ORCCA Multiplication and Division of Rational Expressions

Section 12.2 Multiplication and Division of Rational Expressions

Objectives: PCC Course Content and Outcome Guide

[cross-reference to target(s) "ccog-simplify-rational-functions" missing or not unique]
[cross-reference to target(s) "ccog-operations-on-rational-functions" missing or not unique]

In the last section, we learned some rational function applications. In this section, we will learn how to simplify rational expressions, and how to multiply and divide them.

Figure 12.2.1. Alternative Video Lesson

Subsection 12.2.1 Simplifying Rational Expressions

Consider the two rational functions below. At first glance, which function looks simpler?

\begin{equation*} f(x)=\frac{8x^3-12x^2+8x-12}{2x^3-3x^2+10x-15} \end{equation*}

\begin{equation*} g(x)=\frac{4(x^2+1)}{x^2+5}, \text{ for }x\neq \frac{3}{2} \end{equation*}

It can be argued that the function \(g\) is simpler, at least with regard to the ease with which we can determine its domain, quickly evaluate it, and also determine where its function value is zero. All of these things are considerably more difficult with the function \(f\text{.}\)

These two functions are actually the same function. Using factoring and the same process of canceling that’s used with numerical ratios, we will learn how to simplify the function \(f\) into the function \(g\text{.}\) (The full process for simplifying \(f(x)=\frac{8x^3-12x^2+8x-12}{2x^3-3x^2+10x-15}\) will be shown in Example 8.)

To see a simple example of the process for simplifying a rational function or expression, let’s look at simplifying \(\frac{14}{21}\) and \(\frac{(x+2)(x+7)}{(x+3)(x+7)}\) by canceling common factors:

\begin{align*} \frac{14}{21}\amp=\frac{2\cdot \cancel{7}}{3\cdot \cancel{7}}\amp\frac{(x+2)(x+7)}{(x+3)(x+7)}\amp=\frac{(x+2)\cancel{(x+7)}}{(x+3)\cancel{(x+7)}}\\ \amp=\frac{2}{3}\amp\amp=\frac{x+2}{x+3}, \text{ for }x\neq -7 \end{align*}

The statement “for \(x\neq -7\)” was added when the factors of \(x+7\) were canceled. This is because \(\frac{(x+2)(x+7)}{(x+3)(x+7)}\) was undefined for \(x=-7\text{,}\) so the simplified version must also be undefined for \(x=-7\text{.}\)

Warning 12.2.2. Cancel Factors, not Terms.

It may be tempting to want to try to simplify \(\frac{x+2}{x+3}\) into \(\frac{2}{3}\) by canceling each \(x\) that appears. But these \(x\)’s are terms (pieces that are added with other pieces), not factors. Canceling (an act of division) is only possible with factors (an act of multiplication).

The process of canceling factors is key to simplifying rational expressions. If the expression is not given in factored form, then this will be our first step. We’ll now look at a few more examples.

Example 12.2.3.

Simplify the rational function formula \(Q(x)=\frac{3x-12}{x^2+x-20}\) and state the domain of \(Q\text{.}\)

Explanation.

To start, we’ll factor the numerator and denominator. We’ll then cancel any factors common to both the numerator and denominator.

\begin{align*} Q(x)\amp=\frac{3x-12}{x^2+x-20}\\ Q(x)\amp=\frac{3\cancel{(x-4)}}{(x+5)\cancel{(x-4)}}\\ Q(x)\amp=\frac{3}{x+5}, \text{ for }x\neq 4 \end{align*}

The domain of this function will incorporate the explicit domain restriction \(x\neq 4\) that was stated when the factor of \(x-4\) was canceled from both the numerator and denominator. We will also exclude \(-5\) from the domain as this value would make the denominator zero. Thus the domain of \(Q\) is \(\left\{x\mid x\neq -5,4\right\}\text{.}\)

Warning 12.2.4.

When simplifying the function \(Q\) in Example 3, we cannot simply write \(Q(x)=\frac{3}{x+5}\text{.}\) The reason is that this would result in our simplified version of the function \(Q\) having a different domain than the original \(Q\text{.}\) More specifically, for our original function \(Q\) it held that \(Q(4)\) was undefined, and this still needs to be true for the simplified form of \(Q\text{.}\)

Example 12.2.5.

Simplify the rational function formula \(R(y)=\frac{-y-2y^2}{2y^3-y^2-y}\) and state the domain of \(R\text{.}\)

Explanation.

\begin{align*} R(y)\amp=\frac{-y-2y^2}{2y^3-y^2-y}\\ R(y)\amp=\frac{-2y^2-y}{y(2y^2-y-1)}\\ R(y)\amp=\frac{-\cancelhighlight{y}\secondcancelhighlight{(2y+1)}}{\cancelhighlight{y}\secondcancelhighlight{(2y+1)}(y-1)}\\ R(y)\amp=-\frac{1}{y-1}, \text{ for }y\neq 0, y\neq -\frac{1}{2} \end{align*}

The domain of this function will incorporate the explicit restrictions \(y\neq 0, y\neq -\frac{1}{2}\) that were stated when the factors of \(y\) and \(2y+1\) were canceled from both the numerator and denominator. Since the factor \(y-1\) is still in the denominator, we also need the restriction that \(y\neq 1\text{.}\) Therefore the domain of \(R\) is \(\left\{y\mid y\neq -\frac{1}{2},0,1\right\}\text{.}\)

Example 12.2.6.

Simplify the expression \(\frac{9y+2y^2-5}{y^2-25}\text{.}\)

Explanation.

To start, we need to recognize that \(9y+2y^2-5\) is not written in standard form (where terms are written from highest degree to lowest degree). Before attempting to factor this expression, we’ll re-write it as \(2y^2+9y-5\text{.}\)

\begin{align*} \frac{9y+2y^2-5}{y^2-25}\amp=\frac{2y^2+9y-5}{y^2-25}\\ \amp=\frac{(2y-1)\cancel{(y+5)}}{\cancel{(y+5)}(y-5)}\\ \amp=\frac{2y-1}{y-5}, \text{ for } y\neq -5 \end{align*}

Example 12.2.7.

Simplify the expression \(\frac{-48z+24z^2-3z^3}{4-z}\text{.}\)

Explanation.

To begin simplifying this expression, we will rewrite each polynomial in descending order. Then we’ll factor out the GCF, including the constant \(-1\) from both the numerator and denominator because their leading terms are negative.

\begin{align*} \frac{-48z+24z^2-3z^3}{4-z}\amp=\frac{-3z^3+24z^2-48z}{-z+4}\\ \amp=\frac{-3z(z^2-8z+16)}{-(z-4)}\\ \amp=\frac{-3z(z-4)^2}{-(z-4)}\\ \amp=\frac{-3z(z-4)\cancel{(z-4)}}{-\cancel{(z-4)}}\\ \amp=\frac{3z(z-4)}{1}, \text{ for } z\neq 4\\ \amp=3z(z-4), \text{ for } z\neq 4 \end{align*}

Example 12.2.8.

Simplify the rational function formula \(f(x)=\frac{8x^3-12x^2+8x-12}{2x^3-3x^2+10x-15}\) and state the domain of \(f\text{.}\)

Explanation.

To simplify this rational function, we’ll first note that both the numerator and denominator have four terms. To factor them we’ll need to use factoring by grouping. (Note that if this technique didn’t work, very few other approaches would be possible.) Once we’ve used factoring by grouping, we’ll cancel any factors common to both the numerator and denominator and state the associated restrictions.

\begin{align*} f(x)\amp=\frac{8x^3-12x^2+8x-12}{2x^3-3x^2+10x-15}\\ f(x)\amp=\frac{4(2x^3-3x^2+2x-3)}{2x^3-3x^2+10x-15}\\ f(x)\amp=\frac{4(x^2(2x-3)+(2x-3))}{x^2(2x-3)+5(2x-3)}\\ f(x)\amp=\frac{4(x^2+1)\cancel{(2x-3)}}{(x^2+5)\cancel{(2x-3)}}\\ f(x)\amp=\frac{4(x^2+1)}{x^2+5}, \text{ for }x\neq \frac{3}{2} \end{align*}

In determining the domain of this function, we’ll need to account for any implicit and explicit restrictions. When the factor \(2x-3\) was canceled, the explicit statement of \(x\neq \frac{3}{2}\) was given. The denominator in the final simplified form of this function has \(x^2+5\text{.}\) There is no value of \(x\) for which \(x^2+5=0\text{,}\) so the only restriction is that \(x\neq \frac{3}{2}\text{.}\) Therefore the domain is \(\left\{ x\mid x\neq \frac{3}{2}\right\}\text{.}\)

Example 12.2.9.

Simplify the expression \(\frac{3y-x}{x^2-xy-6y^2}\text{.}\) In this example, there are two variables. It is still possible that in examples like this, there can be domain restrictions when simplifying rational expressions. However since we are not studying functions of more than one variable, this textbook ignores domain restrictions with examples like this one.

Explanation.

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

Subsection 12.2.2 Multiplication of Rational Functions and Expressions

Recall the [cross-reference to target(s) "fact-multiplication-of-fractions" missing or not unique], which states that the product of two fractions is equal to the product of their numerators divided by the product of their denominators. We will use this same rule for multiplying rational expressions.

When multiplying fractions, one approach is to multiply the numerator and denominator, and then simplify the fraction that results by determining the greatest common factor in both the numerator and denominator, like this:

\begin{align*} \frac{14}{9}\cdot\frac{3}{10}\amp=\frac{14\cdot 3}{9\cdot 10}\\ \amp=\frac{42}{90}\\ \amp=\frac{7\cdot \cancel{6}}{15\cdot \cancel{6}}\\ \amp=\frac{7}{15} \end{align*}

This approach works great when we can easily identify that \(6\) is the greatest common factor in both \(42\) and \(90\text{.}\) But in more complicated instances, it isn’t always an easy approach. It also won’t work particularly well when we have \((x+2)\) instead of \(2\) as a factor, as we’ll see shortly.

Another approach to multiplying and simplifying fractions involves utilizing the prime factorization of each the numerator and denominator, like this:

\begin{align*} \frac{14}{9}\cdot\frac{3}{10}\amp=\frac{2\cdot 7}{3^2}\cdot \frac{3}{2\cdot 5}\\ \amp=\frac{\cancelhighlight{2} \cdot 7 \cdot \secondcancelhighlight{3}}{\secondcancelhighlight{3} \cdot 3 \cdot \cancelhighlight{2}\cdot5}\\ \amp=\frac{7}{15} \end{align*}

The method for multiplying and simplifying rational expressions is nearly identical, as shown here:

\begin{align*} \frac{x^2+9x+14}{x^2+6x+9}\cdot \frac{x+3}{x^2+7x+10}\amp=\frac{(x+2)(x+7)}{(x+3)^2}\cdot \frac{x+3}{(x+2)(x+5)}\\ \amp=\frac{\cancelhighlight{(x+2)}(x+7)\secondcancelhighlight{(x+3)}}{\secondcancelhighlight{(x+3)}(x+3)\cancelhighlight{(x+2)}(x+5)}\\ \amp=\frac{(x+7)}{(x+3)(x+5)}, \text{ for } x\neq -2 \end{align*}

This process will be used for both multiplying and dividing rational expressions. The main distinctions in various examples will be in the factoring methods required.

Example 12.2.10.

Multiply the rational expressions: \(\dfrac{x^2-4x}{x^2-4}\cdot\dfrac{4-4x+x^2}{20-x-x^2}\text{.}\)

Explanation.

Note that to factor the second rational expression, we’ll want to re-write the terms in descending order for both the numerator and denominator. In the denominator, we’ll first factor out \(-1\) as the leading term is \(-x^2\text{.}\)

\begin{align*} \frac{x^2-4x}{x^2-4}\cdot\frac{4-4x+x^2}{20-x-x^2}\amp=\frac{x^2-4x}{x^2-4}\cdot\frac{x^2-4x+4}{-x^2-x+20}\\ \amp=\frac{x^2-4x}{x^2-4}\cdot\frac{x^2-4x+4}{-(x^2+x-20)}\\ \amp=\frac{x\cancelhighlight{(x-4)}}{(x+2)\secondcancelhighlight{(x-2)}} \cdot\frac{(x-2)\secondcancelhighlight{(x-2)}}{-(x+5)\cancelhighlight{(x-4)}}\\ \amp=-\frac{x(x-2)}{(x+2)(x+5)}, \text{ for } x\neq 2,x\neq 4 \end{align*}

Example 12.2.11.

Multiply the rational expressions: \(\dfrac{p^2q^4}{3r}\cdot\dfrac{9r^2}{pq^2}\text{.}\) Note this book ignores domain restrictions on multivariable expressions.

Explanation.

We won’t need to factor anything in this example, and can simply multiply across and then simplify.

\begin{align*} \frac{p^2q^4}{3r}\cdot\frac{9r^2}{pq^2}\amp=\frac{p^2q^2\cdot9r^2}{3r\cdot pq^2}\\ \amp=\frac{pq^2\cdot 3r}{1}\\ \amp=3pq^2r \end{align*}

Subsection 12.2.3 Division of Rational Functions and Expressions

We can divide rational expressions using the [cross-reference to target(s) "fact-division-of-fractions" missing or not unique], which simply requires that we change dividing by an expression to multiplying by its reciprocal. Let’s look at a few examples.

Example 12.2.12.

Divide the rational expressions: \(\dfrac{x+2}{x+5}\div \dfrac{x+2}{x-3}\text{.}\)

Explanation.

\begin{align*} \frac{x+2}{x+5}\div \frac{x+2}{x-3}\amp=\frac{\cancel{x+2}}{x+5}\cdot \frac{x-3}{\cancel{x+2}}, \text{ for }x\neq 3\\ \amp=\frac{x-3}{x+5}, \text{ for }x\neq -2, x\neq 3 \end{align*}

Remark 12.2.13.

In the first step of Example 12, the restriction \(x\neq3\) was used. We hadn’t canceled anything yet, so why is there this restriction already? It’s because the original expression \(\frac{x+2}{x+5}\div \frac{x+2}{x-3}\) had \(x-3\) in a denominator, which means that \(3\) is not a valid input. In the first step of simplifying, the \(x-3\) denominator went to the numerator and we lost the information that \(3\) was not a valid input, so we stated it explicitly. Always be sure to compare the restrictions of the original expression with each step throughout the process.

Example 12.2.14.

Simplify the rational expression using division: \(\dfrac{\frac{3x-6}{2x+10}}{\frac{x^2-4}{3x+15}}\text{.}\)

Explanation.

To begin, we’ll note that the larger fraction bar is denoting division, so we will use multiplication by the reciprocal. After that, we’ll factor each expression and cancel any common factors.

\begin{align*} \frac{\frac{3x-6}{2x+10}}{\frac{x^2-4}{3x+15}}\amp=\frac{3x-6}{2x+10}\div\frac{x^2-4}{3x+15}\\ \amp=\frac{3x-6}{2x+10}\cdot\frac{3x+15}{x^2-4}\\ \amp=\frac{3\cancelhighlight{(x-2)}}{2\secondcancelhighlight{(x+5)}}\cdot\frac{3\secondcancelhighlight{(x+5)}}{(x+2)\cancelhighlight{(x-2)}}\\ \amp=\frac{3\cdot 3}{2(x+2)}, \text{ for }x\neq -5, x\neq 2\\ \amp=\frac{9}{2x+4}, \text{ for }x\neq -5, x\neq 2 \end{align*}

Example 12.2.15.

Divide the rational expressions: \(\dfrac{x^2-5x-14}{x^2+7x+10}\div\dfrac{x-7}{x+4}\text{.}\)

Explanation.

\begin{align*} \frac{x^2-5x-14}{x^2+7x+10}\div\frac{x-7}{x+4}\amp=\frac{x^2-5x-14}{x^2+7x+10}\cdot\frac{x+4}{x-7}, \text{ for }x\neq -4\\ \amp=\frac{\cancelhighlight{(x-7)}\secondcancelhighlight{(x+2)}}{(x+5)\secondcancelhighlight{(x+2)}}\cdot\frac{x+4}{\cancelhighlight{x-7}}, \text{ for }x\neq -4\\ \amp=\frac{x+4}{x+5}, \text{ for }x\neq -4, x\neq -2, x\neq 7 \end{align*}

Example 12.2.16.

Divide the rational expressions: \((p^4-16)\div\dfrac{p^4-2p^3}{2p}\text{.}\)

Explanation.

\begin{align*} (p^4-16)\div\frac{p^4-2p^3}{2p}\amp=\frac{p^4-16}{1}\cdot\frac{2p}{p^4-2p^3}\\ \amp=\frac{(p^2+4)(p+2)\cancel{(p-2)}}{1}\cdot\frac{2p}{p^3\cancel{(p-2)}}\\ \amp=\frac{2(p^2+4)(p+2)}{p^2}, \text{ for } p\neq 2 \end{align*}

Note here that we didn’t have to include a restriction in the very first step. That restriction would have been \(p\neq0\text{,}\) but since \(0\) still cannot be inputted into any of the subsequent expressions, we don’t need to explicitly state \(p\neq0\) as a restriction because the expressions tell us that implicitly already.

Example 12.2.17.

Divide the rational expressions: \(\dfrac{3x^2}{x^2-9y^2}\div\dfrac{6x^3}{x^2-2xy-15y^2}\text{.}\) Note this book ignores domain restrictions on multivariable expressions.

Explanation.

\begin{align*} \frac{3x^2}{x^2-9y^2}\div\frac{6x^3}{x^2-2xy-15y^2}\amp=\frac{3x^2}{x^2-9y^2}\cdot\frac{x^2-2xy-15y^2}{6x^3}\\ \amp=\frac{3x^2}{\cancel{(x+3y)}(x-3y)}\cdot\frac{\cancel{(x+3y)}(x-5y)}{6x^3}\\ \amp=\frac{1}{x-3y}\cdot\frac{x-5y}{2x}\\ \amp=\frac{x-5y}{2x(x-3y)} \end{align*}

Example 12.2.18.

Divide the rational expressions: \(\dfrac{m^2n^2-3mn-4}{2mn}\div(m^2n^2-16)\text{.}\) Note this book ignores domain restrictions on multivariable expressions.

Explanation.

\begin{align*} \frac{m^2n^2-3mn-4}{2mn}\div(m^2n^2-16)\amp=\frac{m^2n^2-3mn-4}{2mn}\cdot\frac{1}{m^2n^2-16}\\ \amp=\frac{\cancel{(mn-4)}(mn+1)}{2mn}\cdot\frac{1}{(mn+4)\cancel{(mn-4)}}\\ \amp=\frac{mn+1}{2mn}\cdot\frac{1}{mn+4}\\ \amp=\frac{mn+1}{2mn(mn+4)} \end{align*}

Reading Questions 12.2.4 Reading Questions

1.

What is the difference between a factor and a term?

2.

When canceling pieces of rational function expression to simplify it, what kinds of pieces are the only acceptable pieces to cancel?

3.

When you simplify a rational function expression, you may need to make note of a .

Exercises 12.2.5 Exercises

Review and Warmup.

1.

Multiply: \(\displaystyle{-\frac{20}{7} \cdot \frac{13}{8}}\)

2.

Multiply: \(\displaystyle{-\frac{3}{13} \cdot \frac{11}{27}}\)

3.

Multiply: \(\displaystyle{-\frac{8}{19} \cdot \left(-\frac{9}{14}\right)}\)

4.

Multiply: \(\displaystyle{-\frac{7}{16} \cdot \left(-\frac{4}{7}\right)}\)

5.

Divide: \(\displaystyle{ \frac{3}{5} \div \frac{5}{4} }\)

6.

Divide: \(\displaystyle{ \frac{3}{7} \div \frac{7}{4} }\)

7.

Divide: \(\displaystyle{ \frac{1}{16} \div \left(-\frac{7}{12}\right) }\)

8.

Divide: \(\displaystyle{ \frac{5}{12} \div \left(-\frac{2}{15}\right) }\)

Exercise Group.

9.

Factor the given polynomial.

\({t^{2}-36}=\)

10.

Factor the given polynomial.

\({t^{2}-4}=\)

11.

Factor the given polynomial.

\({x^{2}+15x+56}=\)

12.

Factor the given polynomial.

\({x^{2}+7x+10}=\)

13.

Factor the given polynomial.

\({y^{2}-7y+6}=\)

14.

Factor the given polynomial.

\({y^{2}-18y+80}=\)

15.

Factor the given polynomial.

\({2r^{2}-18r+28}=\)

16.

Factor the given polynomial.

\({5r^{2}-25r+20}=\)

17.

Factor the given polynomial.

\({8r^{8}+24r^{7}+16r^{6}}=\)

18.

Factor the given polynomial.

\({2t^{6}+14t^{5}+24t^{4}}=\)

19.

Factor the given polynomial.

\({144t^{2}-24t+1}=\)

20.

Factor the given polynomial.

\({64x^{2}-16x+1}=\)

Simplifying Rational Expressions with One Variable.

21.

Simplify the following expressions, and if applicable, write the restricted domain on the simplified expression.

\(\displaystyle{{\frac{x+4}{x+4}}=}\)
\(\displaystyle{{\frac{x+4}{4+x}}=}\)
\(\displaystyle{{\frac{x-4}{x-4}}=}\)
\(\displaystyle{{\frac{x-4}{4-x}}=}\)

22.

Simplify the following expressions, and if applicable, write the restricted domain on the simplified expression.

\(\displaystyle{{\frac{y+10}{y+10}}=}\)
\(\displaystyle{{\frac{y+10}{10+y}}=}\)
\(\displaystyle{{\frac{y-10}{y-10}}=}\)
\(\displaystyle{{\frac{y-10}{10-y}}=}\)

23.

Select all correct simplifications, ignoring possible domain restrictions.

\(\displaystyle \frac{x}{7 x}=\frac{1}{7}\)
\(\displaystyle \frac{5}{x+5}=\frac{1}{x}\)
\(\displaystyle \frac{7(x-5)}{x -5}=7\)
\(\displaystyle \frac{x+5}{5}=x\)
\(\displaystyle \frac{x+ 5}{x+5}=1\)
\(\displaystyle \frac{5}{x+5}=\frac{1}{x+1}\)
\(\displaystyle \frac{7 x+5}{x + 5}=7\)
\(\displaystyle \frac{x+5}{x+7}=\frac{5}{7}\)
\(\displaystyle \frac{x+5}{x}=5\)
\(\displaystyle \frac{5 x}{x}=5\)
\(\displaystyle \frac{7 x+5}{7}=x+5\)

24.

Select all correct simplifications, ignoring possible domain restrictions.

\(\displaystyle \frac{x}{7 x}=\frac{1}{7}\)
\(\displaystyle \frac{6}{x+6}=\frac{1}{x+1}\)
\(\displaystyle \frac{6 x}{x}=6\)
\(\displaystyle \frac{x+6}{6}=x\)
\(\displaystyle \frac{x+6}{x}=6\)
\(\displaystyle \frac{7 x+6}{7}=x+6\)
\(\displaystyle \frac{x+6}{x+7}=\frac{6}{7}\)
\(\displaystyle \frac{x+ 6}{x+6}=1\)
\(\displaystyle \frac{7 x+6}{x + 6}=7\)
\(\displaystyle \frac{6}{x+6}=\frac{1}{x}\)
\(\displaystyle \frac{7(x-6)}{x -6}=7\)