What if the constant term is zero?

If the constant term is zero, factor out the greatest common power of $x$ from all terms. This will give $x=0$ as a root with some multiplicity. Apply the Rational Root Theorem to the remaining polynomial factor.

Does the Rational Root Theorem find irrational or complex roots?

No. The theorem only generates a list of possible rational roots. Irrational and complex roots must be found by solving the depressed polynomial (usually a quadratic) after factoring out the rational roots.

What if none of the possible rational roots work?

If none of the candidates yield $P(c) = 0$, the polynomial has no rational roots. All of its roots are either irrational or complex numbers, and other numerical or algebraic methods are required.

How to find all zeros of a polynomial using the rational root theorem

The Rational Root Theorem provides a systematic way to identify potential rational zeros of a polynomial function. It applies to polynomials with integer coefficients, allowing you to test a finite list of fractions to find the actual rational roots.

Once a rational root is found, polynomial division reduces the degree of the polynomial. This process is repeated until a quadratic remains, which can be solved using standard methods like factoring or the quadratic formula.

The setup

Identify a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ with integer coefficients. Ensure $a_n eq 0$ and $a_0 eq 0$ . If $a_0 = 0$ , factor out $x$ until the constant term is non-zero. Let $p$ be an integer factor of the constant term $a_0$ , and $q$ be an integer factor of the leading coefficient $a_n$ .

The steps

List all factors $p$ of the constant term $a_0$ .
List all factors $q$ of the leading coefficient $a_n$ .
Form all possible fractions $\pm \frac{p}{q}$ in simplest form. These are the possible rational roots.
Test the candidates using synthetic division or direct substitution (Evaluate $P(\frac{p}{q})$ ). A candidate $c$ is a root if and only if $P(c) = 0$ .
When a root $c$ is found, use the quotient from the synthetic division (a polynomial of degree $n-1$ ).
Repeat the process on the quotient until you reach a quadratic equation.
Solve the remaining quadratic equation by factoring, completing the square, or the quadratic formula.

Checking the result

Verify the roots by substituting them back into the original polynomial $P(x)$ . Alternatively, multiply the corresponding linear factors $(x - r_1)(x - r_2)\dots$ and any remaining non-linear factors, scaled by the leading coefficient, to ensure the product matches the original polynomial.

Common errors

A frequent mistake is forgetting the $\pm$ sign when listing possible roots. Another error is failing to fully simplify fractions, leading to duplicate testing. Finally, arithmetic mistakes during synthetic division are common; double-check the addition and multiplication steps.

Worked example

Find all zeros of the polynomial $P(x) = 2x^3 - 3x^2 - 11x + 6$ .

The constant term is $a_0 = 6$ . Its factors are $p \in \{1, 2, 3, 6\}$ . The leading coefficient is $a_n = 2$ . Its factors are $q \in \{1, 2\}$ . Possible rational roots $\pm \frac{p}{q}$ : $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$ .

Test $x = 1$ : $P(1) = 2(1)^3 - 3(1)^2 - 11(1) + 6 = 2 - 3 - 11 + 6 = -6 eq 0$ .

Test $x = -2$ : $P(-2) = 2(-8) - 3(4) - 11(-2) + 6 = -16 - 12 + 22 + 6 = 0$ . So, $x = -2$ is a root.

Perform synthetic division with $x = -2$ : $\begin{array}{c|rrrr} -2 & 2 & -3 & -11 & 6 \\ & & -4 & 14 & -6 \\ \hline & 2 & -7 & 3 & 0 \end{array}$

The quotient is the quadratic $2x^2 - 7x + 3$ . Set the quotient to zero: $2x^2 - 7x + 3 = 0$ . Factor the quadratic: $2x^2 - 6x - x + 3 = 0$ $2x(x - 3) - 1(x - 3) = 0$ $(2x - 1)(x - 3) = 0$

The remaining roots are $x = \frac{1}{2}$ and $x = 3$ .

All zeros of $P(x)$ are $x = -2$ , $x = \frac{1}{2}$ , and $x = 3$ .

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References: OpenStax College Algebra, Chapter 5: Polynomial and Rational Functions · Khan Academy: Polynomial zeros · Stewart, Redlin, Watson: Precalculus: Mathematics for Calculus, Chapter 3