# 1.6: Multiplying and Dividing Rational Expressions

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Reducing Rational Expressions
A rational expression is simply an algebraic fraction, and our first consideration will be to reduce these expressions to lowest terms in the same way that we reduce numerical fractions to lowest terms. When we reduce $$\frac{6}{15}$$ to $$\frac{2}{5}$$ by canceling the common factor of three, we are removing a redundant factor of 1 in the form of $$\frac{3}{3}$$
\begin{aligned} \frac{6}{15} &=\frac{3 * 2}{3 * 5}=\frac{3}{3} * \frac{2}{5} \\ &=1 * \frac{2}{5} \\ &=\frac{2}{5} \end{aligned}

Similarly, if there is a common factor that can be factored out of an algebraic fraction, this also can be canceled.
\begin{aligned} \frac{21 x+14}{7 x+7} &=\frac{7(3 x+2)}{7(x+1)} \\ &=\frac{7}{7} * \frac{3 x+2}{x+1} \\ &=1 * \frac{3 x+2}{x+1} \\ &=\frac{3 x+2}{x+1} \end{aligned}
It's important to remeber that only common factors can be canceled. This means that the first priority in each problem will be to identify the factors of the numerator and denominator to see if they share any common factors.

Example
Reduce to lowest terms.
\begin{aligned} \frac{x^{2}+4 x-12}{x^{2}-4} \\ & \frac{x^{2}+4 x-12}{x^{2}-4}=\frac{(x+6)(x-2)}{(x+2)(x-2)} \\ &=\frac{(x+6)\cancel{(x-2)}}{(x+2)\cancel{(x-2)}} \\ &=\frac{x+6}{x+2} \end{aligned}
Notice that we can't cancel the 6 and the 2 in the final answer because they aren't factors. The plus signs in the numerator and denominator prevent us from cancelling the 6 and the 2.

In the previous examples, we saw that cancelling out common factors in the numerator and denominator was actually a process of eliminating a redundant factor of $$1 .$$ In the following example, we'll see a slightly different form of cancelling.
Example
Reduce to lowest terms.
\begin{aligned} \frac{16-x^{2}}{x^{2}+x-20} \\ \frac{16-x^{2}}{x^{2}+x-20} &=\frac{(4+x)(4-x)}{(x+5)(x-4)} \end{aligned}
In this problem, there are no common factors, but we can do some cancelling. We can see that $$(4-x)$$ and $$(x-4)$$ are not the same expression. In the first binomial, the 4 is positive and the $$x$$ is negative, whereas in the second binomial, the 4 is negative and the $$x$$ is positive. So, we know that $$\frac{4-x}{x-4} \neq 1 .$$ However, if we factor $$\mathrm{a}(-1)$$ out of the numerator, we will see an interesting phenomenon:
\begin{aligned} \frac{4-x}{x-4} &=\frac{-1(-4+x)}{x-4} \\ &=\frac{-1(x-4)}{x-4} \\ &=-1 * \frac{x-4}{x-4} \\ &=-1 * 1=-1 \end{aligned}

Therefore, although $$\frac{4-x}{x-4} \neq 1,$$ we can say that $$\frac{4-x}{x-4}=-1 .$$ This will allow us to cancel $$(4-x)$$ and $$(x-4)$$ and replace them with (-1)
\begin{aligned} \frac{16-x^{2}}{x^{2}+x-20} &=\frac{(4+x)(4-x)}{(x+5)(x-4)} \\ &=\frac{(4+x)\cancel{(4-x)}}{(x+5)\cancel{(x-4)}(-1)} \end{aligned}
In the final answer, the (-1) can be placed in the denominator or the numerator, but not both. It can also be placed in front of the fraction.
\begin{aligned} \frac{4+x}{-1(x+5)} &=\frac{-1(4+x)}{x+5} \\ &=-\frac{4+x}{x+5} \end{aligned}

Multiplying and Dividing Rational Expressions
In multiplying and dividing rational expressions, it is often easier to identify and cancel out common factors before multiplying rather than afterwards. Multiplying rational expressions works the same way that multiplying numerical fractions does - multiply straight across the top and straight across the bottom. As a result, any factor in either numerator of the problem will end up in the numerator of the answer. Likewise, any factor in either denominator of the problem will end up in the denominator of the answer. Thus, any factor in either numerator can be cancelled with any factor in either denominator.

Example
\begin{aligned} \frac{x^{2}+5 x+6}{25-x^{2}} * \frac{x^{2}-2 x-15}{x^{2}+6 x+9} & \\ \frac{x^{2}+5 x+6}{25-x^{2}} * \frac{x^{2}-2 x-15}{x^{2}+6 x+9} &=\frac{(x+2)(x+3)}{(5+x)(5-x)} * \frac{(x-5)(x+3)}{(x+3)(x+3)} \\ &=\frac{(x+2)\cancel{(x+3)}}{(5+x)\cancel{(5-x)}(-1)} * \frac{\cancel{(x-5)}\cancel{(x+3)}}{\cancel{(x+3)}\cancel{(x+3)}} \\ &=-\frac{x+2}{x+5} \end{aligned}

Dividing rational expressions works in much the same way that dividing numerical fractions does. We multiply by the reciprocal. There are several ways to demonstrate that this is a valid definition for dividing. First, it is important to understand that the fraction bar is the same as a "divided by" symbol:
$\frac{8}{2}=8 \div 2=4$

The same is true for dividing fractions:
$\frac{1}{3} \div \frac{2}{5}=\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)}$
We can take the complex fraction $$\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)}$$ and multiply it by 1 without changing its value:
\begin{aligned} \frac{1}{3} \div \frac{2}{5} &=\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} * 1 \\ &=\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} \end{aligned}

We can multiply by any form of 1 we want to and not change the value of the result.
$\begin{array}{l} \frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} * 1=\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} \\ \frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} * \frac{9}{9}=\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} \\ \frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} * \frac{12}{12}=\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} \end{array}$
With a carefully chosen form of $$1,$$ we can transform the division problem into a multiplication problem.

\begin{aligned} \frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{5}\right)} * \frac{\left(\frac{5}{2}\right)}{\left(\frac{5}{2}\right)} &=\frac{\frac{1}{3} * \frac{5}{2}}{\frac{2}{5} * \frac{5}{2}} \\ &=\frac{\frac{1}{3} * \frac{5}{2}}{1} \\ &=\frac{1}{3} * \frac{5}{2} \\ &=\frac{5}{6} \end{aligned}
In this way, we can redefine division as multiplication by a reciprocal.

Example
\begin{aligned} \frac{2 x^{2}-x-3}{x^{2}-x-12} \div \frac{x^{2}+5 x+4}{16-x^{2}} \\ \frac{2 x^{2}-x-3}{x^{2}-x-12} \div \frac{x^{2}+5 x+4}{16-x^{2}} &=\frac{2 x^{2}-x-3}{x^{2}-x-12} * \frac{16-x^{2}}{x^{2}+5 x+4} \\ &=\frac{(2 x-3)(x+1)}{(x-4)(x+3)} * \frac{(4+x)(4-x)}{(x+1)(x+4)} \\ &=\frac{(2 x-3)\cancel{(x+1)}}{\cancel{(x-4)}(x+3)} * \frac{\cancel{(4+x)}\cancel{(4-x)}(-1)}{\cancel{(x+1)}\cancel{(x+4)}} \\ &=-\frac{2 x-3}{x+3} \end{aligned}

Exercises 1.6
Reduce each expression to lowest terms.
1) $$\quad \frac{3 x+9}{x^{2}-9}$$
2) $$\quad \frac{4 x^{2}+8 x}{12 x+24}$$
3) $$\quad \frac{x^{2}-2 x}{6-3 x}$$
4) $$\quad \frac{15 x^{2}+24 x}{3 x^{2}}$$
5) $$\quad \frac{24 x^{2}}{12 x^{2}-6 x}$$
6) $$\quad \frac{x^{2}+4 x+4}{x^{2}-4}$$
7) $$\quad \frac{25-y^{2}}{2 y^{2}-8 y-10}$$
8) $$\quad \frac{3 y^{2}-y-2}{3 y^{2}+5 y+2}$$
9) $$\quad \frac{x^{2}+4 x-5}{x^{2}-2 x+1}$$
10) $$\quad \frac{x-x^{2}}{x^{2}+x-2}$$
11) $$\quad \frac{x^{2}+5 x-14}{2-x}$$
12) $$\quad \frac{2 x^{2}+5 x-3}{1-2 x}$$

Multiply or divide the expressions in each problem.
13) $$\quad \frac{3 x+6}{5 x^{2}} * \frac{x}{x^{2}-4}$$
14) $$\quad \frac{4 x^{2}}{x^{2}-16} * \frac{7 x-28}{6 x}$$
15) $$\quad \frac{2 a^{2}-7 a+6}{4 a^{2}-9} * \frac{4 a^{2}+12 a+9}{a^{2}-a-2}$$
16) $$\quad \frac{4 a^{2}-4 a-3}{8 a+4 a^{2}} * \frac{16 a^{2}}{4 a^{2}-6 a}$$
17) $$\quad \frac{x^{2}-y^{2}}{(x+y)^{3}} * \frac{(x+y)^{2}}{(x-y)^{2}}$$
18) $$\quad \frac{2 x^{2}+x-3}{x^{2}-1} * \frac{2 x-2}{2 x^{2}+5 x+3}$$
19) $$\quad \frac{6 x}{x^{2}-4} \div \frac{3 x-9}{2 x+4}$$
20) $$\quad \frac{12 x}{5 x+20} \div \frac{4 x^{2}}{x^{2}-16}$$
21) $$\quad \frac{9 x^{2}+3 x-2}{6 x^{2}-2 x} \div \frac{3 x+2}{6 x^{2}}$$
22) $$\quad \frac{2 a^{2}-5 a-3}{4 a^{2}+2 a} \div \frac{2 a+1}{4 a}$$
23) $$\quad \frac{x^{2}+7 x+6}{x^{2}+x-6} \div \frac{x^{2}+5 x-6}{x^{2}+5 x+6}$$
24) $$\quad \frac{x^{2}+7 x+10}{x^{2}-x-30} \div \frac{3 x^{2}+7 x+2}{9 x^{2}-1}$$
25) $$\quad \frac{2 x^{2}-x-28}{3 x^{2}-x-2} \div \frac{4 x^{2}+16 x+7}{3 x^{2}+11 x+6}$$
26) $$\quad \frac{9 x^{2}+3 x-2}{12 x^{2}+5 x-2} \div \frac{9 x^{2}-6 x+1}{8 x^{2}-10 x-3}$$

This page titled 1.6: Multiplying and Dividing Rational Expressions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Richard W. Beveridge.