Read: Define and Simplify Rational Expressions
Learning Objectives
- Recognize and define a rational expression
- Determine the domain of a rational expression
- Simplify a rational expression
Determine the domain of a rational expression
One sure way you can break math is to divide by zero. Consider the following rational expression evaluated at [latex]x = 2[/latex]:Evaluate [latex]\frac{x}{x-2}[/latex] for [latex]x=2[/latex]
Substitute [latex]x=2[/latex]
[latex]\begin{array}{l}\frac{2}{2-2}\\\text{}\\=\frac{2}{0}\end{array}[/latex]
This means that for the expression [latex]\frac{x}{x-2}[/latex], [latex]x[/latex] cannot be [latex]2[/latex] because it will result in an undefined ratio. In general, finding values for a variable that will not result in division by zero is called finding the domain. Finding the domain of a rational expression or function will help you not break math.Domain of a rational expression or equation
The domain of a rational expression or equation 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]
Example
Identify the domain of the expression. [latex] \frac{x+7}{{{x}^{2}}+8x-9}[/latex]Answer: Find any values for [latex]x[/latex] that would make the denominator equal to [latex]0[/latex] by setting the denominator equal to [latex]0[/latex] and solving the equation.
[latex]x^{2}+8x-9=0[/latex]
Solve the equation by factoring. The solutions are the values that are excluded from the domain.[latex]\begin{array}{c}(x+9)(x-1)=0\\x=-9\,\,\,\text{or}\,\,\,x=1\end{array}[/latex]
Answer
The domain is all real numbers except [latex]−9[/latex] and [latex]1[/latex].Simplify Rational Expressions
Before we dive in to simplifying rational expressions, let's review the difference between a factor, a term, and an expression. This will hopefully help you avoid another way to break math when you are simplifying rational expressions. 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. This distinction is important when you are required to divide. Let's use an example to show why this is important. Simplify: [latex]\large\frac{2x^2}{12x}[/latex] The numerator and denominator of this fraction consist of factors. To simplify it, we can divide without being impeded by addition or subtraction. [latex-display]\begin{array}{cc}\large\frac{2x^2}{12x}\\=\large\frac{2\cdot{x}\cdot{x}}{2\cdot3\cdot2\cdot{x}}\\=\large\frac{\cancel{2}\cdot{\cancel{x}}\cdot{x}}{\cancel{2}\cdot3\cdot2\cdot{\cancel{x}}}\end{array}[/latex-display] We can do this because [latex]\frac{2}{2}=1\text{ and }\frac{x}{x}=1[/latex], so our expression simplifies to [latex]\large\frac{x}{6}[/latex] Compare that to the expression [latex]\large\frac{2x^2+x}{12-2x}[/latex], notice the denominator and numerator consist of two terms connected by addition and subtraction. We have to tip-toe around the addition and subtraction. When asked to simplify it is tempting to want to cancel out like terms as we did when we just had factors. But you can't do that, it will break math!Example
Simplify and state the domain for the expression. [latex] \frac{x+3}{{{x}^{2}}+12x+27}[/latex]Answer: To find the domain (and the excluded values), find the values for which the denominator is equal to [latex]0[/latex]. Factor the quadratic, and apply the zero product principle.
[latex]\begin{array}{c}x+3=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x+9=0\\x=0-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=0-9\\x=-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=-9\\\\x=-3\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=-9\end{array}[/latex]
The domain is all real numbers except [latex]x=-3[/latex] or [latex]x=-9[/latex]. Factor the numerator and denominator. Identify the factors that are the same in the numerator and denominator, and simplify.[latex]\large\begin{array}{c}\frac{x+3}{x^{2}+12x+27}\\\\=\frac{x+3}{\left(x+3\right)\left(x+9\right)}\\\\\frac{\cancel{x+3}}{\cancel{\left(x+3\right)}\left(x+9\right)}\\\\\normalsize=1\cdot\large\frac{1}{x+9}\end{array}[/latex]
Answer
[latex-display] \frac{x+3}{{{x}^{2}}+12x+27}=\frac{1}{x+9}[/latex-display] The domain is all real numbers except [latex]−3[/latex] and [latex]−9[/latex].Example
Simplify and state the domain for the expression. [latex]\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}[/latex]Answer: To find the domain, determine the values for which the denominator is equal to [latex]0[/latex].
[latex]\begin{array}{r}x^{3}-x^{2}-20x=0\\x\left(x^{2}-x-20\right)=0\\x\left(x-5\right)\left(x+4\right)=0\end{array}[/latex]
The domain is all real numbers except [latex]0, 5[/latex], and [latex]−4[/latex]. To simplify, factor the numerator and denominator of the rational expression. Identify the factors that are the same in the numerator and denominator, and simplify.[latex] \large\begin{array}{c}\frac{x^{2}+10x+24}{x^{3}-x^{2}-20x}\\\\=\frac{\left(x+4\right)\left(x+6\right)}{x\left(x-5\right)\left(x+4\right)}\\\\=\frac{\cancel{\left(x+4\right)}\left(x+6\right)}{x\left(x-5\right)\cancel{\left(x+4\right)}}\end{array}[/latex]
Simplify. It is acceptable to either leave the denominator in factored form or to distribute multiplication.[latex]\frac{x+6}{x\left(x-5\right)}\,\,\,\text{or}\,\,\,\frac{x+6}{x^{2}-5x}[/latex]
Answer
[latex-display] \frac{x+6}{x(x-5)}[/latex] or [latex] \frac{x+6}{{{x}^{2}}-5x}[/latex-display] The domain is all real numbers except [latex]0, 5[/latex], and [latex]−4[/latex].Example
Simplify [latex]\frac{{x}^{2}-9}{{x}^{2}+4x+3}[/latex], state the domain.Answer: The special product in the numerator is a difference of squares. [latex-display]\begin{array}\frac{\left(x+3\right)\left(x - 3\right)}{\left(x+3\right)\left(x+1\right)}\hfill & \hfill & \hfill & \hfill & \text{Factor the numerator and the denominator}.\hfill \\ \frac{x - 3}{x+1}\hfill & \hfill & \hfill & \hfill & \text{Cancel common factor }\left(x+3\right).\hfill \end{array}[/latex-display] With the denominator factored it is easier to find the domain of the expression. Determine the values for which the denominator is equal to [latex]0[/latex]. [latex-display]\begin{array}{cc}\left(x+3\right)=0,\left(x+1\right)=0\\x\ne-3,\text{ AND }x\ne-1\end{array}[/latex-display]
Answer
[latex-display]\frac{{x}^{2}-9}{{x}^{2}+4x+3}=\frac{x - 3}{x+1}[/latex], Domain: [latex]x\ne-3,\text{ AND }x\ne-1[/latex-display]Steps for Simplifying a Rational Expression
To simplify a rational expression, follow these steps:- Determine the domain. The excluded values are those values for the variable that result in the expression having a denominator of [latex]0[/latex].
- Factor the numerator and denominator.
- Find common factors for the numerator and denominator and simplify.
Summary
An additional consideration for rational expressions is to determine what values are excluded from the domain. Since division by [latex]0[/latex] is undefined, any values of the variables that result in a denominator of [latex]0[/latex] must be excluded. Excluded values must be identified in the original equation, not from its factored form.Rational expressions are fractions containing polynomials. They can be simplified much like numeric fractions. To simplify a rational expression, first determine common factors of the numerator and denominator, and then remove them by rewriting them as expressions equal to [latex]1[/latex].Licenses & Attributions
CC licensed content, Original
- Screenshot: Breaking Math. Provided by: Lumen Learning License: CC BY: Attribution.
- 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) for Lumen Learning. 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.