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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}=0} with integer coefficients $a_{i}\in \mathbb {Z}$ and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Template:Tmath written in lowest terms (that is, $p$ 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

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

The general cubic equation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Elementary proof

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(x) \ =\ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_0, \ldots, a_n \in \mathbb{Z}, a_0,a_n \neq 0.}

Suppose $P (p / q) = 0$ for some coprime $p, q \in ℤ$ : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 $q n$ : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n p^n + a_{n-1} p^{n-1}q + \cdots + a_1 p q^{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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p \left (a_np^{n-1} + a_{n-1}qp^{n-2} + \cdots + a_1q^{n-1} \right ) = -a_0q^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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q \left (a_{n-1}p^{n-1} + a_{n-2}qp^{n-2} + \cdots + a_0q^{n-1} \right ) = -a_np^n.}

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

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; 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 $P$ .

Examples

First

In the polynomial Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^3+x-1,} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \pm \tfrac{1}{2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \pm1} ; since neither of these equates the polynomial to zero, it has no rational roots.

Second

In the polynomial Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Every rational root of the polynomial Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P=3x^3 - 5x^2 + 5x - 2 } must be one of the 8 numbers Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm 1, \pm2, \pm\tfrac{1}{3}, \pm \tfrac{2}{3} .} These 8 possible values for $x$ can be tested by evaluating the polynomial. It turns out there is exactly one rational root, which is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x=2/3.}

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

Firstly, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x<0,} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle 1, 2, \tfrac{1}{3}, \tfrac{2}{3} .}

One has Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(1)=3-5+5-2=1.} So, $1$ is not a root. Moreover, if one sets $x = 1 + t$ , one gets without computation that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q(t)=P(t+1)} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle \{\pm1, \pm\frac 13 \},} and thus that the rational roots of $P$ satisfy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\textstyle x = 1+t \in \{2, 0, \tfrac{4}{3}, \tfrac{2}{3}\}.} This shows again that any rational root of $P$ is positive, and the only remaining candidates are $2$ and $2/3$ .

To show that $2$ is not a root, it suffices to remark that if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=2,} then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3x^3} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 5x-2} are multiples of $8$ , while Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -5x^2} is not. So, their sum cannot be zero.

Finally, only Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(2/3)} needs to be computed to verify that it is a root of the polynomial.

Fourth

If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a, b} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{a^2}{b} +\tfrac{b^2}{a}} are integers (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\neq 0, b\neq 0} ), then both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{a^2}{b}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{b^2}{a}} must be integer.

Consider the quadratic equation whose roots are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{a^2}{b}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{b^2}{a}} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x-\tfrac{a^2}{b})(x-\tfrac{b^2}{a})=x^2-(\tfrac{a^2}{b} +\tfrac{b^2}{a})x +\tfrac{a^2}{b}\cdot\tfrac{b^2}{a}.}

Simplify the coefficients:

The coefficient of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=-(\tfrac{a^2}{b} +\tfrac{b^2}{a}),}
The constant term is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=\tfrac{a^2}{b}\cdot\tfrac{b^2}{a}=ab.}

Thus, the equation becomes: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2+mx+n=0,} where:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=-(\tfrac{a^2}{b} +\tfrac{b^2}{a})} , obviously integer, as negation of an integer,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=ab} , also integer, as the product of two integers.

Apply the rational root theorem:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a, b} given to be integers (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\neq 0, b\neq 0} ), i.e. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{a^2}{b}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{b^2}{a}} are rational. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{p}{q}} is a rational root of the equation, then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q} is an integer factor of the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2} coefficient, i.e. of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} . Thus, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q=1} . Thus, the rational root is an integer. Thus, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{a^2}{b}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tfrac{b^2}{a}} are integers.

References

↑ 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.

External links

Weisstein, Eric W. "Rational Zero Theorem". MathWorld.
Template: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.

[1]

[2]

Rational root theorem

Page version status

Contents

Application

Cubic equation

Proofs

Elementary proof

Proof using Gauss's lemma

Examples

First

Second

Third

Fourth

See also

References

Further reading

External links

Navigation menu

Rational root theorem

Application

Cubic equation

Proofs

Elementary proof

Proof using Gauss's lemma

Examples

First

Second

Third

Fourth

See also

References

Further reading

External links

Navigation menu

Search