Cubic equation

The general cubic equation

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

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.

Elementary proof

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

Suppose P(p/q) = 0 for some coprime p, q ∈ ℤ:

${\displaystyle P\left({\tfrac {p}{q}}\right)=a_{n}\left({\tfrac {p}{q}}\right)^{n}+a_{n-1}\left({\tfrac {p}{q}}\right)^{n-1}+\cdots +a_{1}\left({\tfrac {p}{q}}\right)+a_{0}=0.}$

To clear denominators, multiply both sides by qⁿ:

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

Shifting the a₀ term to the right side and factoring out p on the left side produces:

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

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

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

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

Reasoning as before, it follows that q divides a_n.^[1]

Proof using Gauss's lemma

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

Given a polynomial of degree n:

${\displaystyle P(x)={a_{n}x^{n}}+{a_{n-1}x^{n-1}}+\cdots +{a_{1}x}+{a_{0}}}$, ${\displaystyle {a_{i}}\in \mathbb {Z} }$

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

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

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

First

In the polynomial

${\displaystyle 2x^{3}+x-1,}$

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

In the polynomial

${\displaystyle x^{3}-7x+6}$

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

${\displaystyle 3x^{3}-5x^{2}+5x-2}$

must be among the numbers

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

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) ≠ 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 t³ remains the same as the coefficient of x³. Applying the rational root theorem thus yields the possible roots ${\displaystyle t=\pm {\tfrac {1}{1,3}}}$, so that

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

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 ≥ 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.