### Excluded Values for Rational Expressions

A simplified rational expression is one where the numerator and denominator have no common factors. In order to simplify an expression to **lowest terms**, we factor the numerator and denominator as much as we can and cancel common factors from the numerator and the denominator.

**Simplifying Rational Expressions**

Reduce each rational expression to simplest terms.

a) \begin{align*}\frac{4x-2}{2x^2+x-1}\end{align*}

\begin{align*}\text{Factor the numerator and denominator completely:} \qquad \frac{2(2x-1)}{(2x-1)(x+1)}\!\\ \\ \text{Cancel the common factor} \ (2x - 1): \qquad \qquad \qquad \qquad \qquad \frac{2}{x+1}\end{align*}

b) \begin{align*}\frac{x^2-2x+1}{8x-8}\end{align*}

\begin{align*}\text{Factor the numerator and denominator completely:} \qquad \frac{(x-1)(x-1)}{8(x-1)}\!\\ \\ \text{Cancel the common factor}\ (x - 1): \qquad \qquad \qquad \qquad \qquad \ \ \frac{x-1}{8}\end{align*}

c) \begin{align*}\frac{x^2-4}{x^2-5x+6}\end{align*}

\begin{align*}\text{Factor the numerator and denominator completely:} \qquad \frac{(x-2)(x+2)}{(x-2)(x-3)}\!\\
\\
\text{Cancel the common factor} (x - 2): \qquad \qquad \qquad \qquad \qquad \quad \frac{x+2}{x-3}\end{align*}

When reducing fractions, you are only allowed to cancel common **factors** from the denominator but NOT common terms. For example, in the expression \begin{align*}\frac{(x+1) \cdot (x-3)}{(x+2) \cdot (x-3)}\end{align*}, we can cross out the \begin{align*}(x - 3)\end{align*} factor because \begin{align*}\frac{(x-3)}{(x-3)}=1\end{align*}. But in the expression \begin{align*}\frac{x^2+1}{x^2-5}\end{align*} we can’t just cross out the \begin{align*}x^2\end{align*} terms.

Why can’t we do that? When we cross out terms that are part of a sum or a difference, we’re violating the order of operations (PEMDAS). Remember, the fraction bar means division. When we perform the operation \begin{align*}\frac{x^2+1}{x^2-5}\end{align*}, we’re really performing the division \begin{align*}(x^2+1) \div (x^2-5)\end{align*} — and the order of operations says that we must perform the operations inside the parentheses before we can perform the division.

Using numbers instead of variables makes it more obvious that canceling individual terms doesn’t work. You can see that \begin{align*}\frac{9+1}{9-5}=\frac{10}{4}=2.5\end{align*} — but if we canceled out the 9’s first, we’d get \begin{align*}\frac{1}{-5}\end{align*}, or -0.2, instead.

**Finding Excluded Values of Rational Expressions**

Whenever there’s a variable expression in the denominator of a fraction, we must remember that the denominator could be zero when the independent variable takes on certain values. Those values, corresponding to the vertical asymptotes of the function, are called **excluded** values. To find the excluded values, we simply set the denominator equal to zero and solve the resulting equation.

Find the excluded values of the following expressions.

a) \begin{align*}\frac{x}{x+4}\end{align*}

\begin{align*}\text{When we set the denominator equal to zero we obtain:} \quad \ \ x+4=0 \Rightarrow x=-4\!\\ \\ \text{So} \ \mathbf{-4} \ \text{is the excluded value.}\end{align*}

b) \begin{align*}\frac{2x+1}{x^2-x-6}\end{align*}

\begin{align*}\text{When we set the denominator equal to zero we obtain:} \qquad x^2-x-6=0\!\\ \\ \text{Solve by factoring:} \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \qquad (x-3)(x+2)=0\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \Rightarrow x=3 \ \text{and}\ x = -2\!\\ \\ \text{So}\ \mathbf{3}\ \mathbf{and}\ \mathbf{-2} \ \text{are the excluded values.}\end{align*}

**Removable Zeros**

Removable zeros are those zeros from the original expression, but is not a zero for the simplified version of the expression. However, we have to keep track of them, because they were zeros in the original expression. This is illustrated in the following examples.

#### Determining Removable Values

1. Determine the removable values of \begin{align*}\frac{4x-2}{2x^2+x-1}\end{align*}.

Notice that in the expressions in Example A, we removed a division by zero when we simplified the problem. For instance, we rewrote \begin{align*}\frac{4x-2}{2x^2+x-1}\end{align*} as \begin{align*}\frac{2(2x-1)}{(2x-1)(x+1)}\end{align*}. The denominator of this expression is zero when \begin{align*}x = \frac{1}{2}\end{align*} or when \begin{align*}x = -1\end{align*}.

However, when we cancel common factors, we simplify the expression to \begin{align*}\frac{2}{x+1}\end{align*}. This reduced form allows the value \begin{align*}x = \frac{1}{2}\end{align*}, so \begin{align*}x = -1\end{align*} is its only excluded value.

Technically the original expression and the simplified expression are not the same. When we reduce a radical expression to its simplest form, we should specify the removed excluded value. In other words, we should write our final answer as \begin{align*}\frac{4x-2}{2x^2+x-1}=\frac{2}{x+1}, x \neq \frac{1}{2}\end{align*}.

2. Determine the removable values of the expressions from parts b and c of the first example.

We should write the answer from the first example, part *b* as \begin{align*}\frac{x^2-2x+1}{8x-8}=\frac{x-1}{8}, x \neq 1\end{align*}.

The answer from the first example, part *c* as \begin{align*}\frac{x^2-4}{x^2-5x+6}=\frac{x+2}{x-3}, x \neq 2\end{align*}.

### Example

#### Example 1

Find the excluded values of \begin{align*}\frac{4}{x^2-5x}\end{align*}.

\begin{align*}\text{When we set the denominator equal to zero we obtain:} \quad \ \ x^2-5x=0\!\\ \\ \text{Solve by factoring:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x(x-5)=0\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \Rightarrow x=0 \ \text{and} \ x = 5\!\\ \\ \text{So} \ \mathbf{0 \ and \ 5}\ \text{are the excluded values.}\end{align*}

### Review

Reduce each fraction to lowest terms.

- \begin{align*}\frac{4}{2x-8}\end{align*}
- \begin{align*}\frac{x^2+2x}{x}\end{align*}
- \begin{align*}\frac{9x+3}{12x+4}\end{align*}
- \begin{align*}\frac{6x^2+2x}{4x}\end{align*}
- \begin{align*}\frac{x-2}{x^2-4x+4}\end{align*}
- \begin{align*}\frac{x^2-9}{5x+15}\end{align*}
- \begin{align*}\frac{x^2+6x+8}{x^2+4x}\end{align*}
- \begin{align*}\frac{2x^2+10x}{x^2+10x+25}\end{align*}
- \begin{align*}\frac{x^2+6x+5}{x^2-x-2}\end{align*}
- \begin{align*}\frac{x^2-16}{x^2+2x-8}\end{align*}
- \begin{align*}\frac{3x^2+3x-18}{2x^2+5x-3}\end{align*}
- \begin{align*}\frac{x^3+x^2-20x}{6x^2+6x-120}\end{align*}

Find the excluded values for each rational expression.

- \begin{align*}\frac{2}{x}\end{align*}
- \begin{align*}\frac{4}{x+2}\end{align*}
- \begin{align*}\frac{2x-1}{(x-1)^2}\end{align*}
- \begin{align*}\frac{3x+1}{x^2-4}\end{align*}
- \begin{align*}\frac{x^2}{x^2+9}\end{align*}
- \begin{align*}\frac{2x^2+3x-1}{x^2-3x-28}\end{align*}
- \begin{align*}\frac{5x^3-4}{x^2+3x}\end{align*}
- \begin{align*}\frac{9}{x^3+11x^2+30x}\end{align*}
- \begin{align*}\frac{4x-1}{x^2+3x-5}\end{align*}
- \begin{align*}\frac{5x+11}{3x^2-2x-4}\end{align*}
- \begin{align*}\frac{x^2-1}{2x^2+x+3}\end{align*}
- \begin{align*}\frac{12}{x^2+6x+1}\end{align*}
- In an electrical circuit with resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance. \begin{align*}\frac{1}{R_c}=\frac{1}{R_1}+\frac{1}{R_2}\end{align*}. If \begin{align*}R_1 = 25 \ \Omega\end{align*} and the total resistance is \begin{align*}R_c = 10 \ \Omega\end{align*}, what is the resistance \begin{align*}R_2\end{align*}?
- Suppose that two objects attract each other with a gravitational force of 20 Newtons. If the distance between the two objects is doubled, what is the new force of attraction between the two objects?
- Suppose that two objects attract each other with a gravitational force of 36 Newtons. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?
- A sphere with radius \begin{align*}R\end{align*} has a volume of \begin{align*}\frac{4}{3} \pi R^3\end{align*} and a surface area of \begin{align*}4 \pi R^2\end{align*}. Find the ratio of the surface area to the volume of a sphere.
- The side of a cube is increased by a factor of 2. Find the ratio of the old volume to the new volume.
- The radius of a sphere is decreased by 4 units. Find the ratio of the old volume to the new volume.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 12.7.