ORCCA Multiplication and Division of Rational Expressions

Section13.2Multiplication and Division of Rational Expressions

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.

Figure13.2.1Alternative Video Lesson

Subsection13.2.1Simplifying 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 13.2.9.)

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 \cancelhighlight{7}}{3\cdot \cancelhighlight{7}}\\ \amp=\frac{2}{3} \end{align*}

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

Remark13.2.2

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

Remark13.2.3

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. These \(x\)'s are terms that are added though, and canceling (a process of dividing) is only possible with factors.

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

Example13.2.4

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

Solution

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\cancelhighlight{(x-4)}}{(x+5)\cancelhighlight{(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 \(\left\{x\mid x\neq -5,4\right\}\text{.}\)

Warning13.2.5

When simplifying the function \(Q\) in Example 13.2.4, 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 \(Q\text{.}\)

Example13.2.6

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

Solution

\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}\cancelhighlight{(2y+1)}}{\cancelhighlight{y}\cancelhighlight{(2y+1)}(y-1)}\\ \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 \(\left\{y\mid y\neq -\frac{1}{2},0,1\right\}\text{.}\)

Example13.2.7

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

Solution

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)\cancelhighlight{(y+5)}}{\cancelhighlight{(y+5)}(y-5)}\\ \amp=\frac{2y-1}{y-5}, \text{ for } y\neq -5 \end{align*}

Example13.2.8

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

Solution

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)\cancelhighlight{(z-4)}}{-\cancelhighlight{(z-4)}}\\ \amp=\frac{3z(z-4)}{1}\\ \amp=3z(z-4), \text{ for } z\neq 4 \end{align*}

Example13.2.9

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

Solution

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)\cancelhighlight{(2x-3)}}{(x^2+5)\cancelhighlight{(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 \(\left\{ x\mid x\neq \frac{3}{2}\right\}\text{.}\)

Example13.2.10

Simplify the expression \(\frac{3y-x}{x^2-xy-6y^2}\text{.}\)

Solution

\begin{align*} \frac{3y-x}{x^2-xy-6y^2}\amp=\frac{-\cancelhighlight{(x-3y)}}{\cancelhighlight{(x-3y)}(x+2y)}\\ \amp=\frac{-1}{x+2y}, \text{ for } x\neq 3y\\ \amp=-\frac{1}{x+2y}, \text{ for } x\neq 3y \end{align*}

Subsection13.2.2Multiplication and Division of Rational Functions and Expressions

Recall the property for multiplying fractions 1.2.14, 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 property 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 \cancelhighlight{6}}{15\cdot \cancelhighlight{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 \cancelhighlight{3}}{\cancelhighlight{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)\cancelhighlight{(x+3)}}{\cancelhighlight{(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.

Example13.2.11

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

Solution

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)\cancelhighlight{(x-2)}} \cdot\frac{(x-2)\cancelhighlight{(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*}

Example13.2.12

Multiply the rational expressions: \(\frac{p^2q^4}{3r}\cdot\frac{9r^2}{pq^2}\)

Solution

Note that we won't need to factor anything in this problem, 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, \text{ for } p\neq 0, q\neq 0, r\neq 0 \end{align*}

We can divide rational expressions using the property for dividing fractions 1.2.16, which simply requires that we change dividing by an expression to multiplying by its reciprocal. Let's look at a few examples.

Example13.2.13

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

Solution

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

Example13.2.14

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

Solution

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\cancelhighlight{(x+5)}}\cdot\frac{3\cancelhighlight{(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*}

Example13.2.15

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

Solution

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

Example13.2.16

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

Solution

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

Example13.2.17

Divide the rational expressions: \(\frac{3x^2}{x^2-9y^2}\div\frac{6x^3}{x^2-2xy-15y^2}\)

Solution

\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}{\cancelhighlight{(x+3y)}(x-3y)}\cdot\frac{\cancelhighlight{(x+3y)}(x-5y)}{6x^3}\\ \amp=\frac{1}{x-3y}\cdot\frac{x-5y}{2x}\\ \amp=\frac{x-5y}{2x(x-3y)}, \text{ for } x\neq -3y, x\neq 5y \end{align*}

Example13.2.18

Divide the rational expressions: \(\frac{m^2n^2-3mn-4}{2mn}\div(m^2n^2-16)\)

Solution

\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{\cancelhighlight{(mn-4)}(mn+1)}{2mn}\cdot\frac{1}{(mn+4)\cancelhighlight{(mn-4)}}\\ \amp=\frac{mn+1}{2mn}\cdot\frac{1}{mn-4}\\ \amp=\frac{mn+1}{2mn(mn-4)}, \text{ for }mn\neq -4 \end{align*}

SubsectionExercises

Simplifying Rational Expressions with One Variable

1

Select all correct simplifications:

\(\frac{8 x+5}{x + 5}=8\)
\(\frac{x+5}{5}=x\)
\(\frac{x+ 5}{x+5}=1\)
\(\frac{5}{x+5}=\frac{1}{x+1}\)
\(\frac{5}{x+5}=\frac{1}{x}\)
\(\frac{x+5}{x+8}=\frac{5}{8}\)
\(\frac{5 x}{x}=5\)
\(\frac{x+5}{x}=5\)
\(\frac{8 x+5}{8}=x+5\)
\(\frac{8(x-5)}{x -5}=8\)
\(\frac{x}{8 x}=\frac{1}{8}\)

2

Select all correct simplifications:

\(\frac{x+ 6}{x+6}=1\)
\(\frac{6 x}{x}=6\)
\(\frac{x+6}{6}=x\)
\(\frac{6}{x+6}=\frac{1}{x}\)
\(\frac{x+6}{x+5}=\frac{6}{5}\)
\(\frac{5 x+6}{5}=x+6\)
\(\frac{6}{x+6}=\frac{1}{x+1}\)
\(\frac{x+6}{x}=6\)
\(\frac{5 x+6}{x + 6}=5\)
\(\frac{5(x-6)}{x -6}=5\)
\(\frac{x}{5 x}=\frac{1}{5}\)

3

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

\(\displaystyle{{\frac{r+2}{r+2}}=}\)
\(\displaystyle{{\frac{r+2}{2+r}}=}\)
\(\displaystyle{{\frac{r-2}{r-2}}=}\)
\(\displaystyle{{\frac{r-2}{2-r}}=}\)

4

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

\(\displaystyle{{\frac{t+8}{t+8}}=}\)
\(\displaystyle{{\frac{t+8}{8+t}}=}\)
\(\displaystyle{{\frac{t-8}{t-8}}=}\)
\(\displaystyle{{\frac{t-8}{8-t}}=}\)