15.1.1: Introduction to Rational Expressions
- Page ID
- 72091
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- Simplify rational expressions.
Introduction
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 in the same way that numeric fractions, also called numerical fractions, are simplified.
Finding the Domain of an Expression
The first step in simplifying a rational expression is to determine the domain, the set of all possible values of the variables. The denominator in a fraction cannot be zero because division by zero is undefined. The reason \(\ \frac{6}{3}=2\) is that when you multiply the answer 2, times the divisor 3, you get back 6. To be able to divide any number \(\ c\) by zero \(\ \left(\frac{c}{0}=?\right)\), you would have to find a number that when you multiply it by 0 you would get back \(\ c\) (what times zero equals \(\ c\)?). There are no numbers that can do this, so we say “division by zero is undefined”. In simplifying rational expressions, you need to 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, so they're called excluded values. Discard them right at the start, before you go any further.
(Note that although the denominator cannot be equivalent to 0, 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 0. Two examples to illustrate finding the domain of an expression are shown below.
Identify the domain of the expression.
\(\ \frac{3 x+2}{x-4}\)
Solution
| \(\ x-4=0\) | Find any values for \(\ x\) that would make the denominator equal 0. |
| \(\ x=4\) | When \(\ x=4\), the denominator is equal to 0. |
The domain is all real numbers, except 4.
You found that \(\ x\) cannot be 4. (Sometimes you may see this idea presented as “\(\ x \neq 4\).”) What happens if you do substitute that value into the expression?
\(\ \begin{array}{c}
\frac{3 x+2}{x-4} \\
\frac{3(4)+2}{(4)-4} \\
\frac{12+2}{0} \\
\frac{14}{0}
\end{array}\)
You find that when \(\ x=4\), the numerator evaluates to 14, but the denominator evaluates to 0. And since division by 0 is undefined, this must be an excluded value.
Let's try one that's a little more challenging.
Identify the domain of the expression.
\(\ \frac{x+7}{x^{2}+8 x-9}\)
Solution
| \(\ x^{2}+8 x-9=0\) | Find any values for \(\ x\) that would make the denominator equal to 0 by setting the denominator equal to 0 and solving the equation. |
| \(\ \begin{array}{c} (x+9)(x-1)=0 \\ x=-9 \text { or } x=1 \end{array}\) |
Solve the equation by factoring. The solutions are the values that are excluded from the domain. |
The domain is all real numbers except -9 and 1.
Find the domain of the rational expression \(\ \frac{5 x}{2 x+8}\).
- all real numbers except -4
- all real numbers except 4
- all real numbers except 0
- all real numbers
- Answer
-
- Correct. When \(\ x=-4\), the denominator is \(\ 2(-4)+8=-8+8=0\). Division by 0 is undefined, so the domain must exclude \(\ x=-4\).
- Incorrect. When \(\ x=4\), the denominator does not equal 0, therefore it is not an excluded value. Set the denominator equal to 0 and solve for \(\ x\). The correct answer is all real numbers except -4.
- Incorrect. When \(\ x=0\), the numerator equals 0 but the denominator does not, therefore it is not an excluded value. Set the denominator equal to 0 and solve for \(\ x\). The correct answer is all real numbers except -4.
- Incorrect. There is one value of \(\ x\) that will make the denominator 0. Set the denominator equal to 0 and solve for \(\ x\). The correct answer is all real numbers except -4.
Simplifying Rational Expressions
Once you've figured out the excluded values, the next step is to simplify the rational expression. To simplify a rational expression, follow the same approach you use to simplify numeric fractions: find common factors in the numerator and denominator. Let’s start by simplifying a numeric fraction.
Simplify. \(\ \frac{15}{27}\)
Solution
| \(\ \frac{15}{27}=\frac{3 \cdot 5}{3 \cdot 3 \cdot 3}\) | Factor the numerator and denominator. |
| \(\ \frac{3 \cdot 5}{3 \cdot 3 \cdot 3}\) | Identify fractions that equal 1, and then pull them out of the fraction. |
| \(\ \begin{array}{l} \frac{5}{3 \cdot 3} \cdot \frac{3}{3} \\ \frac{5}{3 \cdot 3} \cdot 1 \end{array}\) |
In this fraction, the factor 3 is in both the numerator and denominator. Recall that \(\ \frac{3}{3}\) is another name for 1. |
| \(\ \frac{5}{3 \cdot 3}=\frac{5}{9}\) | Simplify. |
\(\ \frac{15}{27}=\frac{5}{9}\)
Now, you could have done that problem in your head, but it was worth writing it all down because that's exactly how you simplify a rational expression.
So let's simplify a rational expression, using the same technique you applied to the fraction \(\ \frac{15}{27}\). Only this time, the numerator and denominator are both monomials with variables.
Simplify. \(\ \frac{5 x^{2}}{25 x}\)
Solution
| \(\ \frac{5 x^{2}}{25 x}=\frac{5 \cdot x \cdot x}{5 \cdot 5 \cdot x}\) | Factor the numerator and denominator. |
| \(\ \begin{array}{r} \frac{5 \cdot x \cdot x}{5 \cdot 5 \cdot x} \\ \frac{x}{5} \cdot \frac{5 \cdot x}{5 \cdot x} \\ \frac{x}{5} \cdot 1 \end{array}\) |
Identify fractions that equal 1, and then pull them out of the fraction. |
| \(\ \frac{x}{5}\) | Simplify. |
\(\ \frac{5 x^{2}}{25 x}=\frac{x}{5}\)
The same steps worked again. Factor the numerator, factor the denominator, identify factors that are common to the numerator and denominator and write as a factor of 1, and simplify.
When simplifying rational expressions, it is a good habit to always consider the domain, and to find the values of the variable (or variables) that make the expression undefined. (This will come in handy when you begin solving for variables a bit later on.)
Identify the domain of the expression.
\(\ \frac{5 x^{2}}{25 x}\)
Solution
| \(\ 25 x=0\) | Find any values for \(\ x\) that would make the denominator equal to 0 by setting the denominator equal to 0 and solving the equation. |
|
\(\ \frac{25 x}{25}=\frac{0}{25}\) \(\ x=0\) |
The values for \(\ x\) that make the denominator equal 0 are excluded from the domain. |
The domain is all real numbers except 0.
Notice that you started with the original expression, and identified values of \(\ x\) that would make \(\ 25x\) equal to 0. Why does this matter? When \(\ \frac{5 x^{2}}{25 x}\) is simplified, it is the fraction \(\ \frac{x}{5}\). Since 5 is the denominator, it seems that no values need to be excluded from the domain. When finding the domain of an expression, you always start with the original expression because variable terms may be factored out as part of the simplification process.
In the examples that follow, the numerator and the denominator are polynomials with more than one term, but the same principles of simplifying will once again apply. Factor the numerator and denominator to simplify the rational expression.
Simplify and state the domain for the expression.
\(\ \frac{x+3}{x^{2}+12 x+27}\)
Solution
| \(\ x^{2}+12 x+27=0\) | |
|
\(\ (x+3)(x+9)=0\) \(\ \begin{array}{llrl} \(\ x=-3\) or \(\ x=-9\) domain is all real numbers except -3 and -9 |
To find the domain (and the excluded values), find the values for which the denominator is equal to 0. Factor the quadratic to find the values. |
| \(\ \frac{x+3}{x^{2}+12 x+27}\) | Factor the numerator and denominator. |
| \(\ \frac{x+3}{(x+3)(x+9)}\) | Identify the factors that are the same in the numerator and denominator. |
| \(\ \begin{array}{r} \frac{x+3}{x+3} \cdot \frac{1}{x+9} \\ 1 \cdot \frac{1}{x+9} \end{array}\) |
Write as separate fractions, pulling out fractions that equal 1. Simplify. |
\(\ \frac{x+3}{x^{2}+12 x+27}=\frac{1}{x+9}\)
The domain is all real numbers except -3 and -9.
Simplify and state the domain for the expression.
\(\ \frac{x^{2}+10 x+24}{x^{3}-x^{2}-20 x}\)
Solution
| \(\ x^{3}-x^{2}-20 x=0\) | |
| \(\ \begin{array}{l} x\left(x^{2}-x-20\right)=0 \\ x(x-5)(x+4)=0 \end{array}\) |
To find the domain, determine the values for which the denominator is equal to 0. |
| domain is all real numbers except 0, 5, and -4 | |
| \(\ \begin{array}{r} \frac{x^{2}+10 x+24}{x^{3}-x^{2}-20 x} \\ \frac{(x+4)(x+6)}{x(x-5)(x+4)} \\ \frac{x+6}{x(x-5)} \cdot \frac{(x+4)}{(x+4)} \end{array}\) |
To simplify, factor the numerator and denominator of the rational expression. Identify the factors that are the same in the numerator and denominator. Write as separate fractions, pulling out fractions that equal 1. |
| \(\ \frac{x+6}{x(x-5)} \text { or } \frac{x+6}{x^{2}-5 x}\) | Simplify. It is acceptable to either leave the denominator in factored form or to distribute with multiplication. |
\(\ \frac{x+6}{x(x-5)} \text { or } \frac{x+6}{x^{2}-5 x}\)
The domain is all real numbers except 0, 5, and -4.
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 0.
- Factor the numerator and denominator.
- Find common factors for the numerator and denominator and simplify.
Simplify the rational expression below.
\(\ \frac{2 x^{2}+13 x+15}{2 x^{2}+23 x+30}\)
[Note: While the domain and excluded values of a rational expression are important, you will not always be asked to find them when simplifying a rational expression. In this expression, the domain is all real numbers except \(\ \frac{3}{2}\) and -10.]
- \(\ \frac{14}{25}\)
- \(\ \frac{x+5}{x+10}\)
- \(\ \frac{2 x}{3}\)
- \(\ \frac{1}{2}\)
- Answer
-
- Incorrect. You must first factor the polynomials in the numerator and the denominator and then express like factors in the numerator and denominator as 1 to simplify. The expression can be factored as \(\ \frac{(2 x+3)(x+5)}{(2 x+3)(x+10)}\), so the correct answer is \(\ \frac{x+5}{x+10}\).
- Correct. The rational expression can be simplified by factoring the numerator and denominator as \(\ \frac{(2 x+3)(x+5)}{(2 x+3)(x+10)}\). Since \(\ \frac{2 x+3}{2 x+3}=1\), simplify the expression to \(\ \frac{x+5}{x+10}\).
- Incorrect. You must first factor the polynomials in the numerator and the denominator and then express like factors in the numerator and denominator as 1 to simplify. The expression can be factored as \(\ \frac{(2 x+3)(x+5)}{(2 x+3)(x+10)}\), so the correct answer is \(\ \frac{x+5}{x+10}\).
- Incorrect. You cannot simplify \(\ \frac{x+5}{x+10}\) to \(\ \frac{1}{2}\) because the \(\ x\) in the numerator and the \(\ x\) in the denominator are not being multiplied, they are being added. The correct answer is \(\ \frac{x+5}{x+10}\).
Summary
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 1.
An additional consideration for rational expressions is to determine what values are excluded from the domain. Since division by 0 is undefined, any values of the variables that result in a denominator of 0 must be excluded. Excluded values must be identified in the original equation, not from its factored form.