Rational root theorem

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.

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 {{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.

Let $P(x)\; \backslash \; =\backslash \; a\_n\; x^n\; +\; a\_\{n-1\}\; x^\{n-1\}\; +\; \backslash cdots\; +\; a\_1\; x\; +\; a\_0$ with $a\_0,\; \backslash ldots,\; a\_n\; \backslash in\; \backslash mathbb\{Z\},\; a\_0,a\_n\; \backslash neq\; 0.$

Suppose {{math|1=P(p/q) = 0}} for some coprime {{math|p, q ∈ ℤ}}:

$$P\backslash left(\backslash tfrac\{p\}\{q\}\backslash right)\; =\; a\_n\backslash left(\backslash tfrac\{p\}\{q\}\backslash right)^n\; +\; a\_\{n-1\}\backslash left(\backslash tfrac\{p\}\{q\}\backslash right)^\{n-1\}\; +\; \backslash cdots\; +\; a\_1\; \backslash left(\backslash tfrac\{p\}\{q\}\backslash right)\; +\; a\_0\; =\; 0.$$

To clear denominators, multiply both sides by {{math|qⁿ}}:

$$a\_n\; p^n\; +\; a\_\{n-1\}\; p^\{n-1\}q\; +\; \backslash cdots\; +\; a\_1\; p\; q^\{n-1\}\; +\; a\_0\; q^n\; =\; 0.$$

Shifting the {{math|a₀}} term to the right side and factoring out {{mvar|p}} on the left side produces:

$$p\; \backslash left\; (a\_np^\{n-1\}\; +\; a\_\{n-1\}qp^\{n-2\}\; +\; \backslash cdots\; +\; a\_1q^\{n-1\}\; \backslash right\; )\; =\; -a\_0q^n.$$

Thus, {{mvar|p}} divides {{math|a₀qⁿ}}. But {{mvar|p}} is coprime to {{mvar|q}} and therefore to {{math|qⁿ}}, so by Euclid's lemma {{mvar|p}} must divide the remaining factor {{math|a₀}}.

On the other hand, shifting the {{math|a_n}} term to the right side and factoring out {{mvar|q}} on the left side produces:

$$q\; \backslash left\; (a\_\{n-1\}p^\{n-1\}\; +\; a\_\{n-2\}qp^\{n-2\}\; +\; \backslash cdots\; +\; a\_0q^\{n-1\}\; \backslash right\; )\; =\; -a\_np^n.$$

Reasoning as before, it follows that {{mvar|q}} divides {{math|a_n}}.{{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 }}

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 {{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 {{math|P}}.

In the polynomial

$$2x^3+x-1,$$

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 ±1/2 and ±1; since neither of these equates the polynomial to zero, it has no rational roots.

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).

Every rational root of the polynomial

$$P=3x^3\; -\; 5x^2\; +\; 5x\; -\; 2$$

must be one of the 8 numbers

$$\backslash pm\; 1,\; \backslash pm2,\; \backslash pm\backslash tfrac\{1\}\{3\},\; \backslash pm\; \backslash tfrac\{2\}\{3\}\; .$$

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 $x=2/3.$

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

Firstly, if 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 $1,\; 2,\; \backslash tfrac\{1\}\{3\},\; \backslash tfrac\{2\}\{3\}\; .$

One has $P(1)=3-5+5-2=1.$ So, {{math|1}} is not a root. Moreover, if one sets {{math|1=x = 1 + t}}, one gets without computation that $Q(t)=P(t+1)$ is a polynomial in {{mvar|t}} with the same first coefficient {{math|3}} and constant term {{math|1}}.{{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 }} The rational root theorem implies thus that a rational root of {{mvar|Q}} must belong to $\backslash \{\backslash pm1,\; \backslash pm\backslash frac\; 13\; \backslash \},$ and thus that the rational roots of {{mvar|P}} satisfy $x\; =\; 1+t\; \backslash in\; \backslash \{2,\; 0,\; \backslash tfrac\{4\}\{3\},\; \backslash tfrac\{2\}\{3\}\backslash \}.$ 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 $x=2,$ then $3x^3$ and $5x-2$ are multiples of {{math|8}}, while $-5x^2$ is not. So, their sum cannot be zero.

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

{{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

{{Reflist}}

{{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}}

• {{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^th 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