# Intro to rational expressions

CCSS Math: HSF.IF.B.5
Learn what rational expressions are and about the values for which they are undefined.

#### What you will learn in this lesson

This lesson will introduce you to rational expressions. You will learn how to determine when a rational expression is undefined and how to find its domain.

## What is a rational expression?

A polynomial is an expression that consists of a sum of terms containing integer powers of $x$, like $3x^2-6x-1$.
A rational expression is simply a quotient of two polynomials. Or in other words, it is a fraction whose numerator and denominator are polynomials.
These are examples of rational expressions:
$\dfrac{1}{x}$, $\quad\dfrac{x+5}{x^2-4x+4}$, $\quad\dfrac{x(x+1)(2x-3)}{x-6}$
Notice that the numerator can be a constant and that the polynomials can be of varying degrees and in multiple forms.

## Rational expressions and undefined values

Consider the rational expression $\dfrac{2x+3}{x-2}$.
We can determine the value of this expression for particular $x$-values. For example, let's evaluate the expression at $\blueD{x}=\blueD1$.
\begin{aligned}\dfrac{2(\blueD{1})+3}{\blueD1-2} &= \dfrac{~~5}{-1}\\ \\ &= \goldD{-5} \\ \end{aligned}
From this, we see that the value of the expression at $\blueD{x}=\blueD{1}$ is $\goldD{-5}$.
Now let's find the value of the expression at $\blueD{x}=\blueD{2}$.
\begin{aligned}\dfrac{2(\blueD{2})+3}{\blueD2-2} &= \dfrac{7}{0}\\ \\ &=\goldD{\text{undefined!}} \\ \end{aligned}
An input of $2$ makes the denominator $0$. Since division by $0$ is undefined, $\blueD x=\blueD 2$ is not a possible input for this expression!

## Domain of rational expressions

The domain of any expression is the set of all possible input values.
In the case of rational expressions, we can input any value except for those that make the denominator equal to $0$ (since division by $0$ is undefined).
In other words, the domain of a rational expression includes all real numbers except for those that make its denominator zero.

### Example: Finding the domain of $\dfrac{x+1}{(x-3)(x+4)}$

Let's find the zeros of the denominator and then restrict these values:
\begin{aligned} &(x-3)(x+4)= 0 \\\\ &x-3=0 \quad \text{or} \quad x+4=0 &&\small{\gray{\text{Zero product property}}}\\\\ &x = 3 \quad\text{or} \quad x=-4 &&\small{\gray{\text{Solve for x}}}\end{aligned}
So we write that the domain is all real numbers except $\textit 3$ and $\textit{-4}$, or simply $x\neq 3, -4$.

## Check your understanding

1) What is the domain of $\dfrac{x+1}{x-7}$?
Choose 1 answer:
Choose 1 answer:
The domain of a rational expression is all real numbers except for those that make the denominator equal to zero.
Let's find the zeros of the denominator:
\begin{aligned} &x-7= 0 \\\\ &x = 7 &&\small{\gray{\text{Solve for x}}}\end{aligned}
Since $7$ makes the denominator $0$, we must exclude this from the domain.
The domain is all real numbers except $7$, or simply $x\neq7$
2) What is the domain of $\dfrac{3x-7}{2x+1}$?
Choose 1 answer:
Choose 1 answer:
The domain of a rational expression is all real numbers except for those that make the denominator equal to zero.
Let's find the zeros of the denominator:
\begin{aligned} 2x+1&=0\\\\ 2x& =-1&&\small{\gray{\text{Subtract 1 from both sides}}}\\\\ x&=- \dfrac{1}{2} &&\small{\gray{\text{Divide both sides by 2}}}\end{aligned}
Since $-\dfrac12$ makes the denominator $0$, we must exclude this from the domain.
The domain is all real numbers except $-\dfrac12$, or simply $x\neq-\dfrac12$.
3) What is the domain of $\dfrac{2x-3}{x(x+1)}$?
Choose 1 answer:
Choose 1 answer:
The domain of a rational expression is all real numbers except for those that make the denominator equal to zero.
Let's find the zeros of the denominator:
\begin{aligned} &x(x+1)= 0 \\\\ &x=0 \quad \text{or} \quad x+1=0 &&\small{\gray{\text{Zero product property}}}\\\\ &x = 0 \quad\text{or} \quad x=-1 &&\small{\gray{\text{Solve for x}}}\end{aligned}
Since the zeros of the denominator are $0$ and $-1$, the domain will exclude these values.
The domain is all real numbers except $0$ and $-1$. In other words, $x\neq0, -1$.

## Challenge problems

4*) What is the domain of $\dfrac{x-3}{x^2-2x-8}$?
Choose 1 answer:
Choose 1 answer:
The domain of a rational expression is all real numbers except for those that make the denominator equal to zero.
In this case, the denominator is not in factored form. So, we must either factor the denominator or use another method to find its zeros.
Notice that $x^2-2x-8=0$ can be easily factored. So we have:
\begin{aligned} &x^2-2x-8= 0 \\\\ &(x-4)(x+2)=0\\\\ &x-4=0 \quad \text{or} \quad x+2=0 &&\small{\gray{\text{Zero product property}}}\\\\ &x = 4 \quad\text{or} \quad x=-2 &&\small{\gray{\text{Solve for x}}}\end{aligned}
Since the zeros of the denominator are $-2$ and $4$, the domain will exclude these values.
The domain is all real numbers except $-2$ and $4$. In other words, $x\neq-2, 4$.
5*) What is the domain of $\dfrac{x+2}{x^2+4}$?
Choose 1 answer:
Choose 1 answer:
The domain of a rational expression is all real numbers except for those that make the denominator equal to zero.
However, in this case the denominator will never be equal to zero. This is because $x^2+4>0$ for all real numbers $x$.
Another way to think about this is to actually try to set the denominator equal to zero and solve for $x$.
\begin{aligned} x^2+4&= 0 \\\\ x^2&=-4 &&\small{\gray{\text{A real number squared is always positive!}}}\\\\ \end{aligned}
But here we see that this is impossible because the square of a real number can never be negative!
Because the denominator cannot be $0$, there are no restrictions on the domain. So, the domain is the set of all real numbers.