01 / 21
\(p\) \(q\)

Rational Zeros
Theorem

Learn how to systematically find every possible rational root of a polynomial, then test them to factor and solve with confidence.

Algebra 2

What Is a Zero of a Polynomial?

A zero (or root) of a polynomial \( f(x) \) is any value \( r \) such that \( f(r) = 0 \). Graphically, zeros are the x-intercepts where the curve crosses the horizontal axis.

x y \( r_1 \) \( r_2 \) \( r_3 \)

Rational vs Irrational Zeros

Rational Zeros
Can be written as a fraction \( \frac{p}{q} \) where \( p, q \) are integers and \( q \neq 0 \). Examples: \( 3, \; -\frac{1}{2}, \; 0 \).
Irrational Zeros
Cannot be expressed as a simple fraction. Examples: \( \sqrt{2}, \; -\sqrt{5}, \; 1 + \sqrt{3} \). These require other methods to find.

Why Does This Matter?

Higher-degree polynomials (degree 3 and above) don't have a simple formula like the quadratic formula. The Rational Zeros Theorem gives us a systematic starting point:

1
Generate a candidate list: All possible rational zeros come from a predictable set.
2
Test efficiently: Use substitution or synthetic division to check each candidate.
3
Reduce the degree: Each confirmed zero factors the polynomial down, making the remaining roots easier to find.

The Rational Zeros Theorem

Possible rational zeros
\( = \pm \dfrac{\text{factors of } a_0}{\text{factors of } a_n} \)

Breaking Down \( \frac{p}{q} \)

Consider \( f(x) = \color{#fbbf24}{2}x^3 + 5x^2 - 11x + \color{#4ade80}{4} \). The leading coefficient is \( a_n = 2 \) and the constant term is \( a_0 = 4 \).

Factors of \( a_0 = 4 \)
\( \pm 1, \; \pm 2, \; \pm 4 \)
Factors of \( a_n = 2 \)
\( \pm 1, \; \pm 2 \)

Possible vs Actual Zeros

The theorem gives you candidates, not answers. Most candidates will not be actual zeros. You must test each one.

Possible Zeros (from theorem) Actual Zeros (confirmed)
Source \( \pm \frac{\text{factors of } a_0}{\text{factors of } a_n} \) Testing via substitution or synthetic division
Count Finite but often large At most \( n \) (degree of polynomial)
Includes Only rational candidates May also have irrational or complex zeros

Step-by-Step Method

1
Identify \( a_0 \) and \( a_n \): The constant term and the leading coefficient.
2
List factors of each: All positive and negative integer divisors of \( a_0 \) and \( a_n \).
3
Form all \( \frac{p}{q} \) combinations: Pair each factor of \( a_0 \) over each factor of \( a_n \). Include \( \pm \).
4
Remove duplicates: Simplify fractions and eliminate repeats.
5
Test candidates: Use synthetic division or direct substitution to find which ones satisfy \( f(r) = 0 \).

Example 1: \( f(x) = x^3 - 6x^2 + 11x - 6 \)

\( a_0 = -6 \)
Factors: \( \pm 1, \pm 2, \pm 3, \pm 6 \)
\( a_n = 1 \)
Factors: \( \pm 1 \)

Example 2: \( g(x) = 2x^3 + 3x^2 - 8x + 3 \)

\( a_0 = 3 \)
Factors: \( \pm 1, \pm 3 \)
\( a_n = 2 \)
Factors: \( \pm 1, \pm 2 \)

Example 3: \( h(x) = 2x^4 + 7x^3 + x^2 - 7x - 3 \)

\( a_0 = -3 \)
Factors: \( \pm 1, \pm 3 \)
\( a_n = 2 \)
Factors: \( \pm 1, \pm 2 \)

Using Synthetic Division to Test

Synthetic division is the fastest way to test a candidate \( r \). If the remainder is 0, then \( r \) is a zero and \( (x - r) \) is a factor.

\(1\)
\(1\)
\(-6\)
\(11\)
\(-6\)
\(1\)
\(-5\)
\(6\)
\(1\)
\(-5\)
\(6\)
\(0\)
 Quotient: \( x^2 - 5x + 6 \) Remainder: \(0\) ✓

Narrowing with Descartes' Rule

Before testing all candidates, Descartes' Rule of Signs tells you the maximum number of positive and negative real zeros. This helps you skip candidates intelligently.

Positive zeros
Count sign changes in \( f(x) \). The number of positive zeros equals that count, or less by an even number.
Negative zeros
Count sign changes in \( f(-x) \). The number of negative zeros equals that count, or less by an even number.

Special Case: \( a_n = 1 \)

When the leading coefficient is 1 (a monic polynomial), the theorem simplifies dramatically:

\( x^3 + 2x^2 - 5x - 6 \)
\( a_n = 1, \; a_0 = -6 \). Candidates: \( \pm 1, \pm 2, \pm 3, \pm 6 \)
\( x^4 - 10x^2 + 9 \)
\( a_n = 1, \; a_0 = 9 \). Candidates: \( \pm 1, \pm 3, \pm 9 \)

Common Mistakes

✕ Swapping \( p \) and \( q \)
Remember: \( p \) divides \( a_0 \) (constant) and \( q \) divides \( a_n \) (leading). Reversing them gives the wrong candidate list.
✕ Forgetting \( \pm \)
Every candidate comes in positive and negative pairs. Omitting negatives means missing half the possible zeros.
✕ Assuming all candidates are zeros
The list is only possible zeros. You must test each one. Most will not make \( f(r) = 0 \).

Summary Table

Step Action Example with \( 2x^3 + 3x^2 - 8x + 3 \)
1. Identify Find \( a_0 \) and \( a_n \) \( a_0 = 3, \; a_n = 2 \)
2. Factor List factors of each \( a_0: \pm1, \pm3 \)   \( a_n: \pm1, \pm2 \)
3. Combine Form \( \pm \frac{p}{q} \) \( \pm1, \pm3, \pm\frac{1}{2}, \pm\frac{3}{2} \)
4. Test Synthetic division \( f(1) = 0 \) ✓
5. Factor Divide and repeat \( (x-1)(2x-1)(x+3) \)

Question 1 of 5

Which list contains all possible rational zeros of \( f(x) = x^3 - 4x^2 + x + 6 \)?

Question 2 of 5

For \( g(x) = 3x^3 + x^2 - 12x - 4 \), which of the following is a possible rational zero according to the theorem?

Question 3 of 5

If \( f(x) = 2x^4 - x^3 + 5x^2 - x - 3 \), how many distinct possible rational zeros does the Rational Zeros Theorem predict?

Question 4 of 5

You test \( x = 2 \) in \( f(x) = x^3 - 3x^2 - 4x + 12 \) using synthetic division and get a remainder of 0. What can you conclude?

Question 5 of 5

Why does the Rational Zeros Theorem require that all coefficients be integers?