## Definition of Rational Expressions

What is the definition of rational expressions then? Well...

A **rational expression **is an expression whose numerator and denominator are polynomials.

From this definition, which of these do you think are rational expressions?

\( (1) \)\[ \frac{2x}{x+1} \]

\( (2) \)\[ \frac{x^3 + 3x^2 + x + 12}{x^2 + 3x + 5} \]

\( (3) \)\[ \frac{\sqrt{3x}}{4x^2} \]

\((1)\) and \( (2) \)! Did you get that right? Number \(3\) is not a rational expression because the numerator is not a polynomial.

Now that we've learned how to recognize Rational Functions, we should know how to categorize them. This isn't too tricky, as there are only **two** categories to remember: **Proper** and **improper** Rational Functions.

Recognize these terms from anywhere? Well, these are two categories of fractions as well!

With fractions, you might remember that a **proper** fraction has a greater degree in the denominator than the numerator, and that an **improper** fraction has a greater numerator than denominator.

\[ \frac{2}{3} \text{ is a Proper Fraction} \]

\[ \frac{3}{2} \text{ is an Improper Fraction} \]

Well, rational expressions are very similar. In fact, a proper rational expression has a greater degree denominator than numerator, and an improper rational expression has a greater* *degree in the numerator than in the denominator.

\[ \frac{2x^2 + 3}{3x^3 + 2x - 1} \text{ is a Proper Rational Expression} \]

\[ \frac{3x^3 + 2x^2 + x + 1}{x^2 + 2x + 4} \text{ is an Improper Rational Expression} \]

The degree of a polynomial is the highest power of any term in the polynomial.

## Simplifying Rational Expressions

Sometimes, you might have a rational expression that is not in its most simple form. In cases like this, it is your job to **simplify** them. This usually entails cancelling common factors of the numerator and denominator.

Let's take, for example, the following rational expression.

\[ \frac{x(x+1)}{x(2x+7)} \]

What common factor do the numerator and denominator share? \( x \) of course! Just like when you are Simplifying Fractions, when you find a common factor between the numerator and denominator, you can take it out and cancel it:

\[ \frac{x(x+1)}{x(2x+7)} = \frac{\cancel{x}(x+1)}{\cancel{x}(2x+7)} .\]

So your simplified rational expression is

\[ \frac{(x+1)}{(2x+7)}. \]

Let's take a look at some more examples.

Simplify the following rational expressions.

**(a) **

\[ \frac{10(3x+2)(x-1)}{5(4x - 7)(x-1)} \]

**(b)**

\[ \frac{(2x-3)(x+4)}{(2x - 3)} \]

**(c)**

\[ \frac{(3x-10)}{2(3x-10)} \]

**Solution:**

**(a)**

The rational expression can be simplified by cancelling the common factors, \(5\) and \((x-1)\). That gives you

\[ \begin{align} \frac{10(3x+2)(x-1)}{5(4x - 7)(x-1)} &= \frac{5 \cdot 2(3x+2)(x-1)}{5(4x-7)(x-1)} \\ &=\frac{\cancel{5} \cdot2 (3x+2)\cancel{(x-1)}}{\cancel{5}(4x - 7)\cancel{(x-1)}}\\ &= \frac{2(3x+2)}{(4x - 7)} .\end{align} \]

**(b)**

The rational expression can be simplified by cancelling the common factor, \((2x-3)\), to get

\[\begin{align} \frac{(2x-3)(x+4)}{(2x - 3)} &= \frac{\cancel{(2x-3)}(x+4)}{\cancel{(2x - 3)}} \\ &= \frac{(x+4)}{1} \\ &= x+4 \end{align} \]

**(c)**

The rational expression can be simplified by cancelling the common factor, \((3x-10)\), to get

\[ \begin{align} \frac{(3x-10)}{2(3x-10)} &= \frac{\cancel{(3x-10)}}{2\cancel{(3x-10)}} \\ &= \frac{1}{2} .\end{align}\]

## Factoring Rational Expressions

The examples above weren't too tricky to simplify. It was just a case of spotting the common factors in the numerator and denominator and cancelling them out. Well, rational expressions aren't always in that nice simple factored form. Luckily, this is something you can do yourself!

If you factor both the numerator and denominator polynomials of a rational expression, often you will find a common term between the two that you can cancel to simplify.

Let's try some examples. If you need your memory jogged on how to factor polynomials first, head over to our Factoring Polynomials explanation!

Factor, then simplify the following rational expressions.

**(a)**

\[ \frac{ x^2 -2x - 8 }{ x^2 + 5x + 6 } \]

**(b)**

\[ \frac{x^2 -2x - 3 }{x^2 -4x +3} \]

**(c)**

\[ \frac{x^3 - x}{ x^2 - x } \]

**Solution:**

**(a)**

By factoring the numerator and denominator, you can find the common factor of \((x+2)\). So

\[ \begin{align} \frac{ x^2 -2x - 8 }{ x^2 + 5x + 6 } &= \frac{(x+2)(x-4)}{(x+2)(x+3)} \\ &= \frac{\cancel{(x+2)}(x-4)}{\cancel{(x+2)}(x+3)} \\&= \frac{x-4}{x+3} .\end{align} \]

**(b)**

By factoring the numerator and denominator, you can find the common factor of \((x-3)\), so

\[ \begin{align} \frac{x^2 -2x - 3 }{x^2 -4x +3} &= \frac{(x+1)(x-3)}{(x-1)(x-3)} \\ &= \frac{(x+1)\cancel{(x-3)}}{(x-1)\cancel{(x-3)}} \\&= \frac{x+1}{x-1} .\end{align} \]

**(c)**

By factoring the numerator and denominator, you can find the common factors, \(x\) and \( (x-1)\). That gives you

\[ \begin{align} \frac{x^3 - x}{ x^2 - x } &= \frac{x(x^2-1)}{x(x - 1)} \\ &= \frac{x(x^2-1)}{x(x - 1)} \\ &= \frac{x(x+1)(x-1)}{x(x - 1)} \\ &= \frac{\cancel{x}(x+1)\cancel{(x-1)}}{\cancel{x}\cancel{(x - 1)}} \\ &= x + 1. \end{align} \]

## Equivalent Rational Expressions

You may remember working with equivalent fractions. That is, fractions with different denominators that are equal in value. For instance

\[\frac{2}{4} = \frac{4}{8}.\]

Starting with one side of the equation, you can rewrite it in steps until you get to the other side. For this fraction you can start with the right hand side and show that

\[\begin{align} \frac{4}{8} &= \frac{2\cdot 2 }{2 \cdot 2 \cdot 2} \\ &= \frac{\cancel{2}\cdot 2}{2\cdot 2\cdot \cancel{2}}\\ &= \frac{ 2}{2 \cdot 2 } \\ &= \frac{2}{4}. \end{align}\]

Notice that you stopped cancelling before you got all the way done cancelling everything. This is because the goal is to make it look like the left hand side of the equation, not to cancel everything.

Equivalent rational expressions function in a very similar way. Start with one side, and work with it until you can get it to look like the other side.

Let's take a look at an example.

Are the following pairs of rational expressions equivalent to one another?

**(a) **\[ \frac{x^2 + 2}{x^2 - 4}, \text{ and } \frac{2x^2 + 4}{2x^2 - 8} \]

**(b)** \[\frac{(x-2)}{(x-2)(x+4)}, \text{ and } \frac{(x+4)}{(x+4)^2}\]

**(c)** \[\frac{x^2 + 2x + 1}{x}, \text{ and } \frac{x^2 + 2x + 1}{3x}\]

**Solution:**

**(a)**

It is a good idea to start with the one that is more complicated looking. Then

\[ \begin{align} \frac{2x^2 + 4}{2x^2 - 8} &=\frac{2(x^2+2)}{2(x^2-4)} \\&= \frac{\cancel{2}(x^2+2)}{\cancel{2}(x^2-4)} . \end{align}\]

Since you have reached the other expression, you are done and the rational expressions are equivalent.

**(b)**

Another way to do this is to simplify both rational expressions and see if you get the same thing. The numerator and denominator of the first rational expression have a common factor of \((x-2)\), so

\[\begin{align} \frac{(x-2)}{(x-2)(x+4)} &= \frac{\cancel{(x-2)}}{\cancel{(x-2)}(x+4)} \\ &= \frac{1}{(x+4)}. \end{align}\]

The numerator and denominator of the second rational expression have a common factor of \((x+4)\), so

\[ \begin{align} \frac{(x+4)}{(x+4)^2} &= \frac{(x+4)}{(x+4)(x+4)} \\ &= \frac{1\cdot \cancel{(x+4)}}{(x+4)\cancel{(x+4)}} \\ &= \frac{1}{(x+4)}. \end{align}\]

Since you got the same thing after simplifying both of them, they are equivalent rational expressions.

**(c)**

The two rational expressions have the same numerator but different denominators, therefore they are not equal and so aren't equivalent rational expressions.

## Examples with Rational Expressions

Let's look at some more examples to practice everything.

Are the following terms rational expressions?

**(a)** \[ \frac{4x^2 - 2}{x} \]

**(b)** \[ \frac{2}{2x - 4} \]

**(c)** \[2x + 5\]

**Solution:**

**(a)** Yes, since the numerator and denominator are polynomials.

**(b)** Yes, since the numerator and denominator are polynomials.

**(c) **Yes, as it can be written as

\[\frac{2x+5}{1}\]

Let's take a look at categorizing rational expressions as proper or improper.

Categorize the following rational expressions as **proper** or **improper.**

**(a) **\[ \frac{2x + 10}{x^2 - x + 20} \]

**(b) **\[ \frac{x^2}{10x} \]

**(c)** \[ \frac{4x^4 + 3x^3 + \frac{1}{2}x^2 + x}{x^2 - 3x + 2} \]

**(d)** \[ \frac{1}{x+3} \]

**Solution:**

**(a)** Proper, since the degree of the numerator is \(1\) which is less than the degree of the denominator which is \(2\).

**(b) **Improper, since the degree of the numerator is greater than the degree of the denominator.

**(****c)** Improper, since the degree of the numerator is greater than the degree of the denominator.

**(d)** Proper, since the degree of the numerator is less than the degree of the denominator.

What about simplifying rational expressions?

Simplify the following rational expressions.

**(a) **\[ \frac{(x-2)(x+3)(x-1)}{x(x-1)(x-2)} \]

**(b)** \[ \frac{x(3x + 5)}{x(4x - 1)} \]

**(c)** \[ \frac{(x-5)(x^3 + 2x^2 + 3)}{x^3 + 2x^2 + 3} \]

**Solution:**

**(a)**

The numerator and denominator have common factors, \((x-1)\) and \((x-2)\). These can be cancelled to simplify, giving you

\[ \begin{align} \frac{(x-2)(x+3)(x-1)}{x(x-1)(x-2)} &=\frac{\cancel{(x-2)}(x+3)\cancel{(x-1)}}{x\cancel{(x-1)}\cancel{(x-2)}} \\ &= \frac{x+3}{x} . \end{align}\]

**(b)**

The numerator and denominator have a common factor, \(x\)). This can be cancelled to simplify, giving you

\[ \begin{align} \frac{x(3x + 5)}{x(4x - 1)} &= \frac{\cancel{x}(3x + 5)}{\cancel{x}(4x - 1)}\\ &= \frac{3x+5}{4x-1}. \end{align}\]

**(c)**

The numerator and denominator have a common factor, \((x^3 + 2x^2 + 3\)). This can be cancelled to simplify, giving you

\[\begin{align} \frac{(x-5)(x^3 + 2x^2 + 3)}{x^3 + 2x^2 + 3} &= \frac{(x-5)\cancel{(x^3 + 2x^2 + 3)}}{\cancel{x^3 + 2x^2 + 3}} \\ &= x-5. \end{align}\]

Let's take a look at another example.

By first factoring them, simplify the following rational expressions.

**(a)** \[ \frac{2x^2 + 3 + 1}{2x+1} \]

**(b)** \[ \frac{x^2+6x+9}{x^2 + x - 6} \]

**(c)** \[ \frac{3x^3 + 8x^2 + 5x}{x^3 + 8x^2 + 7x} \]

**Solutions:**

**(a) **

By factoring the numerator and denominator, you can find the common factor, \((2x+1)\). Obtain the simplified rational expression by cancelling the common factor:

\[\begin{align} \frac{2x^2 + 3 + 1}{2x+1} &= \frac{(2x+1)(x+1)(2x+1)}{2x+1} \\ &= \frac{\cancel{(2x+1)}(x+1)}{\cancel{2x+1}} \\ &=x+1 .\end{align}\]

**(b)**

By factoring the numerator and denominator, you can find the common factors, \(2\) and \((x+3)\). Obtain the simplified rational expression by cancelling the common factor:

\[\begin{align} \frac{x^2+6x+9}{x^2 + x - 6} &= \frac{(x+3)^2}{(x+3)(x-2)} \\ &= \frac{(x+3)^{\cancel{2}}}{\cancel{(x+3)}(x-2)} \\ &= \frac{x+3}{x-2}. \end{align}\]

**(c)**

By factoring the numerator and denominator, you can find the common factors, \(x\) and \((x+1)\). Obtain the simplified rational expression by cancelling the common factor.

\[ \begin{align} \frac{3x^3 + 8x^2 + 5x}{x^3 + 8x^2 + 7x} &= \frac{x(3x + 5)(x+1)}{x(x+7)(x+1)} \\ &= \frac{\cancel{x}(3x + 5)\cancel{(x+1)}}{\cancel{x}(x+7)\cancel{(x+1)}} \\ &=\frac{3x + 5}{x+7} .\end{align} \]

## Rational Expressions - Key takeaways

- Rational expressions are terms with polynomials as the numerator and denominator.
- Proper rational expressions have a lower degree numerator than denominator.
- Improper rational expressions have a higher degree numerator than denominator.
- Rational expressions can be simplified by factorizing the numerator and denominator, then cancelling any common factors.

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##### Frequently Asked Questions about Rational Expressions

How to simplify rational expressions?

Rational expressions can be simplified by factorizing the numerator and denominator, then cancelling any common factors.

What is a rational expression?

A rational expression is a term with polynomials as the numerator and denominator.

How to find the domain of a rational expression?

Rational expressions do not have domains as they are not functions. However, if a rational expression is part of a function, the domain can be found by finding the roots of the denominator. The domain will not include these numbers as it is impossible to divide by zero.

How to find excluded values in rational expressions?

Find the roots of the denominator, i.e. the values of input for which the denominator is equal to 0.

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