Rational root theorem

Template:Short description In algebra, the rational root theorem (or rational root test, rational zero theorem, rational zero test or Template:Math theorem) states a constraint on rational 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 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, Template:Math and Template:Math are relatively prime), satisfies:

Template:Math is an integer factor of the constant term Template:Math, and
Template:Math is an integer factor of the leading coefficient Template:Math.

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 Template:Math.

ApplicationEdit

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 Template:Math is found, a linear polynomial Template:Math 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 equationEdit

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 algebraically uses cube roots. But if the test finds a rational solution Template:Math, then factoring out Template:Math leaves a quadratic polynomial whose two roots, found with the quadratic formula, are the remaining two roots of the cubic, avoiding cube roots.

ProofsEdit

Elementary proofEdit

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}, a_0,a_n \neq 0.</math>

Suppose Template:Math for some coprime Template:Math: <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 Template:Math: <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 Template:Math term to the right side and factoring out Template:Mvar 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, Template:Mvar divides Template:Math. But Template:Mvar is coprime to Template:Mvar and therefore to Template:Math, so by Euclid's lemma Template:Mvar must divide the remaining factor Template:Math.

On the other hand, shifting the Template:Math term to the right side and factoring out Template:Mvar 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 Template:Mvar divides Template:Math.<ref>Template:Cite book</ref>

Proof using Gauss's lemmaEdit

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 Template:Math, then it also factors in Template:Math as a product of primitive polynomials. Now any rational root Template:Math corresponds to a factor of degree 1 in Template:Math of the polynomial, and its primitive representative is then Template:Math, assuming that Template:Math and Template:Math are coprime. But any multiple in Template:Math of Template:Math has leading term divisible by Template:Math and constant term divisible by Template:Math, which proves the statement. This argument shows that more generally, any irreducible factor of Template:Math can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of Template:Math.

ExamplesEdit

FirstEdit

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.

SecondEdit

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).

ThirdEdit

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 Template:Mvar 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 Template:Mvar 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, Template:Math is not a root. Moreover, if one sets Template:Math, one gets without computation that <math>Q(t)=P(t+1)</math> is a polynomial in Template:Mvar with the same first coefficient Template:Math and constant term Template:Math.<ref>Template:Cite journal</ref> The rational root theorem implies thus that a rational root of Template:Mvar must belong to <math display=inline>\{\pm1, \pm\frac 13 \},</math> and thus that the rational roots of Template:Mvar 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 Template:Mvar is positive, and the only remaining candidates are Template:Math and Template:Math.

To show that Template:Math is not a root, it suffices to remark that if <math>x=2,</math> then <math>3x^3</math> and <math>5x-2</math> are multiples of Template:Math, while <math>-5x^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.