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## Precalculus

### Course: Precalculus>Unit 4

Lesson 1: Reducing rational expressions to lowest terms

# Reducing rational expressions to lowest terms

Learn what it means to reduce a rational expression to lowest terms, 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 start fraction, x, plus, 2, divided by, x, plus, 1, end fraction is all real numbers except start text, negative, 1, end text, or x, does not equal, minus, 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 reduce rational expressions to lowest terms by looking at several examples.

## Introduction

A rational expression is reduced to lowest terms if the numerator and denominator have no factors in common.
We can reduce rational expressions to lowest terms in much the same way as we reduce numerical fractions to lowest terms.
For example, start fraction, 6, divided by, 8, end fraction reduced to lowest terms is start fraction, 3, divided by, 4, end fraction. 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} \\\\ &= \dfrac{\tealE{\cancel{2}}\cdot 3}{\tealE{\cancel{2}}\cdot 4} \\\\ &= \dfrac{3}{4} \end{aligned}

## Example 1: Reducing $\dfrac{x^2+3x}{x^2+5x}$start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fraction to lowest terms

Step 1: Factor the numerator and denominator
The only way to see if the numerator and denominator share common factors is to factor them!
start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fraction, equals, start fraction, x, left parenthesis, x, plus, 3, right parenthesis, divided by, x, left parenthesis, x, plus, 5, right parenthesis, end fraction
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 start color #0c7f99, x, does not equal, 0, end color #0c7f99 and start color #7854ab, x, does not equal, minus, 5, end color #7854ab.
start fraction, x, left parenthesis, x, plus, 3, right parenthesis, divided by, start color #0c7f99, x, end color #0c7f99, start color #7854ab, left parenthesis, x, plus, 5, right parenthesis, end color #7854ab, end fraction
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{\tealE x(x+3)}{\tealD x(x+5)}&=\dfrac{\tealE {\cancel {x}}(x+3)}{\tealE{\cancel x}(x+5)} \\\\ &=\dfrac{x+3}{x+5} \end{aligned}
Recall that the original expression is defined for x, does not equal, 0, comma, minus, 5. The reduced expression must have the same restrictions.
Because of this, we must note that x, does not equal, 0. We do not need to note that x, does not equal, minus, 5, since this is understood from the expression.
In conclusion, the reduced form is written as follows:
start fraction, x, plus, 3, divided by, x, plus, 5, end fraction for x, does not equal, 0

### A note about equivalent expressions

Original expressionReduced expression
start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fractionstart fraction, x, plus, 3, divided by, x, plus, 5, end fraction for x, does not equal, 0
The two expressions above are equivalent. This means their outputs are the same for all possible x-values. The table below illustrates this for x, equals, 2.
Original expressionReduced expression
Evaluation at start color #7854ab, x, equals, 2, end color #7854ab\begin{aligned}\dfrac{(\purpleD{2})^2+3(\purpleD{2})}{(\purpleD{2})^2+5(\purpleD{2})}&=\dfrac{10}{14}\\\\&=\dfrac{\purpleD{{2}}\cdot 5}{\purpleD{{2}}\cdot 7}\\\\&=\dfrac{\purpleD{\cancel{2}}\cdot 5}{\purpleD{\cancel{2}}\cdot 7}\\\\&=\dfrac{5}{7}\end{aligned}\begin{aligned}\dfrac{\purpleD{2}+3}{\purpleD{2}+5}&=\dfrac{5}{7}\\\\&\phantom{=\dfrac57}\\\\&\phantom{=\dfrac57}\\\\&\phantom{=\dfrac57}\end{aligned}
NoteThe result is reduced to lowest terms by canceling a common factor of start color #7854ab, 2, end color #7854ab.The result is already reduced to lowest terms because the factor of x (in this case start color #7854ab, x, equals, 2, end color #7854ab), was already canceled when we reduced the expression to lowest terms.
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 start color #7854ab, x, equals, 0, end color #7854ab.
Original expressionReduced expression (without restriction)
Evaluation at start color #7854ab, x, equals, 0, end color #7854ab\begin{aligned}\dfrac{(\purpleD{0})^2+3(\purpleD{0})}{(\purpleD{0})^2+5(\purpleD{0})}&=\dfrac{0}{0}\\\\&=\text{undefined}\end{aligned}\begin{aligned}\dfrac{\purpleD{0}+3}{\purpleD{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, does not equal, 0 for the reduced 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!
start fraction, x, plus, 3, divided by, x, plus, 5, end fraction, does not equal, start fraction, 3, divided by, 5, end fraction
This becomes clear when looking at a numerical example. For example, suppose start color #7854ab, x, equals, 2, end color #7854ab.
start fraction, start color #7854ab, 2, end color #7854ab, plus, 3, divided by, start color #7854ab, 2, end color #7854ab, plus, 5, end fraction, does not equal, start fraction, 3, divided by, 5, end fraction
As a rule, we can only cancel if the numerator and denominator are in factored form!

### Summary of the process for reducing to lowest terms

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

Problem 1
Reduce start fraction, 6, x, plus, 20, divided by, 2, x, plus, 10, end fraction to lowest terms.

Problem 2
Reduce start fraction, x, cubed, minus, 3, x, squared, divided by, 4, x, squared, minus, 5, x, end fraction to lowest terms.
for x, does not equal

## Example 2: Reducing $\dfrac{x^2-9}{x^2+5x+6}$start fraction, x, squared, minus, 9, divided by, x, squared, plus, 5, x, plus, 6, end fraction to lowest terms

Step 1: Factor the numerator and denominator
start fraction, x, squared, minus, 9, divided by, x, squared, plus, 5, x, plus, 6, end fraction, equals, start fraction, left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, divided by, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, end fraction
Step 2: List restricted values
Since division by 0 is undefined, here we see that start color #0c7f99, x, does not equal, minus, 2, end color #0c7f99 and start color #7854ab, x, does not equal, minus, 3, end color #7854ab.
start fraction, left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, divided by, start color #0c7f99, left parenthesis, x, plus, 2, right parenthesis, end color #0c7f99, start color #7854ab, left parenthesis, x, plus, 3, right parenthesis, end color #7854ab, end fraction
Step 3: Cancel common factors
Notice that the numerator and denominator share a common factor of start color #208170, x, plus, 3, end color #208170. This can be canceled out.
\begin{aligned} \dfrac{(x-3)\tealE{(x+3)}}{(x+2)\tealE{(x+3)}}&=\dfrac{(x-3)\tealE{\cancel{(x+3)}}}{(x+2)\tealE{\cancel{(x+3)}}} \\\\ &=\dfrac{x-3}{x+2} \end{aligned}
We write the reduced form as follows:
start fraction, x, minus, 3, divided by, x, plus, 2, end fraction for x, does not equal, minus, 3
The original expression requires x, does not equal, minus, 2, comma, minus, 3. We do not need to note that x, does not equal, minus, 2, since this is understood from the expression.

### Check for understanding

Problem 3
Reduce start fraction, x, squared, minus, 3, x, plus, 2, divided by, x, squared, minus, 1, end fraction to lowest terms.