The Rational Root Theorem (In Depth)

The Rational Root Theorem (also called the Rational Zero Theorem) narrows the search for rational zeros of a polynomial with integer coefficients. Instead of guessing randomly, the theorem gives you a finite list of candidates to check. Combined with synthetic division, it is the primary hand-factoring tool for polynomials of degree 3 and higher.

Statement of the Theorem

If $f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ is a polynomial with integer coefficients and $\frac{p}{q}$ is a rational zero of $f$ (in lowest terms), then:

• $p$ is a factor of the constant term $a_0$
• $q$ is a factor of the leading coefficient $a_n$

In shorthand: every rational zero has the form $\frac{p}{q}$ where $p \mid a_0$ and $q \mid a_n$.

Proof (Derivation)

Suppose $\frac{p}{q}$ is a zero with $\gcd(p, q) = 1$. Then:

$a_n\!\left(\frac{p}{q}\right)^n + a_{n-1}\!\left(\frac{p}{q}\right)^{n-1} + \cdots + a_1\!\left(\frac{p}{q}\right) + a_0 = 0$

Multiply through by $q^n$:

$a_n p^n + a_{n-1}p^{n-1}q + \cdots + a_1 p q^{n-1} + a_0 q^n = 0$

To show $p \mid a_0$: Rearrange so all terms except the last have factor $p$:

$p(a_n p^{n-1} + a_{n-1}p^{n-2}q + \cdots + a_1 q^{n-1}) = -a_0 q^n$

So $p$ divides $a_0 q^n$. Since $\gcd(p, q) = 1$, we have $\gcd(p, q^n) = 1$, so $p \mid a_0$.

To show $q \mid a_n$: Rearrange so all terms except the first have factor $q$:

$a_n p^n = -q(a_{n-1}p^{n-1} + \cdots + a_0 q^{n-1})$

So $q$ divides $a_n p^n$. Since $\gcd(p, q) = 1$, we have $q \mid a_n$.

This elegant argument relies entirely on the divisibility properties of integers and the assumption that $\frac{p}{q}$ is in lowest terms.

Systematic Application

Step-by-Step Process

1. List all factors of the constant term $a_0$ (these are the possible $p$ values)
2. List all factors of the leading coefficient $a_n$ (these are the possible $q$ values)
3. Form all ratios $\pm\frac{p}{q}$ and reduce to eliminate duplicates
4. Test candidates using synthetic division (more efficient than direct substitution)
5. When you find a zero, the synthetic division produces a reduced polynomial of lower degree
6. Repeat on the reduced polynomial until fully factored

Worked Example 1: Leading Coefficient 1

Find all rational zeros of $f(x) = x^3 - 6x^2 + 11x - 6$.

Step 1: $a_0 = -6$, factors: $\pm 1, \pm 2, \pm 3, \pm 6$.

Step 2: $a_n = 1$, factors: $\pm 1$.

Step 3: Candidates: $\pm 1, \pm 2, \pm 3, \pm 6$.

Step 4: Test $x = 1$ by synthetic division:

$\begin{array}{c|cccc} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array}$

Remainder is 0, so $x = 1$ is a zero. The quotient is $x^2 - 5x + 6$.

Step 5: Factor the quotient: $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Complete factorization: $f(x) = (x - 1)(x - 2)(x - 3)$.

Zeros: $x = 1, 2, 3$.

Worked Example 2: Leading Coefficient Greater Than 1

Find all rational zeros of $f(x) = 2x^3 + 3x^2 - 8x + 3$.

Step 1: $a_0 = 3$, factors of $|a_0|$: $1, 3$.

Step 2: $a_n = 2$, factors of $|a_n|$: $1, 2$.

Step 3: Candidates $\frac{p}{q}$: $\pm\frac{1}{1}, \pm\frac{3}{1}, \pm\frac{1}{2}, \pm\frac{3}{2}$, i.e., $\pm 1, \pm 3, \pm\frac{1}{2}, \pm\frac{3}{2}$.

Step 4: Test $x = 1$:

$\begin{array}{c|cccc} 1 & 2 & 3 & -8 & 3 \\ & & 2 & 5 & -3 \\ \hline & 2 & 5 & -3 & 0 \end{array}$

$x = 1$ is a zero. Quotient: $2x^2 + 5x - 3$.

Step 5: Factor: $2x^2 + 5x - 3 = (2x - 1)(x + 3)$.

Zeros of quotient: $x = \frac{1}{2}$ and $x = -3$.

Complete factorization: $f(x) = (x - 1)(2x - 1)(x + 3)$.

Zeros: $x = 1, \frac{1}{2}, -3$.

Worked Example 3: No Rational Zeros

Show that $f(x) = x^3 + x + 1$ has no rational zeros.

Candidates: $\pm 1$.

$f(1) = 1 + 1 + 1 = 3 \neq 0$

$f(-1) = -1 - 1 + 1 = -1 \neq 0$

Neither candidate is a zero. By the Rational Root Theorem, if the polynomial has a rational zero, it must be $\pm 1$. Since neither works, $f$ has no rational zeros. Its real zeros are irrational (there is one real root, approximately $-0.6824$, found by numerical methods).

Synthetic Division Review

Synthetic division is a streamlined method for dividing a polynomial by $(x - c)$. For $f(x) = a_n x^n + \cdots + a_1 x + a_0$ divided by $(x - c)$:

1. Write the coefficients $a_n, a_{n-1}, \ldots, a_0$ in a row
2. Bring down $a_n$
3. Multiply by $c$ and add to the next coefficient
4. Repeat until the last coefficient — that final number is the remainder

If the remainder is 0, then $c$ is a zero and the other numbers give the quotient.

Worked Example 4: Reducing a Degree-4 Polynomial

Factor $f(x) = x^4 - 5x^3 + 5x^2 + 5x - 6$ completely.

Candidates: $\pm 1, \pm 2, \pm 3, \pm 6$.

Test $x = 1$:

$\begin{array}{c|ccccc} 1 & 1 & -5 & 5 & 5 & -6 \\ & & 1 & -4 & 1 & 6 \\ \hline & 1 & -4 & 1 & 6 & 0 \end{array}$

Quotient: $x^3 - 4x^2 + x + 6$. Test $x = 2$ on this:

$\begin{array}{c|cccc} 2 & 1 & -4 & 1 & 6 \\ & & 2 & -4 & -6 \\ \hline & 1 & -2 & -3 & 0 \end{array}$

Quotient: $x^2 - 2x - 3 = (x - 3)(x + 1)$.

Complete factorization: $f(x) = (x - 1)(x - 2)(x - 3)(x + 1)$.

Zeros: $x = -1, 1, 2, 3$.

Complete Factorization Strategy

For a polynomial of degree $n$ with integer coefficients:

1. Check for common factors first (factor out the GCF)
2. Apply the Rational Root Theorem to list candidates
3. Test candidates with synthetic division — start with $\pm 1$ (the easiest to test mentally)
4. Reduce to a lower-degree polynomial each time you find a zero
5. When you reach degree 2, use the quadratic formula to find remaining zeros
6. If no rational zeros exist, the polynomial has only irrational or complex zeros

Worked Example 5: Complete Strategy

Factor $f(x) = 6x^3 + 13x^2 + x - 2$.

$a_0 = -2$: factors $\pm 1, \pm 2$.

$a_n = 6$: factors $\pm 1, \pm 2, \pm 3, \pm 6$.

Candidates: $\pm 1, \pm 2, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{1}{6}$.

Test $x = \frac{1}{3}$:

$\begin{array}{c|cccc} \frac{1}{3} & 6 & 13 & 1 & -2 \\ & & 2 & 5 & 2 \\ \hline & 6 & 15 & 6 & 0 \end{array}$

Quotient: $6x^2 + 15x + 6 = 3(2x^2 + 5x + 2) = 3(2x + 1)(x + 2)$.

Complete factorization: $f(x) = \left(x - \frac{1}{3}\right) \cdot 3(2x + 1)(x + 2) = (3x - 1)(2x + 1)(x + 2)$.

Zeros: $x = \frac{1}{3}, -\frac{1}{2}, -2$.

Real-World Application: Resonant Frequencies

In engineering, the natural frequencies of a vibrating system are found by solving a characteristic polynomial. For a three-mass spring system, the characteristic equation might be:

$\lambda^3 - 6\lambda^2 + 10\lambda - 4 = 0$

Candidates: $\pm 1, \pm 2, \pm 4$.

Testing $\lambda = 2$: $8 - 24 + 20 - 4 = 0$. So $\lambda = 2$ is a root.

Dividing: $\lambda^2 - 4\lambda + 2 = 0 \implies \lambda = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2}$.

The natural frequencies are $\omega = \sqrt{\lambda}$ for each eigenvalue: $\sqrt{2 - \sqrt{2}}$, $\sqrt{2}$, and $\sqrt{2 + \sqrt{2}}$ radians per second.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The Rational Root Theorem says: if $\frac{p}{q}$ is a rational zero, then $p$ divides the constant term and $q$ divides the leading coefficient
• When the leading coefficient is 1, the only candidates are integer factors of the constant term
• Synthetic division efficiently tests candidates and reduces the polynomial’s degree
• Not every polynomial has rational zeros — the theorem only gives candidates, not guarantees
• The complete factoring strategy: list candidates, test with synthetic division, reduce, repeat
• Once you reach degree 2, use the quadratic formula for remaining zeros

Return to College Algebra for more topics in this section.