# Rational Expressions

Hi, and welcome to this video about rational expressions.

Before we talk about what rational expressions are and the operations that can be performed with them, it may be a good idea to review some terminology. A **polynomial** is a group of algebraic or numeric terms that are joined by the operations of addition or subtraction. There are different types of polynomials based on the number of terms that are present:

### Types of Polynomials:

Examples | # of Terms | Type |
---|---|---|

\(4\), \(5x\), \(3x^4\), \(8xy^2\) | 1 | Monomial |

\((x+3)\), \((x^2-1)\), \((3xy+2y)\) | 2 | Binomial |

\((x^2+5x+6)\), \((x^2-2xy+y^2)\) | 3 | Trinomial |

A **rational expression** is nothing more than a ratio of polynomials. As you know from previous practice with **ratios**, you cannot divide by 0. It is important to keep this in mind when dealing with rational expressions because allowing a value of 0 in the denominator would create an expression that is “undefined.”

Using function notation for polynomials, such as \(p(x)\) and \(g(x)\), a rational expression can be defined like this:

\(\frac{p(x)}{g(x)}\)\(\text{ where }g(x)\neq 0\)

Here’s an example:

\(\frac{5x}{x-2}\)

This example shows a rational expression with a monomial, \(5x\), in the numerator and a binomial, \((x-2)\), in the denominator. The value \(x=2\) is the excluded value, as it would result in a denominator of 0. This expression cannot be simplified further.

\(\frac{5x}{x-2}\), \(x\neq 2\)

## Addition and Subtraction of Rational Expressions

Rational expressions cannot be added or subtracted unless they share a common denominator. Algebraic rules allow us to adjust fractions to create common denominators as long as we make the same adjustment to the numerator. Let’s look at an example with fractions:

\(\frac{1}{5}+\frac{3}{7}\)

In order to add \(\frac{1}{5} + \frac{3}{7}\), we must create a common denominator. Specifically, we need to determine the **least common denominator**, meaning the smallest multiple of 5 and 7. In this case, that number is 35. The adjustment to each fraction that needs to be made to create the common denominator is:

\(\frac{1}{5} (\frac{7}{7}) + \frac{3}{7} (\frac{5}{5})\)

\(\frac{7}{35}+\frac{15}{35}\)

\(=\frac{22}{35}\)

We need to multiply the first denominator of the first expression by 7 to get to 35, but we also must multiply the numerator by the same value. Because we have simply created an equivalent fraction to allow us to add. Likewise, the second expression must be multiplied by \(\frac{5}{5}\), in order to create 35 in the denominator. After these adjustments are made and the denominators are the same, simplify the numerators:

\(\frac{7+15}{35} \rightarrow \frac{22}{35}\)

Rational expressions are added and subtracted the same way. Typically, the expressions need to be factored before the least common denominator can be determined and domain restrictions (excluded values) should be noted. Consider this example:

\(\frac{3x}{x-2} + \frac{5}{x^2-x-2}\)

\(\frac{3x}{x-2} + \frac{5}{(x-2)(x+1)}\)

Now we want to determine the lowest common denominator. What is the smallest multiple of \((x-2)\) and \((x-2)(x+1)\)?

Alright now that we have our equations written out, we want to make sure that we don’t have any domains that need to be excluded. Which, we do. Remember we don’t want 0 in the denominator position. So, in these scenarios, we know that \(x\neq 2\), or over here, -1. If \(x=-1\) this would end up being 0, multiplied by another term, still remains 0. The 0 in the denominator, we can’t have that. Over here, if \(x=2\), \(2-2=0\), again, we can’t have a 0 in the denominator, so these are our two terms, our domains that need to be excluded.

Alright, now we need to adjust the first expression by multiplying by the factor needed to match the least common denominator. So if we want our first term here, to match this term over here in the denominator position, we’re going to multiply by \(x+1\) in the numerator and the denominator.

\(\frac{3x}{x-2} \frac{(x+1)}{(x+1)} + \frac{5}{(x-2)(x+1)}\)

Now we’re going to rewrite the expression as a fraction, and simplify the numerator. And now we have our answer:

\(\frac{3x^{2}+3x}{(x+1)(x-2)}+\frac{5}{(x-2)(x+1)}\)\(=\frac{3x^{2}+3x+5}{(x-2)(x+1)}\)

## Multiplication and Division of Rational Expressions

### Multiplying Rational Expressions

Here are the three steps to multiplying rational expressions. Now, remember, when multiplying fractions, numerators and denominators are multiplied straight across.

Step #1:Factor the numerator and denominator of each expression being multiplied.Step #2:Simplify by canceling out common factors from the numerator and the denominator.

Step #3:The final answer is what is left after canceling. You may be asked to include domain restrictions with your solution.

Let’s use these steps to solve an example problem:

\(\frac{3x}{4x-8}\cdot \frac{2x^{2}-4x}{9x}\)

\(\frac{3x}{4(x-2)}\cdot \frac{2x(x-2)}{9x}\)

\(\frac{6x^{2}(x-2)}{36x(x-2)}\)

Now, because we have like terms in the numerator and the denominator position, we’re able to cancel them out. That leaves us with:

\(\frac{6x^2}{36x}\)

But we can simplify this even further, remember, 6 is a **factor**of 36, so let’s simplify:

\(\frac{x^2}{6x}\)

And yet, we can simplify this again, remember, you have an \(x\) in the numerator and an \(x\) in the denominator, so let’s simplify:

\(\frac{x}{6}\)

And now we have our answer, \(\frac{x}{6}\). But that’s not the complete answer. Remember, we have some domain that we have to exclude. Up here, \(x\neq 2\) because \(2\times 4-8=0\). And we can’t have a 0 in the denominator. So 2 is out, \(x\neq 2\). Also, \(x\neq 0\), because \(0\times 9=0\), and again, give us a 0 in the denominator. So the domains we have to exclude from this answer are 2 and 0. So our answer is \(\frac{x}{6},x\neq 2,0\).

### Dividing Rational Expressions

Dividing rational expressions includes one extra step at the beginning of the process. When dividing by a fraction, it is the same as multiplying by the reciprocal of the second fraction. You can remember this rule as, “Keep, Change, and Flip” which translates to keep the first fraction, change the operation to multiplication, and take the reciprocal (or flip) of the second fraction.

Keep in mind that domain restrictions must be considered from both the numerator and denominator of the second fraction because of the “flip” in the division process.

Here’s an example:

\(\frac{9x^2}{x^2+12x+36} \div \frac{12x}{x^2+6x} \)

Now, remember our three steps: *keep* the first fraction, *change* the operation, and then *flip*. Here we go:

\(\frac{9x^2}{x^2+12x+36} \times \frac{x^2+6x}{12x}\), \(x\neq 0\), \(-6\)

Here is now where we multiply, cause we kept the first fraction, we changed to multiplication, and then we flipped the fraction over here. So, time to multiply.

\(\frac{9x^{2}}{(x+6)(x+6)}\cdot \frac{(x+6)}{12}=\frac{3x^{2}}{4(x+6)}\)

So now we have our answer: \(\frac{3x^{2}}{4(x+6)}\).

But remember, that’s not our complete answer if we don’t include our restricted domain, we have \(x\neq 0\), and \(x\neq -6\). Remember we have to make sure that we don’t have a 0 in the denominator or the numerator of our second term.

I hope this review was helpful! See you next time!

## Practice Questions

**Question #1:**

Which polynomial is considered a binomial?

**Answer:**

A polynomial consists of one or more monomials combined by addition or subtraction.

A monomial has one term, such as \(2x\).

A binomial has two terms, such as \(3x−7\).

A trinomial has three terms, such as \(x^2+2x-4\).

**Question #2:**

Add the following polynomials: \(\frac{x+2}{3x}+\frac{x-3}{6x}\)

**Answer:**

**Step 1:** Rewrite the expression with a common denominator. In this case, we need to multiply the numerator and denominator by 2 in the first fraction so that our denominator becomes \(6x\). \(\frac{x+2}{3x}+\frac{x-3}{6x}\) becomes \(\frac{(x+2)(2)}{(3x)(2)}+\frac{(x+3)}{(6x)}\), which simplifies to \(\frac{2x+4}{6x}+\frac{x-3}{6x}\).

**Step 2:** Add the numerators, and combine like terms. \(\frac{2x+4}{6x}+\frac{x-3}{6x}\) becomes \(\frac{3x+1}{6x}\) which is the final answer.

**Question #3:**

What are the domain restrictions for the following expression?

\(\frac{7x+2}{x^2-4}\)

\(x\neq2\) and \(x\neq-2\)

\(x\neq7\) and \(x\neq-4\)

**Answer:**

A rational expression cannot have a zero in the denominator. For the expression \(\frac{7x+2}{x^2-4}\) there are two values for *x* that will produce a zero in the denominator. It can be helpful to break apart the denominator \((x^2-4)\) into \((x+2)(x-2)\) in order to identify these domain restrictions. Now that the denominator is factored, we can see that if *x* is 2, the term on the left will be zero. Similarly, if *x* is -2, the term on the right will be zero. Therefore, *x* cannot be 2 or -2. These are referred to as domain restrictions: \(x\neq2\) and \(x\neq-2\).

**Question #4:**

\(\frac{12x^2}{5y^3}\times\frac{20y^4}{6x^3}\)

**Answer:**

**Step 1:** When multiplying fractions, we simply multiply straight across. Numerator times numerator, and denominator times denominator. \(\frac{12x^2}{5y^3}\times\frac{20y^4}{6x^3}\) now becomes \(\frac{240x^2y^4}{30x^3y^3}\).

**Step 2:** Simplify by canceling out common factors that appear in the numerator and denominator. \(\frac{240x^2y^4}{30x^3y^3}\) now becomes \(\frac{8y}{x}\).

**Question #5:**

Divide the following: \(\frac{x^2-x-12}{3x-15}\div\frac{x^2-9}{24x-72}\)

**Answer:**

**Step 1:** “Keep, Change, Flip”

*Keep* the fraction on the left in its original form. *Change* the division sign to a multiplication sign. *Flip* the second fraction.

\(\frac{x^2-x-12}{3x-15}\div\frac{x^2-9}{24x-72}\) now becomes \(\frac{x^2-x-12}{3x-15}\times\frac{24x-72}{x^2-9}\).

**Step 2:** Factor.

Factoring makes the process of canceling more straight forward later on.

\(\frac{x^2-x-12}{3x-15}\times\frac{24x-72}{x^2-9}\) now becomes \(\frac{(x-4)(x+3)}{3(x-5)}\times\frac{24(x-3)}{(x+3)(x-3)}\).

**Step 3:** Identify the restrictions.

Restrictions are the values for *x* that will produce a zero in the denominator.

*x* cannot be 5 because that would produce a zero in the denominator of the first fraction. *x* cannot be 3 or -3 because that would produce a zero in the denominator of the second fraction. The restriction are \(x≠5,3,\text{ or }-3\)

**Step 4:** Simplify. Cancel out common factors in the numerator and denominator.

\(\frac{(x-4)(x+3)}{3(x-5)}\times\frac{24(x-3)}{(x+3)(x-3)}\) now becomes \(\frac{8(x-4)}{x-5}\).

Final Answer: \(\frac{8(x-4)}{x-5} x ≠ 5,3,-3\).