2.3 Rational Expressions

Learning Objectives

By the end of this section, you will be able to:

• assess the domain of an algebraic expression and describe it with various notations
• add, subtract, multiply, and divide rational expressions
• simplify and cancel legally in rational expressions

A First Look at Domain

A super important question to keep in the back of your mind while looking at an expression (or later, a function) is, "What kinds of values are actually legal to plug in for the variable here?" For example, I've told you two illegal actions already: 1.) Dividing by zero, and 2.) Taking an even root of a negative number. We'll see more legality concerns when we learn about logarithmic functions and trig functions.

The domain of an algebraic expression is the set of all numbers that are legal to plug in for the variable.

When we're reporting back information about domain, we use a variety of notation styles. These are all ways of saying, "The set of all real numbers, except for the number 0."

\[ \{ x \: | \: x \neq 0 \}, \quad \quad \{ x \: : \: x \neq 0 \}, \quad \quad (-\infty, 0) \cup (0, \infty), \quad \quad `` x \in \mathbb{R}, x \neq 0 " \notag \]

Here's a rundown on the symbols involved:

• A pair of braces, \( \{ \quad \} \), is code for "I'm about to give you a bunch of objects, such as a set of numbers." It starts with a nickname for the elements, like the variable name \(x\).
• Inside the braces, a pipe symbol | is code for "such that" and it's telling you the rules for \(x\) are coming next. You might also see a colon : used for this purpose.
• A pair of numbers (or including the \( \infty \) symbol) with parentheses ( , ) or brackets [ , ] or a mix of both like ( , ] or [ , ) are telling you about an interval: a chunk of the number line starting at the first number and ending at the second number. A round parenthesis indicates that the endpoint number is not included, and a square bracket indicates that it is included.
• The symbol \( \cup \) stand for "union" and it means "this set and that set." So \( (-\infty, 0) \cup (0, \infty) \) means, "All real numbers from negative infinity up to but not including 0, and also all the numbers strictly bigger than zero, forever to infinity." If you see the symbol \( \cap\) between two sets, it stands for "intersection" and means "take all the numbers that appear in both of these sets."
• The symbol \( \in \) means "in" and the symbol \( \mathbb{R} \) means "the set of real numbers," so \( x \in \mathbb{R} \) means "\(x\) is in the set of real numbers," aka "\(x\) is any real number."

You can illustrate some sets using a number line. The picture below is showing the set \( (-4, \infty) \).

The next picture shows \( [-2, 4) \).

Make sure to read the next example carefully.

Example \(\PageIndex{1}\)

What is the domain of the expression?

1. \( \frac{1}{x} \)
2. \( \sqrt{x} \)
3. \( x^2 + 2 \)
4. \( \frac{1}{\sqrt{x}} \)
5. \( \frac{1}{x-2} \)
6. \( \sqrt[4]{x + 1} \)

Solution

1. In this expression, if we let \( x \) be 0, terrible things will happen. So we say the domain is "all real numbers except zero," written \( \{x \: | \: x \neq 0 \} \) or even just "\(x \neq 0 \)" if it's clear what we mean.
2. We're not allowed to take square roots of negative numbers, so we can only let \(x\) be 0 or bigger. We write \( [0, \infty) \). You can also write this as \( \{ x \: | \: x \geq 0 \} \).
3. Trick question! Everything is legal for \(x\) because I'm not at risk of dividing by zero or taking a bad root. The domain is "all real numbers," written \(x \in \mathbb{R}\) or \( (-\infty, \infty ) \).
4. Because of the square root, I can't let \(x\) be negative, BUT ALSO, because I'm dividing here, I can't let \(x\) be 0 either. So the domain is "\(x\) strictly greater than 0," written \(\{ x \: | \: x > 0 \} \) or \( (0, \infty) \).
5. If I plug in a number for \(x\) that causes the denominator to be 0, it's illegal! What number would have that effect? Yeah, if \(x\) is 2. So he's not allowed. The domain is \( \{ x \: | \: x \neq 2 \} \). Notice that \( x = 0\) is fine, because that would lead to dividing by \( -2\), which is legal.
6. I can't let whatever is under the fourth root be negative, so I have to only choose \(x\)-values that result in 0 or positive numbers. As long as \(x\) is at least \(-1\), the expression \(x + 1\) is at least 0. So the domain is \( [-1, \infty) \).

Exercise \(\PageIndex{1}\)

1. Translate "the set of real numbers starting at but not including 2, ending at and including 12" into math.
2. Translate \( (-\infty, 18] \) into English.
3. What is the domain of the expression \( \frac{1}{x+4} \)?

Answer

1. \( (2, 12] \) or even \( \{ x \: | \: 2 < x \leq 12 \}\).
2. "The set of all real numbers less than or equal to 18."
3. The only problem child here would be plugging in \(-4\), which would cause division by 0. So \( \{x \: | \: x \neq -4\} \).

Rational Expressions

A rational expression is just a quotient (get it, ratio-nal) of polynomials. We need to be able to deal with these expressions in order to deal with rational functions and simplify stuff later in Calculus. Here are some rational expressions:

\[ \frac{x}{x^2 - 1}, \quad \quad \frac{x^3 - x^2 + 4}{x - 3}, \quad \quad \frac{3a^{12} - 6a^7}{a^4 + a + 1} \notag \]

Something like \( \dfrac{x^{3/2} + 1}{x} \) would not count as a rational expression because the top is not a polynomial (it doesn't have only positive integer powers of \(x\)).

Caution

Be very careful with cancelling in rational expressions. You can only cancel a factor if it is being multiplied by everything, aka the numerator and denominator have a common factor!

(please learn this now—if i see one more calc student do this on an exam i'm gonna jump off a bridge)

Now that I've screamed my head off about simplifying, let's learn multiplication and division. The good news is, this is just a combo of the working with fractions section and the multiplying polynomials section.

To multiply rational expressions, use the usual fraction multiplication rule \( \frac{A}{B} \cdot \frac{C}{D} = \frac{AB}{CD} \), cancelling any common factors. In my class, you can leave the final answer in factored form (do not FOIL top and bottom for no reason).

Example \(\PageIndex{2}\)

Simplify.

1. \( \dfrac{3x^2}{4y} \cdot \dfrac{2y}{5x^3} \)

2. \( \dfrac{1}{a+2} \cdot \dfrac{a-2}{a^2 - 4} \)

3. \( \dfrac{x^2 - 7x + 6}{x-3} \cdot \dfrac{x^2 - 9}{ x-6} \)

Solution

1. Nothing crazy is going on here, just letters involved. We have

\[ \dfrac{3x^2}{4y} \cdot \dfrac{2y}{5x^3} = \dfrac{6x^2 y}{20 x^3y} = \frac{3}{10x} \notag \]

2. Now there are polynomials involved with multiple terms, so we have to be careful with products and cancelling. Factor everything as much as possible (look for shortcuts like "difference of squares") beforehand to see if there are common factors to kill off, and use the heck outta parentheses.

\[ \dfrac{1}{a+2} \cdot \dfrac{a-2}{a^2 - 4} = \dfrac{1}{(a+2)} \cdot \dfrac{(a-2)}{(a +2)(a-2)} = \frac{(a-2)}{(a+2)(a+2)(a-2)} = \frac{1}{(a+2)^2} \notag \]

3. Factor everything beforehand to see if anything magical happens, then multiply and cancel common factors.

\[ \dfrac{x^2 - 7x + 6}{x-3} \cdot \dfrac{x^2 - 9}{ x-6} = \dfrac{(x-1)(x-6)}{(x-3)} \cdot \dfrac{(x-3)(x+3)}{( x-6)} = \frac{(x-1)(x+3)}{1} = (x-1)(x+3) \notag \]

As for dividing rational expressions, if you ever see something like \( \dfrac{x^2 + 2x + 1}{x - 1} \div \dfrac{x+1}{x-1} \) or a compound fraction \( \dfrac{\frac{x^2 + 2x + 1}{x - 1}}{\frac{x+1}{x-1}} \) (sweet sister Frances!), just do the old "multiply by the reciprocal" trick. Get your practice here:

Exercise \(\PageIndex{2}\)

1. Simplify \( \dfrac{x^2 + 2x + 1}{x - 1} \div \dfrac{x+1}{x-1} \).

2. (Also practicing the \( ac\) method...) Simplify \( \dfrac{\frac{y^2 - 36}{2y^2 + 11y - 6}}{\frac{2y^2 - 2y - 60}{8y-4}} \).

Answer

1. Flip over the second fraction and multiply instead:

\[ \dfrac{x^2 + 2x + 1}{x - 1} \cdot \dfrac{x-1}{x+1} = \dfrac{(x+1)^2}{(x-1)}\cdot \dfrac{(x-1)}{(x+1)} = \dfrac{(x+1)}{1} = x+1 \notag \]

2. For this one, take the reciprocal of the downstairs fraction and multiply by that. This is your factoring practice challenge as well.

\[ \frac{(y +6)(y-6)}{(2y-1)(y+6)} \cdot \frac{4(2y-1)}{(2y+10)(y-6)} = \frac{4}{2y+10} = \frac{4}{2(y+5)} = \frac{2}{y+5} \notag \]

To add and subtract rational expressions, first factor everything as much as possible. Get a common denominator by multiplying each fraction top and bottom by any factors needed (use parentheses!). Then use the fraction rule \( \frac{A}{C} + \frac{B}{C} = \frac{A+B}{C} \) and simplify.

Example \(\PageIndex{3}\)

Simplify \( \dfrac{1}{x^2 - 1} - \dfrac{2}{(x+1)^2} \).

Solution

The denominator of the first fraction is a difference of squares and factors as \( (x+1)(x-1) \). The least common denominator for these two expressions must then contain two copies of \( (x+1) \) and one copy of \( (x-1) \). I multiply each term top and bottom by what it's lacking:

\[ \dfrac{1\cdot (x+1)}{(x+1)(x-1)\cdot (x+1)} - \dfrac{2\cdot (x-1)}{(x+1)^2\cdot (x-1)} = \dfrac{(x+1)}{(x+1)^2(x-1)} - \dfrac{2 (x-1)}{(x+1)^2 (x-1)} \notag \]

Now that the denominators match, we subtract (be careful and use parentheses!) and simplify.

\[ \dfrac{(x+1) - 2(x-1) } {(x+1)^2 (x-1)} =\dfrac{x+1 - 2x +2} {(x+1)^2 (x-1)} = \dfrac{-x + 3 } {(x+1)^2 (x-1)} \notag \]

Exercise \(\PageIndex{3}\)

Simplify \( \dfrac{ 8x}{x^2 + 4x + 4} + \dfrac{2}{x-2} \)

Answer

\( \dfrac{ 10x^2 - 8x + 8}{(x+2)^2(x-2)} \) or \( \dfrac{ 2(5x^2 - 4x + 4)}{(x+2)^2(x-2)} \)

Another type of compound fraction you will encounter (esp in Calc I) combines addition or subtraction of rational expressions in the numerator of a giant fraction. There are two ways of dealing with this situation.

Example \(\PageIndex{4}\)

Simplify the difference quotient \( \dfrac{ \dfrac{1}{1+x+h} - \dfrac{1}{1+x}}{h} \).

Solution

Option 1 is to look just at the numerator for the moment and use a common denominator to combine the terms. These denominators are distinct factors, so we have to multiply top and bottom to make them match. I could move the overall division by \(h\) out front of some parentheses to make this easier to write:

\[ \dfrac{1}{h} \left( \dfrac{1\cdot (1+x)}{(1+x+h)\cdot (1+x)} - \dfrac{1 \cdot (1+x+h)}{(1+x)\cdot(1+x+h)} \right) = \dfrac{1}{h} \left( \dfrac{(1+x) - (1+x+h)}{(1+x+h)(1+x)} \right) = \dfrac{1}{h} \left( \dfrac{ 1+x -1-x-h }{(1+x+h) (1+x)} \right) \notag \]

Then I notice that when I simplify the numerator, everything kills each other off but the \(h\), which I can cancel with the \( \frac{1}{h}\) out front. This cleans up nice to

\[ \dfrac{1}{h} \left( \dfrac{ h }{(1+x+h)\cdot (1+x)} \right) = \dfrac{1}{(1+x+h)\cdot (1+x)}. \notag \]

Option 2 is to use a technique called clearing denominators. To do this, you multiply the compound fraction top and bottom by whatever would be needed to cancel every denominator involved. This is the least common multiple of all the denominators you see. Watch what happens when we distribute.

\[ \dfrac{\left( \dfrac{1}{1+x+h} - \dfrac{1}{1+x} \right)}{h} \cdot \dfrac{ (1+x+h)(1+x)}{ (1+x+h)(1+x)} =\dfrac{\left( \dfrac{(1+x+h)(1+x)}{(1+x+h)} - \dfrac{(1+x+h)(1+x)}{(1+x)} \right)}{h} = \dfrac{ (1+x) - (1+x+h)}{h(1+x+h)(1+x)} \notag \]

This is the same expression we found during Option 1, and simplifies the same way! Clearing denominators can be a great way to clean up compound fractions in general.

Rationalizing the Denominator

This isn't about "rational expressions" per se, but while we're talking about fractiony stuff, let me show you a special technique. Any time you see an expression of the form \( A + B \sqrt{C}\), "something plus somebody with a square root," there is a trick we can do to make the square root go away. Multiplying \( A + B \sqrt{C}\) by his conjugate radical, \( A - B \sqrt{C} \), causes something magical to happen. FOIL this out...

\[ ( A + B \sqrt{C})(A - B \sqrt{C}) = A^2 + AB \sqrt{C} - AB\sqrt{C} - (B\sqrt{C})^2 = A^2 - B^2C \notag \]

Et voilà, no radicals in sight. To make the conjugate of an expression, just copy it but change the sign in the middle. We describe this in math like so: The conjugate of \(A \pm B\sqrt{C} \) is \( A \mp B\sqrt{C} \). This is a trick that comes in handy in Calc I when you're evaluating limits, usually applied to a numerator or denominator of a fractional expression. Just make sure to multiply on top and bottom, so you're not illegally changing the expression!

Example \(\PageIndex{5}\)

Rationalize the numerator of \( \dfrac{\sqrt{9+h} - 3}{h} \).

Solution

The conjugate of the numerator is \( \sqrt{9+h} \textcolor{red}{+} 3 \), so we multiply top and bottom by that:

\[ \dfrac{(\sqrt{9+h} - 3)}{h} \cdot \frac{( \sqrt{9+h} + 3)}{( \sqrt{9+h} + 3)} = \frac{(\sqrt{9+h})^2 - 3 \sqrt{9+h} + 3\sqrt{9+h} - 3^2}{h(\sqrt{9+h} + 3)} \notag \]

Now simplify...

\[ (\frac{\sqrt{9+h})^2 - 9}{h(\sqrt{9+h} + 3)} = \frac{(9+h) - 9}{h(\sqrt{9+h} + 3)} = \frac{h}{h(\sqrt{9+h} + 3)} = \frac{1}{(\sqrt{9+h} + 3)} \notag \]

Exercise \(\PageIndex{4}\)

Rationalize the denominator of \( \dfrac{3}{x + \sqrt{1-x}} \).

Answer

\( \dfrac{3(x-\sqrt{1-x})}{x^2 + x- 1} \)

To wrap up this section, I'd like to call out some mistakes I see students make ALL THE TIME so you can avoid them from the beginning.

YES	NO!!!!!
\( \textcolor{green}{ A - (B+C) = A - B - C } \)	\( \textcolor{red}{ A - (B+C) \neq A - B + C }\)
\( \textcolor{green}{(AB)^2 = A^2 B^2} \)	\( \textcolor{red}{ (A+B)^2 \neq A^2 + B^2 }\)
\( \textcolor{green}{\dfrac{A+B}{C} = \dfrac{A}{C} + \dfrac{B}{C} } \)	\( \textcolor{red}{\dfrac{A}{B+C} \neq \dfrac{A}{B} + \dfrac{A}{C}} \)
\(\textcolor{green}{ \sqrt{AB} = \sqrt{A}\sqrt{B}} \)	\( \textcolor{red}{\sqrt{A+B} \neq \sqrt{A} + \sqrt{B} } \)
\( \textcolor{green}{\dfrac{AB}{AC} = \dfrac{B}{C} } \)	\( \textcolor{red}{ \dfrac{A+B}{A} \neq B }\)
\( \textcolor{green}{ \sqrt{a^2b^2} = \sqrt{(ab)^2} = ab, \quad \text{ for } a,b \geq 0 } \)	\( \textcolor{red}{ \sqrt{a^2 + b^2} \neq a + b } \)

And one last time, for the people in the back...