Will I need this in real life?

Maybe. Rational functions appear quite often in business and economics applications. Additionally, understanding rational functions will help you to understand more complex functions that can model other "real life" situations. Most people never notice when they can model a situation using rational function (or other mathematical functions) because they never attempt to do so. Hopefully you will keep sincerely asking questions like this and looking for ways to use this knowledge to model things in the real world. Such an outlook will serve you well.

This is very confusing? How will this help us in real life?

It probably, most likely, won't. although some parts of the business and trade use this math, although I don't know any, there might be some.

I don't understand much the restricted values. Can someone help me to understand it please?

If we have an equation (x^2 + 3x) / (x^2 + 5x) , we can simplify it to (x+3) / (x+5) . At first glance, the two equations seem to be equal, but they are actually not! Remember that x can be any value - let just try the values -1, 1 and 0 for x in the two equations above. For x = -1, (x^2 + 3x) / (x^2 + 5x) = (1 - 3) / (1 - 5) = -2/-4 = 0.5 (x+3) / (x+5) = (-1 + 3)/ (-1 + 5) = 2/4 = 0.5 Both equations are equivalent as both of them give the value 0.5 For x = 1, (x^2 + 3x) / (x^2 + 5x) = (1 + 3) / (1 + 5) = 4/6 = 2/3 (x+3) / (x+5) = (1 + 3)/ (1 + 5) = 4/6 = 2/3 Both equations are equivalent as both of them give the value 2/3 For x = 0, (x^2 + 3x) / (x^2 + 5x) = (0 + 0) / (0 + 0) = undefined ! (x+3) / (x+5) = (0 + 3)/ (0 + 5) = 3/5 ! Both equations are NOT equivalent as both of them give different values in the case of x =0 So from the above, whenever the denominator results in a value of 0, we get an undefined value for the expression. Therefore we need to put the restriction x not equal to 0 in this case on the equation (x+3) / (x+5). With this restriction, the simplified equation is now equivalent to the original rational expression.

How do I remember all of these steps?

Practice, practice, practice. Eventually, you will do it enough that you don’t even think about it. You could come up with an acronym to help you, but the final goal is that you won’t need it.

In #3, why can't x be equal to 1? I understand why it can't equal -1, but I don't see where 1 causes the expression to be undefined.

Factor both the numerator and the denominator. Numerator = (x-2)(x-1) Denominator = (x-1)(x+1) The factor of (x-1) in the denominator would cause the an undefined if x=1 Even though this factor cancels out, you need to maintain the restriction that x can't = 1 to be consistent with the original expression. Hope this helps.

what makes an expression undefined?

With rational expressions (fractions), they become undefined if the denominator = 0. If a denominator = 0, then we are dividing by 0, which is undefined.

Why is x not equal to -3? And why is that stated if we found that out in the middle of the problem?

Your simplified fraction must be equivalent to the original fraction. The original fraction is undefined for x=-3 and x=2. Once the fraction is reduced, you can no longer tell that the new fraction needs to be undefined for x=-3, so it must be explicitly stated. If you don't state it, you omit one of the restrictions that existed in the original fraction so the new fraction is no longer equivalent to the original. Hope this helps.

In Example 1 if x= -5 at any step this would make the whole expression undefined giving a zero in the denominator. Why is only x can not equal 0 listed and not x can not equal -5?

You are correct. But Sal is concerned about the now missing (after x/x is eliminated) x in the denominator, and not (x+5), which is still there to see.

Main content

College Algebra

Course: College Algebra > Unit 8

Lesson 3: Reducing rational expressions to lowest terms

Reducing rational expressions to lowest terms

Google Classroom

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

\frac{x + 2}{x + 1}

is all real numbers except $-1$ ‍, 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 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,

\frac{6}{8}

reduced to lowest terms is

\frac{3}{4}

. Notice how we canceled a common factor of

2

from the numerator and the denominator:

\begin{aligned} \frac{6}{8} & = \frac{2 \cdot 3}{2 \cdot 4} \\ = \frac{2 \cdot 3}{2 \cdot 4} \\ = \frac{3}{4} \end{aligned}

Example 1: Reducing $\frac{x^{2} + 3 x}{x^{2} + 5 x}$ ‍ 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!

\frac{x^{2} + 3 x}{x^{2} + 5 x} = \frac{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

x \neq 0

and

x \neq - 5

\frac{x (x + 3)}{x (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} \frac{x (x + 3)}{x (x + 5)} & = \frac{x (x + 3)}{x (x + 5)} \\ = \frac{x + 3}{x + 5} \end{aligned}

Step 4: Final answer

Recall that the original expression is defined for

x \neq 0, - 5

. The reduced 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.

[Can you elaborate here?]

In conclusion, the reduced form is written as follows:

\frac{x + 3}{x + 5}

for

x \neq 0

A note about equivalent expressions

Original expression	Reduced expression
$\frac{x^{2} + 3 x}{x^{2} + 5 x}$ ‍	$\frac{x + 3}{x + 5}$ ‍ for $x \neq 0$ ‍

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 = 2

	Original expression		Reduced expression
Evaluation at $x = 2$ ‍	$\begin{aligned} \frac{(2)^{2} + 3 (2)}{(2)^{2} + 5 (2)} & = \frac{10}{14} \\ = \frac{2 \cdot 5}{2 \cdot 7} \\ = \frac{2 \cdot 5}{2 \cdot 7} \\ = \frac{5}{7} \end{aligned}$ ‍		$\begin{aligned} \frac{2 + 3}{2 + 5} & = \frac{5}{7} \end{aligned}$ ‍
Note	The result is reduced to lowest terms by canceling a common factor of $2$ ‍.		The result is already reduced to lowest terms because the factor of $x$ ‍ (in this case $x = 2$ ‍), 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

x = 0

	Original expression		Reduced expression (without restriction)
Evaluation at $x = 0$ ‍	$\begin{aligned} \frac{(0)^{2} + 3 (0)}{(0)^{2} + 5 (0)} & = \frac{0}{0} \\ = undefined \end{aligned}$ ‍		$\begin{aligned} \frac{0 + 3}{0 + 5} & = \frac{3}{5} \end{aligned}$ ‍

Because the two expressions must be equivalent for all possible inputs, we must require

x \neq 0

for the reduced expression.

Misconception alert

Note that we cannot cancel the

x

's in the expression below. This is because these are terms rather than factors of the polynomials!

\frac{x + 3}{x + 5} \neq \frac{3}{5}

This becomes clear when looking at a numerical example. For example, suppose

x = 2

\frac{2 + 3}{2 + 5} \neq \frac{3}{5}

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.

Check your understanding

Problem 1

Reduce $\frac{6 x + 20}{2 x + 10}$ ‍ to lowest terms.

Problem 2

Reduce $\frac{x^{3} - 3 x^{2}}{4 x^{2} - 5 x}$ ‍ to lowest terms.

for

x \neq

Example 2: Reducing $\frac{x^{2} - 9}{x^{2} + 5 x + 6}$ ‍ to lowest terms

Step 1: Factor the numerator and denominator

\frac{x^{2} - 9}{x^{2} + 5 x + 6} = \frac{(x - 3) (x + 3)}{(x + 2) (x + 3)}

[Can you show the factorization, please?]

Step 2: List restricted values

Since division by

0

is undefined, here we see that

x \neq - 2

and

x \neq - 3

\frac{(x - 3) (x + 3)}{(x + 2) (x + 3)}

Step 3: Cancel common factors

Notice that the numerator and denominator share a common factor of

x + 3

. This can be canceled out.

\begin{aligned} \frac{(x - 3) (x + 3)}{(x + 2) (x + 3)} & = \frac{(x - 3) (x + 3)}{(x + 2) (x + 3)} \\ = \frac{x - 3}{x + 2} \end{aligned}

Step 4: Final answer

We write the reduced form as follows:

\frac{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

Problem 3

Reduce $\frac{x^{2} - 3 x + 2}{x^{2} - 1}$ ‍ to lowest terms.

Problem 4

Reduce $\frac{x^{2} - 2 x - 15}{x^{2} + x - 6}$ ‍ to lowest terms.

for

x \neq

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Mary
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Will I need this in real life?
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- Paul Miller
  Posted 8 years ago. Direct link to Paul Miller's post “Maybe. Rational functio...”
  Maybe.
  Rational functions appear quite often in business and economics applications.
  
  Additionally, understanding rational functions will help you to understand more complex functions that can model other "real life" situations.
  Most people never notice when they can model a situation using rational function (or other mathematical functions) because they never attempt to do so.
  
  Hopefully you will keep sincerely asking questions like this and looking for ways to use this knowledge to model things in the real world. Such an outlook will serve you well.
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This is very confusing? How will this help us in real life?
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- Larryisdaboss
  Posted 6 years ago. Direct link to Larryisdaboss's post “It probably, most likely,...”
  It probably, most likely, won't. although some parts of the business and trade use this math, although I don't know any, there might be some.
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Makado
Posted 7 years ago. Direct link to Makado's post “I don't understand much t...”
I don't understand much the restricted values. Can someone help me to understand it please?
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(6 votes)
Answer
- Timothy Soh
  Posted 7 years ago. Direct link to Timothy Soh's post “If we have an equation (x...”
  If we have an equation (x^2 + 3x) / (x^2 + 5x) , we can simplify it to (x+3) / (x+5) . At first glance, the two equations seem to be equal, but they are actually not!
  Remember that x can be any value - let just try the values -1, 1 and 0 for x in the two equations above.
  For x = -1,
  (x^2 + 3x) / (x^2 + 5x) = (1 - 3) / (1 - 5) = -2/-4 = 0.5
  (x+3) / (x+5) = (-1 + 3)/ (-1 + 5) = 2/4 = 0.5
  Both equations are equivalent as both of them give the value 0.5
  For x = 1,
  (x^2 + 3x) / (x^2 + 5x) = (1 + 3) / (1 + 5) = 4/6 = 2/3
  (x+3) / (x+5) = (1 + 3)/ (1 + 5) = 4/6 = 2/3
  Both equations are equivalent as both of them give the value 2/3
  For x = 0,
  (x^2 + 3x) / (x^2 + 5x) = (0 + 0) / (0 + 0) = undefined !
  (x+3) / (x+5) = (0 + 3)/ (0 + 5) = 3/5 !
  Both equations are NOT equivalent as both of them give different values in the case of x =0
  So from the above, whenever the denominator results in a value of 0, we get an undefined value for the expression. Therefore we need to put the restriction x not equal to 0 in this case on the equation (x+3) / (x+5). With this restriction, the simplified equation is now equivalent to the original rational expression.
  Button navigates to signup page
  (21 votes)
Caligula as a Girl
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How do I remember all of these steps?
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- Beaniebopbunyip
  Posted 5 years ago. Direct link to Beaniebopbunyip's post “Practice, practice, pract...”
  Practice, practice, practice. Eventually, you will do it enough that you don’t even think about it. You could come up with an acronym to help you, but the final goal is that you won’t need it.
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arkais88
Posted 6 years ago. Direct link to arkais88's post “In #3, why can't x be equ...”
In #3, why can't x be equal to 1? I understand why it can't equal -1, but I don't see where 1 causes the expression to be undefined.
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- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “Factor both the numerator...”
  Factor both the numerator and the denominator.
  Numerator = (x-2)(x-1)
  Denominator = (x-1)(x+1)
  The factor of (x-1) in the denominator would cause the an undefined if x=1
  Even though this factor cancels out, you need to maintain the restriction that x can't = 1 to be consistent with the original expression.
  Hope this helps.
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  (7 votes)
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um im really confused
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what makes an expression undefined?
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- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “With rational expressions...”
  With rational expressions (fractions), they become undefined if the denominator = 0. If a denominator = 0, then we are dividing by 0, which is undefined.
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Leif
Posted 3 months ago. Direct link to Leif's post “Why is x not equal to -3?...”
Why is x not equal to -3? And why is that stated if we found that out in the middle of the problem?
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- Kim Seidel
  Posted 3 months ago. Direct link to Kim Seidel's post “Your simplified fraction ...”
  Your simplified fraction must be equivalent to the original fraction. The original fraction is undefined for x=-3 and x=2. Once the fraction is reduced, you can no longer tell that the new fraction needs to be undefined for x=-3, so it must be explicitly stated. If you don't state it, you omit one of the restrictions that existed in the original fraction so the new fraction is no longer equivalent to the original.
  
  Hope this helps.
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  (4 votes)
Ryan O'Connor
Posted 7 years ago. Direct link to Ryan O'Connor's post “In Example 1 if x= -5 at ...”
In Example 1 if x= -5 at any step this would make the whole expression undefined giving a zero in the denominator. Why is only x can not equal 0 listed and not x can not equal -5?
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- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “You are correct. But Sal ...”
  You are correct. But Sal is concerned about the now missing (after x/x is eliminated) x in the denominator, and not (x+5), which is still there to see.
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