evaluating $P (r)$ , for example using Horner's method. It turns out there is exactly one with $P (r) = 0$ .

This process may be made more efficient: if $P (r) \neq 0$ , it can be used to shorten the list of remaining candidates.^[2] For example, $x = 1$ does not work, as $P (1) = 1$ . Substituting $x = 1 + t$ yields a polynomial in $t$ with constant term $P (1) = 1$ , while the coefficient of $t 3$ remains the same as the coefficient of $x 3$ . Applying the rational root theorem thus yields the possible roots $t=\pm {\tfrac {1}{1,3}}$ , so that

x=1+t=2,0,{\tfrac {4}{3}},{\tfrac {2}{3}}.

True roots must occur on both lists, so list of rational root candidates has shrunk to just $x = 2$ and $x = 2/3$ .

If $k \geq 1$ rational roots are found, Horner's method will also yield a polynomial of degree $n - k$ whose roots, together with the rational roots, are exactly the roots of the original polynomial. If none of the candidates is a solution, there can be no rational solution.

Notes edit

^ Arnold, D.; Arnold, G. (1993). Four unit mathematics. Edward Arnold. pp. 120–121. ISBN 0-340-54335-3.
^ King, Jeremy D. (November 2006). "Integer roots of polynomials". Mathematical Gazette. 90: 455–456. doi:10.1017/S0025557200180295.

References edit

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
Phillip S. Jones, Jack D. Bedient: The historical roots of elementary mathematics. Dover Courier Publications 1998, ISBN 0-486-25563-8, pp. 116–117 (online copy, p. 116, at Google Books)
Ron Larson: Calculus: An Applied Approach. Cengage Learning 2007, ISBN 978-0-618-95825-2, pp. 23–24 (online copy, p. 23, at Google Books)

External links edit

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

[1] Arnold, D.; Arnold, G. (1993). Four unit mathematics. Edward Arnold. pp. 120–121. ISBN 0-340-54335-3.

[2] King, Jeremy D. (November 2006). "Integer roots of polynomials". Mathematical Gazette. 90: 455–456. doi:10.1017/S0025557200180295.

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