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# Simplifying Rational Expressions

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Practice Simplifying Rational Expressions
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Rational Expression Simplification

How could you use factoring to help simplify the following rational expression?

$\frac{3x^2-27}{x^2+7x+12}$

### Guidance

A rational number is any number of the form $\frac{a}{b}$ , where $b \ne 0$ . A rational expression is any algebraic expression of the form $\frac{a(x)}{b(x)}$ , where $b \ne 0$ . An example of a rational expression is: $\frac{4x^2+20x+24}{2x^2+8x+8}$ .

Consider that any number or expression divided by itself is equal to 1. For example, $\frac{2}{2}=1$ and $\frac{(x+2)}{(x+2)}=1$ . This fact allows you to simplify rational expressions that are in factored form by looking for "1's". Consider the following rational expression:

$\frac{4x^2+20x+24}{2x^2+8x+8}$

Factor both the numerator and denominator completely:

$\frac{4(x+2)(x+3)}{2(x+2)(x+2)}$

Notice that there is one factor of $x+2$ in both the numerator and denominator. These factors divide to make 1, so they "cancel out" (the second factor of $(x+2)$ in the denominator will remain there).

$\frac{4\cancel{(x+2)}(x+3)}{2\cancel{(x+2)}(x+2)}$

Also, the $\frac{4}{2}$ reduces to just $2$ . The simplified expression is:

$\frac{2(x+3)}{x+2}$

Keep in mind that you cannot "cancel out" common factors until both the numerator and denominator have been factored.

A rational expression is like any other fraction in that it is said to be undefined if the denominator is equal to zero. Values of the variable that cause the denominator of a rational expression to be zero are referred to as restrictions and must be excluded from the set of possible values for the variable. For the original expression above, the restriction is $x\ne -2$ because if $x=-2$ then the denominator would be equal to zero. Note that to determine the restrictions you must look at the original expression before any common factors have been cancelled.

#### Example A

Simplify the following and state any restrictions on the denominator.

$\frac{x-2}{x^2-10x+16}$

Solution: To begin, factor both the numerator and the denominator:

$\frac{x-2}{(x-8)(x-2)}$

Cancel out the common factor of $x-2$ to create the simplified expression:

$\frac{\cancel{(x-2)}}{(x-8)\cancel{(x-2)}}$

$\frac{1}{x-8}$

The restrictions are $x\ne 2$ and $x\ne 8$ because both of those values for $x$ would have made the denominator of the original expression equal to zero.

#### Example B

Simplify the following and state any restrictions on the denominator.

$\frac{x^2+7x+12}{x^2-16}$

Solution: To begin, factor both the numerator and the denominator:

$\frac{(x+4)(x+3)}{(x-4)(x+4)}$

Cancel out the common factor of $x+4$ to create the simplified expression:

$\frac{\cancel{(x+4)}(x+3)}{(x-4)\cancel{(x+4)}}$

$\frac{x+3}{x-4}$

The restrictions are $x\ne 4$ and $x\ne -4$ because both of those values for $x$ would have made the denominator of the original expression equal to zero.

#### Example C

Simplify the following and state any restrictions on the denominator.

$\frac{3x^2-7x-6}{4x^2-13x+3}$

Solution: To begin, factor both the numerator and the denominator:

$\frac{(x-3)(3x+2)}{(x-3)(4x-1)}$

Cancel out the common factor of $x-3$ to create the simplified expression:

$\frac{\cancel{(x-3)}(3x+2)}{\cancel{(x-3)}(4x-1)}$

$\frac{3x+2}{4x-1}$

The restrictions are $x\ne 3$ and $x\ne \frac{1}{4}$ because both of those values for $x$ would have made the denominator of the original expression equal to zero.

#### Concept Problem Revisited

$&\frac{3x^2-27}{x^2+7x+12}\\=&\frac{3(x^2-9)}{(x+3)(x+4)}\\=&\frac{3(x+3)(x-3)}{(x+3)(x+4)}\\=&\frac{3\cancel{(x+3)}(x-3)}{\cancel{(x+3)}(x+4)}\\=&\frac{3(x-3)}{x+4}$

where $x\ne -3$ and $x\ne -4$

### Vocabulary

Rational Expression
A rational expression is an algebraic expression that can be written in the form $\frac{a(x)}{b(x)}$ where $b \ne 0$ .
Restriction
Any value of the variable in a rational expression that would result in a zero denominator is called a restriction on the denominator.

### Guided Practice

Simplify each of the following and state the restrictions.

1. $\frac{m^2-9m+18}{4m^2-24m}$

2. $\frac{2x^2-8}{4x+8}$

3. $\frac{c^2+4c-5}{c^2-2c-35}$

1. $\frac{m^2-9m+18}{4m^2-24m}=\frac{(m-6)(m-3)}{(4m)(m-6)}=\frac{(m-3)}{4m}$ , $m \ne 0$ ; $m \ne 6$

2. $\frac{2x^2-8}{4x+8}=\frac{(2)(x-2)(x+2)}{(4)(x+2)}=\frac{(x-2)}{2}$ , $x \ne -2$

3. $\frac{c^2+4c-5}{c^2-2c-35}=\frac{(c+5)(c-1)}{(c-7)(c+5)}=\frac{(c-1)}{(c-7)}$ , $c \ne -5$ ; $c \ne 7$

### Practice

For each of the following rational expressions, state the restrictions.

1. $\frac{7}{x+4}$
2. $\frac{-3}{x-5}$
3. $\frac{5x+1}{5x-1}$
4. $\frac{6}{4x-3}$
5. $\frac{(x+1)}{x^2-4}$
6. $\frac{x-8}{x^2+3x+2}$
7. $\frac{x+6}{x^2-5x-24}$
8. $\frac{5x+2}{2x^2+5x+2}$

Simplify each of the following rational expressions and state the restrictions.

1. $\frac{4}{4x+12}$
2. $\frac{4c^2}{8c^2-4c}$
3. $\frac{10x+5}{2x+1}$
4. $\frac{x-4}{x^2-16}$
5. $\frac{y+1}{y^2+5y+4}$
6. $\frac{c+2}{c^2-5c-14}$
7. $\frac{(b-3)^2}{b^2-6b+9}$
8. $\frac{3n^2-27}{6n+18}$
9. $\frac{6k^2+7k-20}{12k^2-19k+4}$
10. $\frac{4x^2-4x-3}{2x^2+3x-9}$