# Intro to simplifying rational expressions

CCSS Math: HSA.APR.D.6
Learn what it means to simplify a rational expression, and how it's done!

#### What you should be familiar with before taking this lesson

A rational expression is a ratio of two polynomials. The domain of a rational expression is all real numbers except those that make the denominator equal to zero.
For example, the domain of the rational expression $\dfrac{x+2}{x+1}$ is all real numbers except $\textit{-1}$, or $x\neq -1$.
If this is new to you, we recommend that you check out our intro to rational expressions.
You should also know how to factor polynomials for this lesson.

#### What you will learn in this lesson

In this article, we will learn how to simplify rational expressions by looking at several examples.

## Introduction

A rational expression is considered simplified if the numerator and denominator have no factors in common.
We can simplify rational expressions in much the same way as we simplify numerical fractions.
For example, the simplified version of $\dfrac 68$ is $\dfrac{3}{4}$. Notice how we canceled a common factor of $2$ from the numerator and the denominator:
\begin{aligned} \dfrac68&= \dfrac{2\cdot 3}{2\cdot 4}&&\small{\gray{\text{Factor}}} \\\\ &= \dfrac{\tealD{\cancel{2}}\cdot 3}{\tealD{\cancel{2}}\cdot 4}&&\small{\gray{\text{Cancel common factors}}} \\ \\ &= \dfrac{3}{4} &&\small{\gray{\text{Simplify}}} \end{aligned}

## Example 1: Simplifying $\dfrac{x^2+3x}{x^2+5x}$

Step 1: Factor the numerator and denominator
The only way to see if the numerator and denominator share common factors is to factor them!
$\dfrac{x^2+3x}{x^2+5x}=\dfrac{ x(x+3)}{ x(x+5)}$
Step 2: List restricted values
At this point, it is helpful to notice any restrictions on $x$. These will carry over to the simplified expression.
Since division by $0$ is undefined, here we see that $\blueD{x\neq0}$ and $\purpleC{x\neq -5}$.
$\dfrac{ x(x+3)}{ \blueD x\purpleC{(x+5)}}$
Step 3: Cancel common factors
Now notice that the numerator and denominator share a common factor of $x$. This can be canceled out.
\begin{aligned}\dfrac{\tealD x(x+3)}{\tealD x(x+5)}&=\dfrac{\tealD {\cancel {x}}(x+3)}{\tealD{\cancel x}(x+5)}\\ \\ &=\dfrac{x+3}{x+5} \end{aligned}
Recall that the original expression is defined for $x\neq 0,-5$. The simplified expression must have the same restrictions.
Because of this, we must note that $x\neq 0$. We do not need to note that $x\neq -5$, since this is understood from the expression.
Original expression:
$\dfrac{x^2+3x}{x^2+5x}=\dfrac{ x(x+3)}{ \blueD x\maroonD{(x+5)}}$
Simplified expression:
$\dfrac{x+3}{\maroonD{x+5}}$
Notice how $\maroonC{x=-5}$ makes both denominators $0$, and so it is not a valid input. However, this is implied by each expression.
In contrast, if we didn't indicate $x\neq0$ for the simplified expression, then $\blueD{x=0}$ would be a valid input, since $\dfrac{\blueD0+3}{\blueD0+5}=\dfrac{3}{5}$.
Because the expression evaluated at $\blueD{x=0}$ should be undefined (see original expression), we must restrict it from the domain.
In conclusion, the simplified form is written as follows:
$\dfrac{x+3}{x+5}$ for $x\neq 0$

### A note about equivalent expressions

Original expression$\quad$Simplified expression
$\dfrac{x^2+3x}{x^2+5x}$$\quad$$\dfrac{x+3}{x+5}$ for $x\neq 0$

The two expressions above are equivalent. This means their output values are the same for all possible $x$-values. The table below illustrates this for $x=2$.
Original expression$\quad$Simplified expression
Evaluation at $\purpleC{x=2}$\begin{aligned}\dfrac{(\purpleC{2})^2+3(\purpleC{2})}{(\purpleC{2})^2+5(\purpleC{2})}&=\dfrac{10}{14}\\\\&=\dfrac{\purpleC{{2}}\cdot 5}{\purpleC{{2}}\cdot 7}\\\\&=\dfrac{\purpleC{\cancel{2}}\cdot 5}{\purpleC{\cancel{2}}\cdot 7}\\\\&=\dfrac{5}{7}\end{aligned}\begin{aligned}\dfrac{\purpleC{2}+3}{\purpleC{2}+5}&=\dfrac{5}{7}\\\\&\phantom{=\dfrac57}\\\\&\phantom{=\dfrac57}\\\\&\phantom{=\dfrac57}\end{aligned}
NoteThe result is simplified by canceling a common factor of $\purpleC 2$.The result is already simplified because the factor of $x$ $($in this case $\purpleC{x=2})$, was already canceled during the simplification process.

For this reason, the two expressions have the same value for the same input. However, values that make the original expression undefined often break this rule. Notice how this is the case with $\purpleC{x=0}$.
Original expression$\quad$Simplified expression (without restriction)
Evaluation at $\purpleC{x=0}$\begin{aligned}\dfrac{(\purpleC{0})^2+3(\purpleC{0})}{(\purpleC{0})^2+5(\purpleC{0})}&=\dfrac{0}{0}\\\\&=\text{undefined}\end{aligned}\begin{aligned}\dfrac{\purpleC{0}+3}{\purpleC{0}+5}&=\dfrac{3}{5}\\\\\\\\&\phantom{\text{undefined}}\end{aligned}

Because the two expressions must be equivalent for all possible inputs, we must require $x\neq 0$ for the simplified expression.

Note that we cannot cancel the $x$'s in the expression below. This is because these are terms rather than factors of the polynomials!
$\dfrac{x+3}{x+5}~~\Large{\goldD{\neq}}$ $~\dfrac{3}{5}$
This becomes clear when looking at a numerical example. For example, suppose $\purpleC{x=2}$.
$\dfrac{\purpleC2+3}{\purpleC2+5}~~\Large{\goldD{\neq}}$$~\dfrac{3}{5}$
As a rule, we can only cancel if the numerator and denominator are in factored form!

### Summary of the simplification process

• Step 1: Factor the numerator and the denominator.
• Step 2: List restricted values.
• Step 3: Cancel common factors.
• Step 4: Simplify and note any restricted values not implied by the expression.

1) Simplify $\dfrac{6x+20}{2x+10}$.
Step 1: Factor the numerator and denominator
$\dfrac{6x+20}{2x+10}=\dfrac{2(3x+10)}{2(x+5)}$
Step 2: List restricted values
Here we see that $\blueD{x\neq{-5}}$.
$\dfrac{2(3x+10)}{2\blueD{(x+5)}}$
Step 3: Cancel common factors
\begin{aligned}\dfrac{\tealD2(3x+10)}{\tealD{{2}}(x+5)}&=\dfrac{\tealD{\cancel{2}}(3x+10)}{\tealD{\cancel{2}}(x+5)}\\ \\ &=\dfrac{3x+10}{x+5} \end{aligned}
We write the simplified form as follows:
$\dfrac{3x+10}{x+5}$
The original expression requires $x\neq-5$. The simplified expression also requires $x\neq-5$. We don't need to restrict any additional values.
2) Simplify $\dfrac{x^3-3x^2}{4x^2-5x}$.
for $x\neq$
Step 1: Factor the numerator and denominator
$\dfrac{x^3-3x^2}{4x^2-5x}=\dfrac{x^2(x-3)}{x(4x-5)}$
Step 2: List restricted values
Here we see that $\blueD{x\neq0}$ and $\purpleC{x\neq \dfrac54}$.
$\dfrac{x^2(x-3)}{\blueD x\purpleC{(4x-5)}}$
Step 3: Cancel common factors
\begin{aligned}\dfrac{x^2(x-3)}{x(4x-5)}&=\dfrac{\tealD x\cdot x(x-3)}{\tealD x(4x-5)}\\ \\ &=\dfrac{\tealD{\cancel{ x}}\cdot x(x-3)}{\tealD{\cancel{ x}}(4x-5)}\\ \\ &=\dfrac{x(x-3)}{4x-5} \end{aligned}
We write the simplified form as follows:
$\dfrac{x(x-3)}{4x-5}$ for $x\neq 0$
The original expression requires $x\neq 0,\dfrac54$. We do not need to note that $x\neq \dfrac54$, since this is understood from the expression.

## Example 2: Simplifying $\dfrac{x^2-9}{x^2+5x+6}$

Step 1: Factor the numerator and denominator
$\dfrac{x^2-9}{x^2+5x+6}=\dfrac{(x-3)(x+3)}{({x+2})({x+3})}$
To factor the numerator, we can use the difference of squares pattern:
$\blueD{ a}^2-\greenD {b}^2=(\blueD {a}+\greenD b)(\blueD {a}-\greenD{ b})$
So we have the following:
\begin{aligned}x^2-9&=\blueD{x}^2-\greenD{3}^2\\ \\ &=(\blueD{x}+\greenD3)(\blueD{x}-\greenD3) \end{aligned}
We can factor the denominator using the sum-product pattern.
$(x+a)(x+b)=\ x^2 +\goldD{(a+b)}\ x + \purpleC{ab}$
To factor $x^2+5x+6$, we can find factors of $6$ that add up to $5$. Since $2\cdot 3=6$ and $2+3=5$, the polynomial factors to $(x+2)(x+3)$.
Step 2: List restricted values
Since division by $0$ is undefined, here we see that $\blueD{x\neq-2}$ and $\purpleC{x\neq -3}$.
$\dfrac{(x-3)(x+3)}{\blueD{(x+2)}\purpleC{(x+3)}}$
Step 3: Cancel common factors
Notice that the numerator and denominator share a common factor of $\tealD{x+3}$. This can be canceled out.
\begin{aligned}\dfrac{(x-3)\tealD{(x+3)}}{(x+2)\tealD{(x+3)}}&=\dfrac{(x-3)\tealD{\cancel{(x+3)}}}{(x+2)\tealD{\cancel{(x+3)}}}\\ \\ &=\dfrac{x-3}{x+2} \end{aligned}
We write the simplified form as follows:
$\dfrac{x-3}{x+2}$ for $x\neq -3$
The original expression requires $x\neq-2,-3$. We do not need to note that $x\neq -2$, since this is understood from the expression.

### Check for understanding

3) Simplify $\dfrac{x^2-3x+2}{x^2-1}$.
Step 1: Factor the numerator and denominator
$\dfrac{x^2-3x+2}{x^2-1}=\dfrac{(x-2)(x-1)}{(x+1)(x-1)}$
Step 2: List restricted values
Here we see that $\blueD{x\neq-1}$ and $\purpleC{x\neq1}$.
$\dfrac{(x-2)(x-1)}{\blueD{(x+1)}\purpleC{(x-1)}}$
Step 3: Cancel common factors
\begin{aligned}\dfrac{(x-2)\tealD{(x-1)}}{(x+1)\tealD{(x-1)}}&=\dfrac{(x-2)\tealD{\cancel{(x-1)}}}{(x+1)\tealD{\cancel{(x-1)}}}\\ \\ &=\dfrac{x-2}{x+1} \end{aligned}
We write the simplified form as follows:
$\dfrac{x-2}{x+1}$ for $x\neq 1$
The original expression requires $x\neq\pm1$. We do not need to note that $x\neq -1$, since this is understood from the expression.
4) Simplify $\dfrac{x^2-2x-15}{x^2+x-6}$.
for $x\neq$
Step 1: Factor the numerator and denominator
$\dfrac{x^2-2x-15}{x^2+x-6}=\dfrac{(x-5)(x+3)}{(x+3)(x-2)}$
Step 2: List restricted values
Here we see that $\blueD{x\neq-3}$ and $\purpleC{x\neq 2}$.
$\dfrac{(x-5)(x+3)}{\blueD{(x+3)}\purpleC{(x-2)}}$
Step 3: Cancel common factors
\begin{aligned}\dfrac{(x-5)\tealD{(x+3)}}{\tealD{(x+3)}(x-2)}&=\dfrac{(x-5)\tealD{\cancel{(x+3)}}}{\tealD{\cancel{(x+3)}}(x-2)}\\ \\ &=\dfrac{x-5}{x-2} \end{aligned}
$\dfrac{x-5}{x-2}$ for $x\neq -3$
The original expression requires $x\neq-3,2$. We do not need to note that $x\neq 2$, since this is understood from the expression.