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Study Guides > College Algebra CoRequisite Course

Define and Simplify Rational Expressions

Learning Outcomes

  • Recognize and define a rational expression
  • Determine the domain of a rational expression
  • Simplify a rational expression
Rational expressions are fractions that have a polynomial in the numerator, denominator, or both. Although rational expressions can seem complicated because they contain variables, they can be simplified using the techniques used to simplify expressions such as [latex]\dfrac{4x^3}{12x^2}[/latex]. Because rational expressions often contain a variable in the denominator, though, there are certain values of the variable that will would cause the denominator to equal zero. When the denominator of a fraction is zero, the fraction is said to be undefined. When working with rational expressions, it is important to identify values that would cause the denominator to equal zero and state them explicitly as numbers that can't be included in the set of values for which the expression may be evaluated. We call the set of values for which the expression is defined the domain of the expression.

Determine the domain of a rational expression

Consider the following rational expression evaluated at [latex]x = 2[/latex]:

Evaluate  [latex]\dfrac{x}{x-2}[/latex] for [latex]x=2[/latex]

Substitute [latex]x=2[/latex]

[latex]\begin{array}{l}\dfrac{2}{2-2}\\\text{}\\=\dfrac{2}{0}\end{array}[/latex]

But we have stated that a fraction with a zero in the denominator is not defined. This means that for the expression [latex]\dfrac{x}{x-2}[/latex], [latex]x[/latex] cannot be [latex]2[/latex] because it will result in an undefined fraction. In general, finding values for a variable that will not result in division by zero allows us to identify all the values for which the expression is defined. We call the set of all such values the domain of the expression (later we'll apply them to equations and functions as well).

Domain of a rational expression

The domain of a rational expression is a collection of the values for the variable that will not result in an undefined mathematical operation such as division by zero.  For a = any real number, we can notate the domain in the following way:

 [latex]x[/latex] is all real numbers where [latex]x\neq{a}[/latex]

We cannot divide a number [latex]c[/latex] by zero [latex] \left( \dfrac{c}{0}\,\,=\,\,? \right)[/latex] because if we could, there would have to exist some number that, when multiplied by [latex]0[/latex], would recover [latex]c \,\left( ?\,\,\cdot \,\,0\,\,=\,\,c \right)[/latex]. There are no numbers that can do this, so we say that division by zero is undefined. In simplifying rational expressions, pay attention to what values of the variable(s) in the expression would make the denominator equal zero. These values cannot be included in the domain. We call them excluded values. (Note that although the denominator cannot be equivalent to [latex]0[/latex], the numerator can—this is why you only look for excluded values in the denominator of a rational expression.) For rational expressions, the domain will exclude values for which the value of the denominator is [latex]0[/latex]. The following example illustrates finding the domain of an expression.

Example

Identify the domain of the expression. [latex] \dfrac{x+7}{x+9}[/latex]

Answer: We can see that if [latex]x = -9[/latex], the denominator would equal zero because [latex]-9 + 9 = 0.[/latex] So we exclude [latex]-9[/latex] from the set of possible values in the domain of [latex] \dfrac{x+7}{x+9}[/latex]. The domain is all real numbers except [latex]-9[/latex].

Simplify Rational Expressions

Before we begin to discuss simplifying rational expressions, let's review the difference between a factor,  a term,  and an expression. Factors are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: [latex]2[/latex] and [latex]10[/latex] are factors of [latex]20[/latex], as are [latex]4, 5, 1, 20[/latex]. Terms are single numbers, or variables and numbers connected by multiplication. [latex]-4, 6x[/latex] and [latex]x^2[/latex] are all terms. Expressions are groups of terms connected by addition and subtraction.  [latex]2x^2-5[/latex] is an expression. Simplify: [latex]\dfrac{2x^2}{12x}[/latex] Just as we did when we simplified fractions containing only numbers, we'll write the numerator and denominator each as a product of its smallest factors. Then, we can cancel common factors in the numerator and denominator by dividing them. [latex-display]\begin{array}{cc}\dfrac{2x^2}{12x}\\=\dfrac{2\cdot{x}\cdot{x}}{2\cdot3\cdot2\cdot{x}}\\=\dfrac{\cancel{2}\cdot{\cancel{x}}\cdot{x}}{\cancel{2}\cdot3\cdot2\cdot{\cancel{x}}}=\dfrac{x}{6}\end{array}[/latex-display] We can do this because [latex]\dfrac{2}{2}=1\text{ and }\dfrac{x}{x}=1[/latex], so our expression simplifies to [latex]\dfrac{x}{6}[/latex] Compare that to the expression [latex]\dfrac{2x^2+x}{2x+2}[/latex]. Notice the numerator and denominator in this rational expression consist of polynomials (sums or differences of terms). We would have to write those polynomials as products, a process called factoring, before canceling any terms or variables. We'll see how to do that later in this module. The polynomial above, written in factored form is [latex]\dfrac{2x(x+1)}{2(x+1)}[/latex]. Note the common factors of [latex]2[/latex] and [latex](x+1)[/latex] in the numerator and denominator. Since these terms are being multiplied together, we can cancel the common factors. [latex]\begin{array}{cc}\dfrac{2x(x+1)}{2(x+1)}\\ =\dfrac{2\cdot{x}\cdot(x+1)}{2\cdot(x+1)}\\ =\dfrac{\cancel{2}\cdot{x}\cdot\cancel{(x+1)}}{\cancel{2}\cdot\cancel{(x+1)}}\\ =\dfrac{x}{1}=x\end{array}[/latex]. We'll discover how to write the numerator and denominator of a rational expression as products later in the module. For now, consider the steps to simplify a rational expression.

How To: Given a rational expression, simplify it

  1. Write the numerator and denominator as products of factors.
  2. Cancel any common factors.

Example

Simplify the rational expression [latex]\dfrac{x(x-7)(x+2)}{x(x+3)(x+2)}[/latex] and state its domain.

Answer: [latex-display]\dfrac{x(x-7)(x+2)}{x(x+3)(x+2)} =\dfrac{\cancel{x}(x-7)\cancel{(x+2)}}{\cancel{x}(x+3)\cancel{(x+2)}}=\dfrac{(x-7)}{(x+3)}[/latex-display] Since the unsimplified expression would have been undefined for [latex]x = 0, -3, \text{ or } -2,[/latex] the domain is all real numbers except [latex]-3, -2[/latex] and [latex]0[/latex].

 

Licenses & Attributions

CC licensed content, Original

  • Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
  • Simplify and Give the Domain of Rational Expressions. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.

CC licensed content, Shared previously

  • Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at : http://cnx.org/contents/[email protected]:1/Preface.