### Division of Rational Expressions

Just as with ordinary fractions, we first rewrite the division problem as a multiplication problem and then proceed with the multiplication as outlined in the previous section.

**Note:** Remember that \begin{align*}\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}\end{align*}*second* fraction. Do not fall into the common trap of flipping the first fraction.

*Divide \begin{align*}\frac{4x^2}{15} \div \frac{6x}{5}\end{align*} 4x215÷6x5.*

First convert into a multiplication problem by flipping the second fraction and then simplify as usual:

\begin{align*}\frac{4x^2}{15} \div \frac{6x}{5} = \frac{4x^2}{15} \cdot \frac{5}{6x} = \frac{2x}{3} \cdot \frac{1}{3} = \frac{2x}{9}\end{align*}

**Dividing a Rational Expression by a Polynomial**

When we divide a rational expression by a whole number or a polynomial, we can write the whole number (or polynomial) as a fraction with denominator equal to one, and then proceed the same way as in the previous examples.

Divide \begin{align*}\frac{9x^2-4}{2x-2} \div (21x^2-2x-8)\end{align*}

Rewrite the expression as a division of fractions, and then convert into a multiplication problem by taking the reciprocal of the divisor:

\begin{align*}\frac{9x^2-4}{2x-2} \div \frac{21x^2-2x-8}{1} = \frac{9x^2-4}{2x-2} \cdot \frac{1}{21x^2-2x-8}\end{align*}

Then factor and solve:

\begin{align*}\frac{9x^2-4}{2x-2} \cdot \frac{1}{21x^2-2x-8} = \frac{(3x-2)(3x+2)}{2(x-1)} \cdot \frac{1}{(3x-2)(7x+4)} = \frac{(3x+2)}{2(x-1)} \cdot \frac{1}{(7x+4)} = \frac{3x+2}{14x^2-6x-8}\end{align*}

**Solve Applications Involving Multiplication and Division of Rational Expressions**

Suppose Marciel is training for a running race. Marciel’s speed (in miles per hour) of his training run each morning is given by the function \begin{align*}x^3-9x\end{align*}

\begin{align*}\text{time} = \frac{\text{distance}}{\text{speed}}\!\\
\\
\text{time} = \frac{3x^2-9x}{x^3-9x} = \frac{3x(x-3)}{x(x^2-9)} = \frac{3x(x-3)}{x(x+3)(x-3)}\!\\
\\
\text{time} = \frac{3}{x+3}\!\\
\\
\text{If} \ x = 5, \ \text{then}\!\\
\\
\text{time} = \frac{3}{5+3}=\frac{3}{8}\end{align*}

**Marciel will run for \begin{align*}\frac{3}{8}\end{align*} of an hour.**

### Example

#### Example 1

Divide \begin{align*}\frac{3x^2-15x}{2x^2+3x-14} \div \frac{x^2-25}{2x^2+13x+21}\end{align*}.

\begin{align*}\frac{3x^2-15x}{2x^2+3x-14} \cdot \frac{2x^2+13x+21}{x^2-25} = \frac{3x(x-5)}{(2x+7)(x-2)} \cdot \frac{(2x+7)(x+3)}{(x-5)(x+5)} = \frac{3x}{(x-2)} \cdot \frac{(x+3)}{(x+5)} = \frac{3x^2+9x}{x^2+3x-10}\end{align*}

### Review

Divide the rational functions and reduce the answer to lowest terms.

- \begin{align*}2xy \div \frac{2x^2}{y}\end{align*}
- \begin{align*}\frac{2x^3}{y} \div 3x^2\end{align*}
- \begin{align*}\frac{3x+6}{y-4} \div \frac{3y+9}{x-1}\end{align*}
- \begin{align*}\frac{x^2}{x-1} \div \frac{x}{x^2+x-2}\end{align*}
- \begin{align*}\frac{a^2+2ab+b^2}{ab^2-a^2b} \div (a+b)\end{align*}
- \begin{align*}\frac{3-x}{3x-5} \div \frac{x^2-9}{2x^2-8x-10}\end{align*}
- \begin{align*}\frac{x^2-25}{x+3} \div (x-5)\end{align*}
- \begin{align*}\frac{2x+1}{2x-1} \div \frac{4x^2-1}{1-2x}\end{align*}
- \begin{align*}\frac{3x^2+5x-12}{x^2-9} \div \frac{3x-4}{3x+4}\end{align*}
- \begin{align*}\frac{x^2+x-12}{x^2+4x+4} \div \frac{x-3}{x+2}\end{align*}
- \begin{align*}\frac{x^4-16}{x^2-9} \div \frac{x^2+4}{x^2+6x+9}\end{align*}
- Maria’s recipe asks for \begin{align*}2 \frac{1}{2}\end{align*} times as much flour as sugar. How many cups of flour should she mix in if she uses \begin{align*}3 \frac{1}{3}\end{align*} cups of sugar?
- George drives from San Diego to Los Angeles. On the return trip he increases his driving speed by 15 miles per hour. In terms of his initial speed, by what factor is the driving time decreased on the return trip?
- Ohm’s Law states that in an electrical circuit \begin{align*}I = \frac{V}{R_{tot}}.\end{align*} The total resistance for resistors placed in parallel is given by: \begin{align*}\frac{1}{R_{tot}} = \frac{1}{R_1} + \frac{1}{R_2}.\end{align*} Write the formula for the electric current in terms of the component resistances: \begin{align*}R_1\end{align*} and \begin{align*}R_2.\end{align*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 12.9.