# Rational Root Theorem Overview & Examples

## What is the Rational Root Theorem?

The **Rational Root Theorem** is used in math to find the possible **rational roots** of a polynomial function, most specifically when the function is not factorable. These rational roots can also be called: x-intercepts, zeros or solutions. To solve for them set the polynomial equal to 0, as y=0 on our x-axis. In order to use the Rational Root Theorem, it is important to understand the basics of a polynomial function. A **polynomial function ** is defined as a function involving only positive integer powers of x with terms combined with addition and subtraction. A polynomial is in **standard form** if the terms are in order from highest exponent to lowest. For example, {eq}3x^3 + 2 - 4x^5 {/eq} in standard form is {eq}-4x^5 + 3x^3 +2 {/eq}. The leading coefficient in a polynomial is the first term *only if the polynomial is in standard form*, if not, then look to the term with the highest exponent. The constant term of a polynomial is the only term that does not have a variable. In the example before, -4 is the leading coefficient and 2 is the constant. Lastly, a **rational number **is a number that can be expressed as the quotient or fraction of two integers. This means that numbers such as {eq}\pi {/eq} are not rational because they are repeating decimals.

The Rational Root Theorem states that for a polynomial equation with integer coefficients to have a rational solution, that solution must be represented by the a factor of the constant (denoted by the variable p) divided by a factor of the leading coefficient (denoted by the variable q). This means, the rational root theorem results in finding all *possible* solutions from a polynomial by listing all of the factors of both the leading coefficient and constant and representing them as a fraction. This is represented by listing the possible roots as {eq}\frac{+}{-} \frac{p}{q} {/eq}

### Understanding the Rational Root Theorem

The Rational Root Theorem (RRT) is used to help in a few different situations

- Solving a polynomial with more than 4 terms
- Solving a polynomial with a degree 3 or higher that is not factorable
- Finding out where to start on a problem

Note that the RRT only provides the **possible solutions** to a polynomial. There are still more steps needed to be able to confirm a root.

## Rational Root Theorem Examples

Here are a few examples to show how the Rational Root Theorem is used.

### Example 1: Finding Rational Roots

Using the polynomial {eq}f(x) = x^3 + x^2 + x - 3 {/eq} answer the following questions.

1) Find the possible rational roots of the polynomial.

First identify that our constant (p) is -3 and our leading coefficient (q) is 1. The factors of 1 are just {eq}\frac{+}{-} 1 {/eq} and the factors of 3 are {eq}\frac{+}{-} 1, 3 {/eq}. This means all of the possible roots of the rational function {eq}f(x) = x^3 + x^2 + x - 3 {/eq} are {eq}\frac{+}{-} 1 and \frac{+}{-} 2 {/eq}.

2) Find the exact roots of the polynomial provided.

In order for a possible root to be a solution, when substituting the value into the function, the output should be zero.

{eq}f(1) = (1)^3 + (1)^2 + (1) - 3 = 0 ; f (-1) = (-1)^3 + (-1)^2 + (-1) - 3 = -4 ; f(3) = (3)^3 + (3)^2 + (3) - 3 = 36 ; f(-3) = (-3)^3 + (-3)^2 + (-3) - 3 = -24 ; {/eq} Since f(1) is the only input that results in 0, the only rational zero of our polynomial is x = 1. Since f(-1), f(3) and f(-3) have output values other than 0, these are other points on our graph *not on the x-axis*, and therefore not solutions.

This can be seen on the graph of the polynomial function used, where it only crosses the x-axis when x = 1.

### Example 2: Using Rational Root Theorem

Do the exact same check for a polynomial with a higher degree and more terms: {eq}h(x) = 3x^4 - 10x^3 -24x^2 - 6x + 5 {/eq}

Here the leading coefficient is 3 and the constant is 5. This results in {eq}\frac{+}{-} 1, 5, \frac{1}{3}, \frac{5}{3} {/eq} as our possible roots.

Going through an evaluating {eq}h(x) {/eq} for all of our roots a remainder is found for every value except {eq}x = -1, \frac{1}{3}, 5 {/eq}. Therefore the function has 3 rational solutions at {eq}(-1,0) ( \frac{1}{3}, 0) (5,0) {/eq}. All other x-values are other points on our graph with y-values (outputs) other than 0.

This can be see on the graph of this polynomial, where the function crosses the x-axis three times at {eq}(-1,0) ; (\frac{1}{3},0) ; (5,0) {/eq}

## Lesson Summary

The **Rational Root Theorem** is a useful tool to help solve polynomials, whether or not they are factorable. In order to find the possible solutions to a polynomial, take the factors of the constant and divide them by the factors of the leading coefficient. This provides a list of the positive and negative values that could possibly work as solutions to the polynomial. Then, substitute those values into the polynomial to eliminate the values that do not work. If the output is zero when a possible root is substituted in, that confirms the **rational root** is a solution to the original polynomial. If when a possible root is substituted into the polynomial, any number other than zero occurs, that root is not an official solution.

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#### What is the importance of the rational root theorem?

It is used to solve any polynomial with rational roots, whether or not that polynomial is factorable. It can save a lot of time when struggling to know where to start.

#### What does the rational root theorem say?

All possible rational roots of a polynomial can be found by dividing the positive and negative values of your constant by the leading coefficient.

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