Fun with Rational Expressions

Published: Aug 17th, 2020

Learning Objectives
By the end of this section, you should be able to:
1. Describe a rational expression and state restrictions on its variables
2. Multiply, divide, add, and subtract rational expressions

The title of this lesson is inspired by the hit series "The Big Bang Theory"; you might already have heard of it. I probably shouldn't be promoting less successful brands than this one, but I'll give Warner Bros. a shout-out this time around... 

A wise man once said: "Why make new graphic when old graphic do trick?"

Jokes aside, in this lesson we will explore the world of rational functions. Our journey will first begin with a reintroduction to the rational functions, followed by an overview of their operations, namely: addition, subtraction, multiplication, and division.

Getting Acquainted

In the lesson "Four Basic Functions", we talked about the quadratic, square root, reciprocal, and inverse functions. The reciprocal function is, in fact, a type of rational function! In math terms, we define a rational expression as the quotient of two functions, where the denominator functional cannot equal zero:

\(\large{\frac{f(x)}{g(x)}}\), where \(g(x)\neq0\)

A couple examples of the rational expression include \(\frac{1}{x}\)\(\frac{1}{2x-3}\)\(\frac{3x^2+3x}{x+y}\), and so on. Because the denominator cannot equal zero, this represents a restriction on the function of the domain. However, unlike with the reciprocal function, where the numerator was a constant, the reciprocal function often has a variable for the denominator, so its range can include the number zero!

The main type of problem you will be asked to solve involving the rational expression is "what are the function's restrictions on the independent variable?". Let's try a couple examples, shall we? 

What restrictions should be placed on the variable in \(\frac{2x+5}{5x-3}\)?

First things first. You may be tempted to look at the numerator for restrictions. This is not the case here; it's perfectly fine to have our numerator equal to zero. The only thing we have a problem with is the denominator being zero. This is because dividing by zero is a big no-no, as demonstrated in the illustration below.

A graphic illustrating what happens when you divide by zero.

With that cleared up, we are looking for values where \(5x-3\neq0\). Solving for \(x\), we get \(x \neq \frac{3}{5}\); our only restriction on the domain. Thank you, next!

Determine the restrictions on the domain of \(f(x) = \frac{x^2+6x+9}{x^2-3x-18}\)

Once again, we are looking for restrictions of the denominator. In this case, the only way we can find these restrictions (apart from tedious trial and error) is by factoring! If you've been following along, this should be a cinch, but we'll review anyway. We are looking for two numbers that multiply to -18 and add to -3. -6 and 3 come to mind: 

\(= \frac{x^2+6x+9}{(x+3)(x-6)}\)

Solving the denominator for zero, we get \(x+3 \neq 0\) and \(x-6 \neq 0\). Isolating for \(x\), we get 6 and -3. Therefore, our restrictions are \(x \neq 6,-3\). Now, something different

Simplify \(f(x) = \frac{x^2+6x+9}{x^2-3x-18}\) (reuse, reduce, RECYCLE, am I right?!)

As we saw in the previous example, we can factor this expression's denominator to get \(\frac{x^2+6x+9}{(x+3)(x-6)}\). But, did you notice that we can also factor the numerator? We need two numbers that multiply to 9 and add to 6: 3 and 3!

Factoring, we get \(\frac{(x+3)(x+3)}{(x+3)(x-6)}\). What's that I see?... could it be, a COMMON FACTOR? Indeed, it is. We can divide out \((x+3)\) factor to yield a more simplified expression: \(\frac{(x+3)}{(x-6)}\). Note that just because we divided out the factor does NOT mean it is no longer a restriction. It simply means that knowing \(x+3\neq0\), the function can be more simply expressed as \(\frac{(x+3)}{(x-6)}\). Please keep this in mind. Find your restrictions before you simplify rather than after to avoid confusion. Let's try another example:

Simplify and find restrictions on \(x\) for \(f(x) = \frac{2x^2+7x-4}{x+4}\)

Aha, an \(a\neq1\) example. In this case, we are looking for four numbers \((ax+b)(cx+d)\), where \(a \times b = 2\)\(b \times d = -4\), and \((a\times d) + (c\times b) = 7\). By trial and error, we find \((2x-1)(x+4)\) to meet our criteria. We can therefore rewrite our function as \(f(x) = \frac{(2x-1)(x+4)}{x+4}\). At this point—before simplifying—we should state the restrictions. In this case, our only restriction is that \(x+4\neq0\). Solving for \(x\), we find that \(x\neq -4\). Now, let's simplify. Common factoring \((x+4)\), we get \(f(x) = 2x-1\). Who would have thought that \(f(x) = \frac{2x^2+7x-4}{x+4}\) was just a linear function (given the restriction \(x+4\neq0\), of course)? We did, that's who.

A graph of \(f(x) = 2x-1\), \(x\neq -4\). Note the discontinuity! 

Multiplying and Dividing Rational Expressions

This one's short because there's not much to know, but a LOT to practice devil

Multiply and state restrictions for \(\frac{x-2}{x^2-4} \times \frac{x+2}{x^2-4x-4}\).

First, we factor. The left expression has a difference of squares in the denominator which simplifies to \((x+2)(x-2)\). For the denominator of the right expression, we need two numbers that multiply to -4 and add to -4; -2 and -2. In factored form, we have: 

\(\frac{x-2}{(x+2)(x-2)} \times \frac{x+2}{(x-2)(x-2)}\) . Wow, that's a lot of common factors; let's highlight them in red: 

\(\frac{\color{red}{x-2}}{\color{red}{(x+2)(x-2)}} \times \frac{\color{red}{x+2}}{(x-2)(x-2)}\). You might notice that \((x+2)\) is highlighted although it's not common to the denominator of the expression. That's because when we multiply rational expressions, we can common factor diagonally! WHAAAAAT?! That's right, diagonally; between the denominator of the left expression and numerator of the right. Before we simplify, however, we should look at restrictions of both functions. We find two: \(x+2\neq 0\) and \(x-2\neq 0\), or \(x\neq 2\) and \(x \neq -2\).

Now, the fun part. Simplifying, we get 

\(\frac{1}{(x-2)(x-2)}\)\(x\neq -2,2\)

How about division? It's about the same, except for one small difference. Let's see:

Simplify \(\frac{x^2-12x+18}{3x^2-9x}\div \frac{x-3}{x^2-x-30}\) and state any restrictions. 

First, factor the expression (see "Factoring Polynomials" for review)

\(\frac{(x-6)(x-3)}{3x(x-3)}\div \frac{x-3}{(x-6)(x+5)}\)

Now, the special part about dividing rational expressions. To divide a rational expression, we have to flip the second expression and convert the division to multiplication. At this point, you might be saying "WHAT?!" Don't worry; this isn't much different from what you already do when you divide. For example, if I asked you to divide 36 by 9, we can represent the division as \(\frac{36}{9}\), or \(36 \times \frac{1}{9}\). In this case, we're doing the same thing but on an algebraic expression level. 

\(\frac{(x-6)(x-3)}{3x(x-3)} \times \frac{(x-6)(x+5)}{x-3}\)

Before we do anything, let's state our restrictions. We have only two: \(3x \neq 0\) and \(x-3 \neq 0\). Solving for \(x\), we get \(x \neq 0,3\).

The only term we can factor out is \((x-3)\). There are two ways we can show this common factoring (\((x-3)\) being factored from the first expression, or the diagonal factoring) and both are correct.

Now, the main course. Upon factoring and simplifying, we get: 


Not the prettiest thing in the world, but at least it's simpler than the original function. 

Adding and Subtracting Rational Expressions

If you've ever dealt with fractions, you already know how this works. For example, if we were asked to add \(\frac{1}{4} + \frac{1}{12}\), we would do so by finding the lowest common multiple (LCM) of the denominator for which the numerator is an integer. In this case, this is 12. To get 12 in the denominator of \(\frac{1}{4}\), we multiply the numerator and denominator by 3. This gives \(\frac{3}{12} + \frac{1}{12}\), or \(\frac{4}{12} = \frac{1}{3}\). Rational expressions are doing the same thing, but with algebraic expressions. 

Simplify \(\frac{2x}{x-3} - \frac{x}{2x-6}\)

This one's straightforward. The lowest common multiple between the two expressions is \(2x-6\). We can achieve this in the first term by multiplying the numerator and denominator by 2, giving \(\frac{4x}{2x-6} - \frac{x}{2x-6}\). Subtracting, we get \(\frac{3x}{2x-6}\). That was too easy, right? How about something a bit more challenging?

Simplify \(\frac{x-2}{x^2-x-2}+\frac{2x-6}{x^2-5x+6}\).

As always, we start by factoring: 

\(\frac{x-2}{(x-2)(x+1)}+\frac{2(x-3)}{(x-2)(x-3)}\). Because we aren't asked to state restrictions, we can simplify right away. In the first term, \((x-2)\) can be factored, and in the second, \((x-3)\). Simplifying, we get: 

\(=\frac{1}{(x+1)}+\frac{2}{(x-2)}\). The lowest common multiple between these two terms is the product of the two terms: \((x+1)(x-2)\). Let's make this the denominator for both terms:


\(= \frac{x-2+2x+2}{(x-2)(x+1)}\)

\(= \frac{3x}{(x-2)(x+1)}\)

Ouf, that sure was long. With that, however, we've covered the basics. Use the flashcards to test your understanding of the lesson's content. 



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