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Precalculus

Course: Precalculus > Unit 4

Lesson 1: Reducing rational expressions to lowest terms
Math
Precalculus
Rational functions
Reducing rational expressions to lowest terms
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Intro to rational expressions

Google Classroom
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 3, x, squared, minus, 6, x, minus, 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:
  • start fraction, 1, divided by, x, end fraction
  • start fraction, x, plus, 5, divided by, x, squared, minus, 4, x, plus, 4, end fraction
  • start fraction, x, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, 2, x, minus, 3, right parenthesis, divided by, x, minus, 6, end fraction
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 start fraction, 2, x, plus, 3, divided by, x, minus, 2, end fraction.
We can determine the value of this expression for particular x-values. For example, let's evaluate the expression at start color #11accd, x, end color #11accd, equals, start color #11accd, 1, end color #11accd.
2(1)+31−2=5−1=−5\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 start color #11accd, x, end color #11accd, equals, start color #11accd, 1, end color #11accd is start color #e07d10, minus, 5, end color #e07d10.
Now let's find the value of the expression at start color #11accd, x, end color #11accd, equals, start color #11accd, 2, end color #11accd.
2(2)+32−2=70=undefined!\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, start color #11accd, x, end color #11accd, equals, start color #11accd, 2, end color #11accd 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 start fraction, x, plus, 1, divided by, left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, plus, 4, right parenthesis, end fraction

Let's find the zeros of the denominator and then restrict these values:
(x−3)(x+4)=0↙↘x−3=0orx+4=0x=3orx=−4\begin{aligned} (x-3)&(x+4)= 0 \\\\ \swarrow \quad&\quad \searrow \\\\ x-3=0 \quad &\text{or} \quad x+4=0 \\\\ x = 3 \quad&\text{or} \quad x=-4 \end{aligned}
So we write that the domain is all real numbers except start text, 3, end text and start text, negative, 4, end text, or simply x, does not equal, 3, comma, minus, 4.

Check your understanding

Problem 1
What is the domain of start fraction, x, plus, 1, divided by, x, minus, 7, end fraction?
Choose 1 answer:

Problem 2
What is the domain of start fraction, 3, x, minus, 7, divided by, 2, x, plus, 1, end fraction?
Choose 1 answer:

Problem 3
What is the domain of start fraction, 2, x, minus, 3, divided by, x, left parenthesis, x, plus, 1, right parenthesis, end fraction?
Choose 1 answer:

Problem 4
What is the domain of start fraction, x, minus, 3, divided by, x, squared, minus, 2, x, minus, 8, end fraction?
Choose 1 answer:

Problem 5
What is the domain of start fraction, x, plus, 2, divided by, x, squared, plus, 4, end fraction?
Choose 1 answer:

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  • duskpin sapling style avatar for user Jenn Vo
    Posted 7 years ago. Direct link to Jenn Vo's post “I have a question about #...”
    I have a question about #5 under the Check your understanding section. So for the denominator in that fraction, can I use the method "the different of 2 squares" to factor it out to (x+2) (x-2) and solve for x from there? Can you explain more about it? I didn't get the last part in the explanation. Thanks!
    (22 votes)
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    • primosaur ultimate style avatar for user Mr. Brownridge
      Posted 7 years ago. Direct link to Mr. Brownridge's post “Difference refers to subt...”
      Good Answer
      Difference refers to subtraction. x^2+4 is a sum. Therefore, it is a "sum of two squares." If you graph the function you will see that it is an upward facing parabola with a y-intercept of 4. It has no solutions. So the expression will never equal zero (unless we use a different set of numbers called complex numbers).
      (22 votes)
  • duskpin ultimate style avatar for user Jedidiah Poh
    Posted 5 months ago. Direct link to Jedidiah Poh's post “this is so fun”
    this is so fun
    (32 votes)
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  • blobby green style avatar for user Dolfo Aaron
    Posted 5 months ago. Direct link to Dolfo Aaron's post “Help me please my brain i...”
    Help me please my brain is hurting :(
    (11 votes)
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  • blobby green style avatar for user MatthewS
    Posted 3 years ago. Direct link to MatthewS's post “I don't have a good under...”
    I don't have a good understanding of how exactly you find the domain, and what "all real numbers" means.
    (6 votes)
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    • duskpin ultimate style avatar for user Victor
      Posted 3 years ago. Direct link to Victor's post “Domain means that you are...”
      Domain means that you are trying to find all possible values of x. Domain's are usually written in this format: {xeR} where xeR means that for every real number, x is a solution. All real numbers mean any number that exists, and they may be irrational, rational, negative, positive, etc. However, they cannot be undefinable values such as √-1, which is i in short. In order to find the domain, you'll have to find what can't be in the denominator usually by factoring, and you'll be able to find out what x cannot be. If you have a specific question you'd like me to walk you through, don't hesitate to ask!
      (9 votes)
  • leafers tree style avatar for user Vishwa Patel
    Posted 2 months ago. Direct link to Vishwa Patel's post “For Problem 5, why can’t ...”
    For Problem 5, why can’t x= +- 2i ?
    (4 votes)
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    • stelly blue style avatar for user Kim Seidel
      Posted 2 months ago. Direct link to Kim Seidel's post “We define domain and rang...”
      Good Answer
      We define domain and range using the set of real numbers. The domain in problem 5 is all real numbers. There is no value of x that makes the denominator = 0, so there are no values to exclude from the domain.

      You are asking about imaginary numbers. They are outside the set of real numbers, so they are no considered.

      Hope this helps.
      (10 votes)
  • duskpin sapling style avatar for user aaliyahmariecerveny
    Posted 5 years ago. Direct link to aaliyahmariecerveny's post “Why is number 5, _all rea...”
    Why is number 5, all real numbers shouldn't it be +/- 2 since x^2=+4, factors out to (x+2)(x-2)?
    (7 votes)
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    • stelly blue style avatar for user Kim Seidel
      Posted 5 years ago. Direct link to Kim Seidel's post “The denominator is: x^2+4...”
      The denominator is: x^2+4. You changed it into x^2-4.
      x^2+4 is not factorable. Any real number squared will create a positive value. That positive value plus 4 creates an even larger positive value. There is no value that you can use for X that would cause the denominator to become 0. This is why the answer is that the domain = all real numbers.

      Hope this helps.
      (1 vote)
  • blobby green style avatar for user zunnunam
    Posted 2 months ago. Direct link to zunnunam's post “explain why domain of a r...”
    explain why domain of a rational expression is all real numbers except for those that make the denominator equal to zero.
    (3 votes)
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  • blobby green style avatar for user Mrs. Head
    Posted 3 years ago. Direct link to Mrs. Head's post “Why do you use the term "...”
    Why do you use the term "cancel"? I know a lot of teachers use it and that was what my teachers called it when I was in school. But is this really a mathematically correct term?
    I spend a great deal of time correcting students who just want to "cancel" terms just because they are alike, without understanding that in order for terms to be removed from an expression you have to use a mathematical operation, division or subtraction. Therefore terms can only be "divided out" or "subtracted out". Students will often times cross out or as you say "cancel out" terms that are both in numerators when multiplying terms or both in the denominators. To help resolve this issue my students are only allowed to use correct mathematical operations when simplifying expressions (divide out or subtract out).
    (4 votes)
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  • blobby green style avatar for user jher4900
    Posted a month ago. Direct link to jher4900's post “what is the equation for ...”
    what is the equation for a rational function?
    (1 vote)
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  • leaf orange style avatar for user hwang
    Posted 3 years ago. Direct link to hwang's post “In rational expression wh...”
    In rational expression why is domain all real number?
    (3 votes)
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    • female robot grace style avatar for user loumast17
      Posted 3 years ago. Direct link to loumast17's post “rational expressions depe...”
      rational expressions depend on the denominator for domain. If you know how to find vertical asymptotes and holes, those are what would limit the domain of a rational function. The only time a rational function has a domain of all reals is if the denominator is just 1.

      EDIT

      Thanks to Hecretary Bird for his correction. denominator just has to be a constant, other than 0 still though.
      (2 votes)