Did you know that Pluto is no longer considered a planet? Even so, we can still calculate the distance between it and the Sun. All we would have to know is the force of attraction between Pluto and the Sun, the masses of the two objects, and the value of the gravitation constant. In this Concept, you'll learn how to simplify rational expressions, including real-life expressions such as the one we would encounter here.

### Guidance

You have gained experience working with rational functions so far. In this Concept, you will continue simplifying rational expressions by factoring.

To **simplify** a rational expression means to reduce the fraction into its lowest terms.

To do this, you will need to remember a property about multiplication.

For all real values \begin{align*}a \text{ and } b\end{align*}

#### Example A

*Simplify \begin{align*}\frac{x^2-2x+1}{8x-8}\end{align*} x2−2x+18x−8.*

**Solution:** Factor both pieces of the rational expression and reduce.

\begin{align*}& \frac{x^2-2x+1}{8x-8} \rightarrow \frac{(x-1)(x-1)}{8(x-1)}\\
& \frac{x^2-2x+1}{8x-8}=\frac{x-1}{8}\end{align*}

**Finding Excluded Values of Rational Expressions**

As stated in a previous Concept, excluded values are also called **points of discontinuity.** These are the values that make the denominator equal to zero and are not part of the domain.

#### Example B

*Find the excluded values of \begin{align*}\frac{2x+1}{x^2-x-6}\end{align*} 2x+1x2−x−6.*

**Solution:** Factor the denominator of the rational expression.

\begin{align*}\frac{2x+1}{x^2-x-6}=\frac{2x+1}{(x+2)(x-3)}\end{align*}

Find the value that makes each factor equal zero.

\begin{align*}x=-2, x=3\end{align*}

These are excluded values of the domain of the rational expression.

**Real-Life Rational Expressions**

*The gravitational force between two objects in given by the formula* \begin{align*}F=\frac{G(m_1m_2)}{(d^2)}\end{align*}

#### Example C

*What is the distance between the Earth and the Moon?*

**Solution:**

\begin{align*}&\text{Let's start with the Law of Gravitation formula}. && F =G\frac {m_1m_2}{d^2} \\
&\text{Now plug in the known values}. && 2.0\times 10^{20} N =6.67\times 10^{-11} \frac {N \cdot m^2}{kg^2}.\frac {(5.97\times 10^{24}kg)(7.36\times 10^{22}kg)}{d^2} \\
&\text{Multiply the masses together}. && 2.0\times 10^{20} N =6.67\times 10^{-11} \frac {N \cdot m^2}{kg^2}.\frac {4.39\times 10^{47}kg^2}{d^2} \\
&\text{Cancel the}\ kg^2\ \text{units}. && 2.0\times 10^{20} N = 6.67 \times 10^{-11} \frac{N \cdot m^2} {\cancel{kg^2}} \cdot \frac{4.39 \times 10^{47} \cancel{kg^2}} {d^2} \\
&\text{Multiply the numbers in the numerator}. && 2.0\times 10^{20} N \frac {2.93\times 10^{37}}{d^2}N \cdot m^2 \\
&\text{Multiply both sides by}\ d^2. && 2.0\times 10^{20} N \cdot d^2 =\frac {2.93\times 10^{37}}{d^2} \cdot d^2 \cdot N \cdot m^2 \\
&\text{Cancel common factors}. && 2.0\times 10^{20} N \cdot d^{2} = \frac{2.93 \times 10^{37}} {\cancel{d^2}} \cdot \cancel{d^2} \cdot N \cdot m^2 \\
&\text{Simplify}. && 2.0\times 10^{20} N \cdot d^2 = 2.93\times 10^{37}N \cdot m^2 \\
&\text{Divide both sides by}\ 2.0 \times 10^{20} N. && d^2 =\frac {2.93 \times 10^{37}}{2.0\times 10^{20}}\frac {N \cdot m^2}{N} \\
&\text{Simplify}. && d^2 =1.465\times 10^{17} m^2 \\
&\text{Take the square root of both sides}. && d =3.84 \times 10^8m\\\end{align*}

### Guided Practice

*Find the excluded values by simplifying \begin{align*}\frac{4x-2}{2x^2+x-1}\end{align*}.*

**Solution:**

Both the numerator and denominator can be factored using methods learned in previous Concepts.

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

The expression \begin{align*}(2x-1)\end{align*} appears in both the numerator and denominator and can be canceled. The expression becomes:

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

Since \begin{align*}x+1=0\Rightarrow x=-1\end{align*}, then \begin{align*}x=-1\end{align*} is an excluded value.

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Simplifying Rational Expressions (15:22)

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^3+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 the sphere.
- The side of a cube is increased by a factor of two. Find the ratio of the old volume to the new volume.
- The radius of a sphere is decreased by four units. Find the ratio of the old volume to the new volume.

**Mixed Review**

- Name \begin{align*}4p^6+7p^3-9\end{align*}.
- Simplify \begin{align*}(4b^2+b+7b^3 )+(5b^2-6b^4+b^3)\end{align*}. Write the answer in standard form.
- State the Zero Product Property.
- Why can’t the Zero Product Property be used in this situation: \begin{align*}(5x+1)(x-4)=2\end{align*}?
- Shelly earns $4.85 an hour plus $15 in tips. Graph her possible total earnings for one day of work.
- Multiply and simplify: \begin{align*}(-4x^2+8x-1)(-7x^2+6x+8)\end{align*}.
- A rectangle’s perimeter is 65 yards. The length is 7 more yards than its width. What dimensions would give the largest area?