The rational root theorem: proof, candidates, and worked cubic equations

What is a rational root?

A rational root is a solution that can be expressed as a ratio of integers. For a polynomial with integer coefficients, the rational root theorem reduces the search to a finite set of candidates.

Suppose

$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$

where the coefficients are integers and $a_n\ne0$. If a nonzero rational root is written in lowest terms as $x=q/p$, with $p>0$, then

$$\boxed{q\mid a_0,\qquad p\mid a_n.}$$

Thus its numerator divides the constant term and its denominator divides the leading coefficient. These are necessary conditions, not a guarantee that a candidate is a root.

If $a_0=0$, first remove the obvious factor $x$ and apply the theorem to the remaining polynomial. If the leading coefficient is $1$ or $-1$, every rational root is an integer. An integer root must divide the constant term.

Example of a candidate list

For $3x^3+2x^2+x+4=0$, the possible rational roots are

$$\pm1,\ \pm2,\ \pm4,\ \pm\frac13,\ \pm\frac23,\ \pm\frac43.$$

Checking this list decides whether rational roots exist; it does not exclude irrational or nonreal roots.

Proof of the theorem

Let $q/p$ be a root in lowest terms. Multiplying $P(q/p)=0$ by $p^n$ gives

$$a_nq^n+a_{n-1}q^{n-1}p+\cdots+a_1qp^{n-1}+a_0p^n=0.$$

All terms except the last are divisible by $q$, so $q$ divides $a_0p^n$. Since $q$ and $p^n$ are coprime, $q\mid a_0$.

Similarly, every term except the first is divisible by $p$, so $p$ divides $a_nq^n$. Since $p$ and $q^n$ are coprime, $p\mid a_n$. This proves both claims.

The coprimality assumption is essential. An unreduced fraction could have numerator and denominator containing arbitrary common factors.

Problem 1: a monic polynomial with an integer parameter

Problem. Let $k$ be an integer. Show that every rational root of

$$x^3+kx^2+7=0$$

is an integer, and find all $k$ for which a positive integer root exists.

The polynomial is monic, so the denominator of a reduced rational root must divide $1$. Hence it is an integer. A positive integer root must divide $7$, leaving only $x=1$ and $x=7$.

• For $x=1$, the equation is $1+k+7=0$, giving $k=-8$.
• For $x=7$, it is $343+49k+7=0$, giving $k=-50/7$, which is not an integer.

Conversely, for $k=-8$, substitution of $x=1$ works. Therefore

$$\boxed{k=-8.}$$

A direct denominator argument

If the root is $q/p$ in lowest terms, multiplication by $p^3$ gives

$$q^3+kq^2p+7p^3=0.$$

Thus $p\mid q^3$, forcing $p=1$. This is the theorem's proof specialized to the given equation.

Problem 2: prove that a cubic has no rational root

Problem. Prove that $3x^3+2x+4=0$ has no rational root.

The candidate list is the same twelve values given earlier. Positive candidates cannot work because every term in $3x^3+2x+4$ is then positive.

For the six negative candidates, direct substitution gives

$$\begin{aligned} P(-1)&=-1,&P(-2)&=-24,&P(-4)&=-196,\\ P(-1/3)&=29/9,&P(-2/3)&=16/9,&P(-4/3)&=-52/9. \end{aligned}$$

None is zero, so no rational root exists. This is a finite exhaustion of all candidates, which completes the proof.

The polynomial still has a real root: for instance $P(-1)<0<P(0)$. The theorem says that this root is irrational, not that there are no roots at all.

Problem 3: solve a cubic with a noninteger rational root

Problem. Solve

$$5x^3+8x^2-14x+4=0.$$

Possible rational roots have a numerator dividing $4$ and a denominator dividing $5$. This suggests checking $2/5$ even if no integer root has been found. Substitution gives zero, and division by $5x-2$ yields

$$5x^3+8x^2-14x+4=(5x-2)(x^2+2x-2).$$

The quadratic factor gives $x=-1\pm\sqrt3$. Therefore the complete solution is

$$\boxed{x=\frac25,\quad-1+\sqrt3,\quad-1-\sqrt3.}$$

Expanding the factorization verifies that no coefficient was lost in division.

A practical workflow

Clear rational coefficient denominators if necessary, remove any factor $x$, list the reduced numerator/denominator candidates, and use signs or simple congruences to shorten the list. Once a root is found, factor it out and solve the remaining polynomial. When proving that no rational root exists, make the exhaustion explicit.

For equations whose coefficients themselves involve irrational numbers, see equating rational and irrational parts. For integer solutions in several variables, see four approaches to Diophantine equations.