Rational Zeros Calculator

The Rational Root Theorem

Finding exact solutions (roots or zeros) of higher-degree polynomial equations is a cornerstone problem in classical algebra. The Rational Root Theorem (also known as the Rational Zero Theorem) provides a systematic method for identifying all possible rational numbers that could be zeros of a polynomial with integer coefficients.

Consider a polynomial of degree $n \ge 1$ with integer coefficients:

$$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$$

Where $a_n \ne 0$ and $a_0 \ne 0$. If $\frac{p}{q}$ is a rational zero of $P(x)$ expressed in lowest terms ($\gcd(p, q) = 1$), then:

• $p$ must be an integer factor of the constant term $a_0$.
• $q$ must be an integer factor of the leading coefficient $a_n$.

Consequently, every potential rational root must take the form:

$$\text{Possible Rational Zeros} = \pm \frac{\text{Factors of } |a_0|}{\text{Factors of } |a_n|}$$

Step-by-Step Procedure to Find Rational Zeros

1. List Factors of $a_0$ ($p$): Find all positive and negative integer divisors of the constant term.
2. List Factors of $a_n$ ($q$): Find all positive integer divisors of the leading coefficient.
3. Form Candidates ($\pm p/q$): Divide each factor $p$ by each factor $q$. Simplify fractions and discard duplicates to form the candidate pool.
4. Test Each Candidate: Evaluate $P(p/q)$ directly or apply synthetic division. If $P(p/q) = 0$, then $p/q$ is an actual rational zero.
5. Factor and Depress the Polynomial: When a zero $r$ is confirmed, divide $P(x)$ by $(x - r)$ to obtain a lower-degree quotient polynomial, repeating the process until fully factored.

Worked Example

Find all rational zeros of the cubic polynomial:

$$P(x) = 2x^3 + x^2 - 13x + 6$$

• The constant term is $a_0 = 6$. Its factors are $p \in \{1, 2, 3, 6\}$.
• The leading coefficient is $a_3 = 2$. Its factors are $q \in \{1, 2\}$.
• The candidate rational roots $\pm p/q$ are: $$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 6$$
• Testing candidates:
  • $P(1) = 2(1)^3 + (1)^2 - 13(1) + 6 = -4 \ne 0$
  • $P(2) = 2(8) + 4 - 13(2) + 6 = 16 + 4 - 26 + 6 = 0$ (Root confirmed!)
  • $P(-3) = 2(-27) + 9 - 13(-3) + 6 = -54 + 9 + 39 + 6 = 0$ (Root confirmed!)
  • $P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) + \frac{1}{4} - 13\left(\frac{1}{2}\right) + 6 = \frac{1}{4} + \frac{1}{4} - \frac{13}{2} + 6 = 0$ (Root confirmed!)

The complete set of rational zeros is $x = -3, \frac{1}{2}, 2$, yielding the complete factorization $P(x) = (x - 2)(x + 3)(2x - 1)$.

What About Non-Rational Roots?

The Rational Root Theorem only tests for rational numbers (fractions and integers). If a polynomial has real irrational roots (such as $\sqrt{2}$) or complex conjugate roots (such as $1 \pm 2i$), they will not appear in the candidate list. However, once all rational zeros are extracted via synthetic division, the quadratic formula or numerical methods can readily solve the remaining depressed polynomial.

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