Lesson Objectives
  • Demonstrate an understanding of how to factor a polynomial
  • Learn how to identify a rational expression
  • Learn how to find the restricted value(s) for a rational expression
  • Learn how to simplify a rational expression

What is a Rational Expression?


Since pre-algebra, we have seen real numbers placed into certain categories. We know the whole numbers start with zero and increase in increments of one forever:
Whole Numbers: {0,1,2,3,...}
The integers allow us to work with negative values. An integer includes any whole number and their opposite.
Integers: {...,-2,-1,0,1,2,...}
When we want to work with numbers between two integers, we generally deal with fractions or decimals. When we work with fractions, we have the quotient of one integer over another integer. We can refer to a fraction or the quotient of one integer over another as a rational number. Some examples of a rational number: $$\frac{1}{2}, \frac{7}{3}, \frac{11}{2}$$ Similarly, the quotient of two polynomials with a non-zero denominator is referred to as a rational expression. We will see this in our textbook as: $$\frac{P}{Q} \hspace{.5em} Q ≠ 0$$ Where P and Q each represent a polynomial and Q is not equal to zero. Let's look at a few examples of a rational expression: $$\frac{x + 5}{2x + 10}, \frac{4x}{10x^2 - 3}$$

Finding the Restricted Value for a Rational Expression

At this point, we should know that we can never divide by zero. When we encounter division by zero, the problem is said to be "undefined". When we work with rational expressions, our numerator can be any value including zero. Our denominator, however, will have restrictions based on not dividing by zero. We will restrict any value that replaces our variable and creates a denominator of zero. To find the restricted values of a rational expression:
  • Set the denominator equal to zero
  • Solve the equation
  • The solution or solutions are the restricted values
Let's look at an example.
Example 1: Find the restricted values for each rational expression $$\frac{x + 4}{x - 3}$$ Step 1) Set the denominator equal to zero
x - 3 = 0
Step 2) Solve the equation
x = 3
Step 3) The solution (3) is the restricted value
We can see in this case if we plug a 3 in for x, the denominator would become zero. This is where the rational expression is undefined.
Example 2: Find the restricted values for each rational expression $$\frac{x^2 + 11x + 10}{x^2 - 6x - 7}$$ Step 1) Set the denominator equal to zero
x2 - 6x - 7 = 0
Step 2) Solve the equation
(x - 7)(x + 1) = 0
x = -1,7
Step 3) The solutions of -1 and 7 are the restricted values.
If we plug in a -1 or a 7 for x in the rational expression, the denominator would become zero. This is where the rational expression is undefined.

Simplifying Rational Expressions

At this point, we should know how to simplify fractions. Essentially we factor the numerator and denominator and cancel any common factors other than 1. let's look at an example: $$\frac{120}{165}$$ We can factor both numerator and denominator: $$\frac{2 \cdot 2 \cdot 2 \cdot 3 \cdot 5}{3 \cdot 5 \cdot 11}$$ We can cancel a common factor of 3 and a common factor of 5 between the numerator and denominator: $$\require{cancel}\frac{2 \cdot 2 \cdot 2 \cdot \cancel{3} \cdot \cancel{5}}{\cancel{3} \cdot \cancel{5} \cdot 11} = \frac{8}{11}$$ We can apply the same thought process when we simplify a rational expression. To simplify a rational expression, we factor the numerator and denominator and then we cancel any common factors. Let's take a look at a few examples.
Example 3: Simplify each rational expression $$\frac{2x + 20}{x^2 + 12x + 20}$$ Step 1) Factor the polynomial in the numerator and denominator: $$\frac{2(x + 10)}{(x + 10)(x + 2)}$$ Step 2) Cancel common factors:
Here the quantity (x + 10) is a common factor. We can cancel this between the numerator and the denominator: $$\frac{2\cancel{(x + 10)}}{\cancel{(x + 10)}(x + 2)}$$ Our simplified result: $$\frac{2}{x + 2}$$ Example 4: Simplify each rational expression $$\frac{5x^2 - 27x - 56}{10x - 70}$$ Step 1) Factor the polynomial in the numerator and the denominator: $$\frac{(5x + 8)(x - 7)}{10(x - 7)}$$ Step 2) Cancel common factors:
Here the quantity (x - 7) is a common factor. We can cancel this between the numerator and the denominator: $$\frac{(5x + 8)\cancel{(x - 7)}}{10\cancel{(x - 7)}}$$ Our simplified result: $$\frac{5x + 8}{10}$$