Introduction to Rational Expressions

Introduction

Just like rational numbers are fractions made of integers, rational expressions are fractions made of polynomials. Before we dive into simplifying and evaluating these expressions, it is important to first understand what rational expressions are and how they behave.

In this part of the lesson, we will be using factoring techniques to identify what specific values of the rational expression will make it undefined.

#1 Determine the values of x that will make \bf {\dfrac{14}{3x - 12}} undefined

To determine the desired values of x , we will only equate the denominator to 0. Thus we have

We then solve for 3x - 12 = 0.

Therefore, \boxed{x =4} will make \dfrac{14}{3x - 12} undefined.

#2 Determine the values of x that will make \bf{\dfrac{6x - 2}{(x + 3)(2x - 30}} undefined

To determine the desired values of x , we only need to set the denominator equal to 0 since division by 0 is undefines. Thus, we have

We then solve for (x + 3(2x - 3) = 0 using the Zero Product Property.\

Therefore, \boxed{x = -3} \text{ and } \boxed{x = \dfrac{3}{2}} will make \dfrac{6x - 2}{(x + 3)(2x - 3)} undefined.

#3 Determine the values of x that will make \bf{\dfrac{x^2 + 1}{2x^2 - x - 3}} undefined

To determine the values of x that will make the expression undefines, we set the **denominator** equal to 0:

We factor the quadratic expression. (Note: Factoring of 2x^2 - x - 3 can be done by grouping)

Now we solve for (2x + 3)(x - 1) = 0 using the Zero Product Property.

Therefore, \boxed{x =-\dfrac{3}{2}} and \boxed{x = 1} will make \dfrac{x^2 + 1}{2x^2 - x - 3} undefined.

#4 Determine the values of x that will make \bf{\dfrac{3x + 4}{x^3 - 8}} undefined

To determine the values of x that will make \dfrac{3x + 4}{x^3 - 8} undefined, we set the denominator equal to 0:

We recognize that x^3 - 8 is a difference of cubes, so we factor it:

Now we solve for (x - 2)9x^2 +2x + 4) = 0 using the Zero Product Property.

\textcolor{red}{\textit{Note: } x^2 + 2x + 4 \textit{ has no real factors since it is not factorable.}}

Since x² + 2x + 4 has no real solutions, the only real value of x that will make \dfrac{3x + 4}{x^3 - 8} undefines is \boxed{x = 2}.

#5 Determine the values of x that will make \bf{\dfrac{3x^2 + x + 1}{x^2 + 4}} undefined

To determine the values of x that will make the expression undefined, we equate the denominator to 0 and solve:

We then solve for x^2 + 4 = 0.

Since x = \sqrt{-4} is not a real number, there is no real value of x that will make x^2 + 4 = 0.

Therefore, \boxed{\dfrac{3x^2 + x + 1}{x^2 + 4} \space \text{ is defined for all real values of } \space x}.