7.4: Multiplication and Division of Rational Functions

Learning Outcomes

Simplify rational expressions
Determine restrictions on a variable
Perform multiplication and division of rational expressions
Perform multiplication and division of rational functions
Determine the domain of a product or quotient function

Earlier we defined a rational expression as [latex]\dfrac{P(x)}{Q(x)}[/latex] where [latex]P(x)[/latex] and [latex]Q(x)[/latex] are polynomials. Just as we can multiply and divide fractions, we can multiply and divide rational expressions. In fact, we use the same processes for multiplying and dividing rational expressions as we use for multiplying and dividing fractions. The process is the same even though the expressions look different.

Simplifying Rational Expressions

Rational expressions that have common factors on the numerator and denominator can be simplified by cancelling out the common factors to 1. To determine if common factors exist, the numerator and denominators may have to be factored. However, we cannot divide by 0, so we must make sure to state any restrictions on the variable. These restrictions will appear when the denominator equals zero.

Example 1

State the restrictions on the variable of the rational expression:

1. [latex]\dfrac{4x(2x-3)}{(x+4)(x-5)}[/latex]

2. [latex]\dfrac{7x+1}{5x(3x-2)}[/latex]

3. [latex]\dfrac{4x^2-3x+5}{(x^2-9)(10x^2-7x-12)}[/latex]

Solution

Restrictions occur in a rational expression when the denominator equals zero.

[latex]\begin{aligned}(x+4)(x-5)&=0\\x+4=0\text{ or }x-5&=0\\x=-4,\;x&=5\end{aligned}[/latex]

Consequently, [latex]x\neq -4,\,5[/latex]

[latex]\begin{aligned}5x(3x-2)&=0\\5x=0\text{ or }3x-2&=0\\x=0,\;x&=\frac{2}{3}\end{aligned}[/latex]

Consequently, [latex]x\neq 0,\,\dfrac{2}{3}[/latex]

3. Often, we have to factor the denominator to determine the restrictions.

[latex]\begin{aligned}(x^2-9)(10x^2-7x-12)&=0\\(x-3)(x+3)(5x+4)(2x-3)&=0\\x-3=0\text{ or }x+3=0\text{ or }5x+4=0\text{ or }2x-3&=0\\x=3,\;x=-3,\;x=-\frac{4}{5},\;x&=\frac{3}{2}\end{aligned}[/latex]

Consequently, [latex]x\neq \pm 3,\,-\dfrac{4}{5},\,\dfrac{3}{2}[/latex]

Try It 1

State the restrictions on the variable of the rational expression:

1. [latex]\dfrac{(2x-5)}{(x+2)(x-1)}[/latex]

2. [latex]\dfrac{3x+1}{5x^2(x-2)}[/latex]

3. [latex]\dfrac{x^2}{(x^2-16)(x^2-7x-8)}[/latex]

Show Answer

Example 2

State the restricted values and simplify.

1. [latex]\dfrac{4x^2}{8x^3+4x^2}[/latex]

2. [latex]\dfrac{x^2-9}{x^2-2x-15}[/latex]

3. [latex]\dfrac{6x^2-7x-20}{2x^2+9x-35}[/latex]

Solution

We start by factoring the numerator and denominator to determine if there are any common factors. We make note of any restrictions on the variable, then cancel common factors.

[latex]\begin{aligned}\dfrac{4x^2}{8x^3+4x^2}&=\dfrac{4x^2}{4x^2(2x+1)}&&\text{Factor. Restrictions: }x\neq 0,\,-\frac{1}{2}\\\\&=\dfrac{\cancel{4x^2}}{\cancel{4x^2}(2x+1)}&&\text{Cancel: }\dfrac{4x^2}{4x^2}=1\\\\&=\dfrac{1}{2x+1}\end{aligned}[/latex]

[latex]\begin{aligned}\dfrac{x^2-9}{x^2-2x-15}&=\dfrac{(x-3)(x+3)}{(x-5)(x+3)}&&\text{Factor. Restrictions: }x\neq 5,\,-3\\\\&=\dfrac{(x-3)\cancel{(x+3)}}{(x-5)\cancel{(x+3)}}&&\text{Cancel: }\dfrac{x+3}{x+3}=1\\\\&=\dfrac{x-3}{x-5}\end{aligned}[/latex]

[latex]\begin{aligned}\dfrac{6x^2-7x-20}{2x^2+9x-35}&=\dfrac{(3x+4)(2x-5)}{(2x-5)(x+7)}&&\text{Factor. Restrictions: }x\neq \frac{5}{2},\,-7\\\\&=\dfrac{(3x+4)\cancel{(2x-5)}}{\cancel{(2x-5)}(x+7)}&&\text{Cancel: }\dfrac{2x-5}{2x-5}=1\\\\&=\dfrac{3x+4}{x+7}\end{aligned}[/latex]

Try It 2

State restricted values, then simplify.

1. [latex]\dfrac{9x^2}{18x^3+27x^2}[/latex]

2. [latex]\dfrac{x^2-16}{3x^2+11x-4}[/latex]

3. [latex]\dfrac{6x^2+19x+10}{2x^2+13x+20}[/latex]

Show Answer

Multiplying Rational Expressions

We multiply rational expressions the same way we multiply fractions: [latex]\dfrac{A}{B}\cdot \dfrac{C}{D}=\dfrac{A\cdot C}{B\cdot D}[/latex]. We can simplify by cancelling common factors, and we must state any restrictions on the variable.

Example 3

Multiply: [latex]\displaystyle \frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}}[/latex]

Solution

[latex]\begin {aligned}\dfrac{5a^{2}}{14}\cdot\dfrac{7}{10a^{3}}&=\dfrac{5a^2\cdot 7}{14\cdot 10a^3}&&\text{Multiply straight across}\\\\&=\dfrac{5\cdot a^2\cdot 7}{7\cdot 2\cdot 5\cdot 2\cdot a^2\cdot a}&&\text{Factor. Restrictions: }a\neq 0\\\\&=\dfrac{\cancel{5}\cdot \cancel{a^2}\cdot \cancel{7}}{\cancel{7}\cdot 2\cdot\cancel{5}\cdot 2\cdot\cancel{a^2}\cdot a}&&\text{Cancel}\\\\&=\dfrac{1}{4a}\end{aligned}[/latex]

Example 4

State the restrictions. Multiply: [latex]\dfrac{{{a}^{2}}-a-2}{5a}\cdot \dfrac{10a}{a+1}[/latex]. State the restrictions.

Solution

Factor the numerators and denominators, then state the restrictions:

[latex]\dfrac{{{a}^{2}}-a-2}{5a}\cdot \dfrac{10a}{a+1}=\dfrac{\left(a-2\right)\left(a+1\right)}{5\cdot{a}}\cdot\dfrac{5\cdot2\cdot{a}}{\left(a+1\right)}\;\;\;\;a\neq 0,\,-1[/latex]

Simplify common factors:

[latex]\begin{aligned}&\dfrac{\left(a-2\right)\cancel{\left(a+1\right)}}{\cancel{5}\cdot{\cancel{a}}}\cdot\dfrac{\cancel{5}\cdot2\cdot{\cancel{a}}}{\cancel{\left(a+1\right)}}\\\\&=\dfrac{a-2}{1}\cdot\dfrac{2}{1}\end{aligned}[/latex]

Multiply simplified rational expressions. This expression can be left with the numerator in factored form or multiplied out.

[latex]\begin{aligned}&\dfrac{\left(a-2\right)}{1}\cdot\dfrac{2}{1}\\\\&=2\left(a-2\right)\end{aligned}[/latex]

Try It 3

State the restrictions on the variable, then multiply the rational expressions.

1. [latex]\dfrac{x+2}{x^2}\cdot\dfrac{5x}{x^2-4}[/latex]

2. [latex]\dfrac{4x-3}{x^2-2x+1}\cdot\dfrac{x-1}{16x^2-9}[/latex]

3. [latex]\dfrac{x^2-3x-4}{4x}\cdot \dfrac{2x^2+2x}{x^2+7x+6}[/latex]

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Multiplying Rational Functions

Now, we can apply the methods we learned multiplying rational expressions to multiply rational functions.

The product function

The product function [latex]\left(f\cdot g\right)(x)=f(x)\cdot g(x)[/latex]

Its domain is all real numbers except the combined restrictions of [latex]f(x)[/latex] and [latex]g(x)[/latex].

Example 5

State the domain of the product function, then multiply the two rational functions [latex]f(x)=\dfrac{4x^5}{5x^2-125}[/latex] and [latex]g(x)=\dfrac{x^2-3x-10}{20x^8}[/latex].

Solution

Before we perform the multiplication, we first need to discuss the domain of the product. The denominators of the two functions cannot be zero because division by zero is undefined. The domain of the product of two functions has restrictions equal to those of each function. In this case, [latex]5x^2-125=5(x^2-25)=5(x-5)(x+5)[/latex], which when set equal to zero produces the restrictions [latex]x=\pm 5[/latex]. Also, [latex]20x^8=0[/latex] when [latex]x=0[/latex]. Consequently, the domain of [latex]\left(f\cdot g\right)(x)=\{x\;|\;x\in \mathbb{R},\;x\neq 0,\,\pm 5\}[/latex].

[latex]\begin{aligned}(f \cdot g)(x)&=f(x)\cdot g(x)\\\\&=\dfrac{4x^5}{5x^2-125} \times \dfrac{x^2-3x-10}{20x^8}\\\\&=\dfrac{4x^5}{5(x^2-25)} \times \dfrac{(x-5)(x+2)}{20x^8}&&\text{Factor}\\\\&=\dfrac{\cancel{4}\cancel{x^5}}{5(x+5)\cancel{(x-5)}} \times \dfrac{\cancel{(x-5)}(x+2)}{\cancel{4} \cdot 5 \cdot \cancel{x^5} \cdot x^3}&&\text{Cancel common factors to }1\\\\&=\dfrac{1}{5(x+5)} \times \dfrac{(x+2)}{5x^3}&&\text{Multiply}\\\\&= \dfrac{(x+2)}{25x^3(x+5)}\end{aligned}[/latex]

Try It 4

State the domain of the product, then multiply the two rational functions [latex]f(x)=\dfrac{4x^5}{5x^2-125}[/latex] and [latex]g(x)=\dfrac{x^2-3x-10}{20x^8}[/latex].

Show Answer

Dividing Rational Expressions

We have seen that we multiply rational expressions as we multiply numerical fractions. It should come as no surprise that we also divide rational expressions the same way we divide numerical fractions. Specifically, to divide rational expressions, keep the first rational expression, change the division sign to multiplication, and then take the reciprocal of the second rational expression. [latex]\dfrac{A}{B}\div\dfrac{C}{D}=\dfrac{A}{B}\cdot\dfrac{D}{C}[/latex].

We still need to think about restrictions on the variable, specifically, the variable values that would make either denominator equal zero. But there is a new consideration this time—because we divide by multiplying by the reciprocal of the second rational expression, we also need to find the values that would make the
numerator of that expression equal zero.

Example 6

Divide: [latex]\displaystyle\frac{5x^{2}}{9}\div\frac{15x^{3}}{27}[/latex]. State the restrictions on the variable.

Solution

To determine the restricted values, first notice the two denominators, [latex]9[/latex] and [latex]27[/latex], can never equal [latex]0[/latex]. This means there are no restrictions from those denominators.

However, because [latex]15x^{3}[/latex] becomes the denominator in the reciprocal of [latex]\displaystyle \frac{15{{x}^{3}}}{27}[/latex], we must find the values of [latex]x[/latex] that would make [latex]15x^{3}=0[/latex].

[latex]\begin{aligned}15x^{3}&=0\\x&=0\end{aligned}[/latex]

Therefore, [latex]x=0[/latex] is a restricted value. i.e. [latex]x\neq 0[/latex].

To divide we rewrite division as multiplication by the reciprocal:

[latex]\begin{aligned}\dfrac{5x^{2}}{9}\div\dfrac{15x^{3}}{27}&=\dfrac{5x^{2}}{9}\cdot\frac{27}{15x^{3}}\\\\&=\dfrac{5\cdot x^2\cdot 9\cdot 3}{9\cdot 5 \cdot 3\cdot x^2\cdot x}&&\text{Factor}\\\\&=\dfrac{\cancel{5}\cdot\cancel{x^2}\cdot\cancel{9}\cdot\cancel{3}}{\cancel{9}\cdot \cancel{5} \cdot \cancel{3}\cdot \cancel{x^2}\cdot x}&&\text{Cancel}\\\\&=\dfrac{1}{x}&&\text{Simplify}\end{aligned}[/latex]

Example 7

Divide: [latex]\displaystyle \frac{3x^{2}}{x+2}\div\frac{6x^{4}}{\left(x^{2}+5x+6\right)}[/latex]. State the restrictions.

Solution

We can determine the restricted values as we work through the division.We will highlight the denominators where restrictions may lie in blue to make determining the restrictions on the domain more obvious.

[latex]\begin{aligned}\dfrac{3x^2}{\color{blue}{x+2}}\div\frac{6x^4}{\left(\color{blue}{x^2+5x+6}\right)}&=\dfrac{3x^2}{x+2}\cdot\dfrac{\left(\color{blue}{x^2+5x+6}\right)}{\color{blue}{6x^4}}&&\text{Factor}\\\\&=\dfrac{3x^2}{\color{blue}{x+2}}\cdot\dfrac{\color{blue}{(x+3)(x+2)}}{\color{blue}{6x^4}}&&\text{Determine restricted values: }x\neq 0,\,-2,\,-3\\\\&=\dfrac{3\cdot x^2}{x+2}\cdot\dfrac{(x+3)(x+2)}{2\cdot 3\cdot x^2\cdot x^2}&&\text{Factor}\\\\&=\dfrac{\cancel{3}\cdot\cancel{x^2}}{\cancel{(x+2)}}\cdot\dfrac{(x+3)\cancel{(x+2)}}{2\cdot\cancel{3}\cdot x^2\cdot\cancel{x^2}}&&\text{Cancel}\\\\&=\dfrac{x+3}{2x^2}&&\text{Simplify}\end{aligned}[/latex]

Notice that once we rewrite the division as multiplication by a reciprocal, we follow the same process we use to multiply rational expressions.

Try It 5

Divide: [latex]\dfrac{2x^3}{x^2-16}\div\dfrac{2x^3+4x^2}{x^2-2x-8}[/latex]. State the restrictions on the variable.

Show Answer

Dividing Rational Functions

Now, we can apply the methods we learned dividing rational expressions to divide rational functions.

The quotient function

The quotient function [latex]\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}[/latex].

Its domain is all real numbers except the combined restricted values of [latex]f(x)[/latex] and [latex]g(x)[/latex], and the restricted values from the numerator of [latex]g(x)[/latex] since it becomes the denominator of the reciprocal.

Example 8

Divide the rational functions [latex]f(x)=\dfrac{x^2-6x+9}{4x^2-4x-24}[/latex] and [latex]g(x)=\dfrac{5x^2-45}{20x^3}[/latex]. State the domain of the quotient function.

Solution

Before we perform the division, we first need to discuss the domain of the quotient. The denominators of the two functions cannot be zero because division by zero is undefined. The domain of the quotient of two functions has restrictions equal to those of each function, as well as any restrictions on the denominator of the reciprocal of the second function as we change division to multiplication of the reciprocal.

We will highlight the denominators where restrictions may lie in blue to make determining the restrictions on the domain more obvious.

[latex]\begin{aligned}\left(\dfrac{f}{g}\right)(x)&=f(x)\div g(x)\\\\&=\dfrac{x^2-6x+9}{\color{blue}{4x^2-4x-24}} \div \dfrac{5x^2-45}{\color{blue}{20x^3}}\\\\&=\dfrac{x^2-6x+9}{\color{blue}{4x^2-4x-24}} \cdot \dfrac{\color{blue}{20x^3}}{\color{blue}{5x^2-45}}&&\text{Multiply by the reciprocal}\\\\&=\dfrac{x^2-6x+9}{\color{blue}{4(x^2-x-6)}} \cdot \dfrac{\color{blue}{20x^3}}{\color{blue}{5(x^2-9)}}&&\text{Factor}\\\\&=\dfrac{(x-3)(x-3)}{\color{blue}{4(x-3)(x+2)}} \cdot \dfrac{\color{blue}{20x^3}}{\color{blue}{5(x-3)(x+3)}}&&\text{Factor. Restrictions: }x\neq \pm 3,\,-2,\,0\\\\&=\dfrac{\cancel{(x-3)}\cancel{(x-3)}}{\cancel{4}\cancel{(x-3)}(x+2)} \cdot \dfrac{\cancel{5} \cdot \cancel{4} \cdot x^3}{\cancel{5}(x+3)\cancel{(x-3)}}&&\text{Cancel}\\\\&=\dfrac{x^3}{(x+2)(x+3)}&&\text{Simplify}\end{aligned}[/latex]

Domain of [latex]\left(\dfrac{f}{g}\right)(x)=\{x\;|\;x\in\mathbb{R},\;x\neq\pm 3,\,-2,\,0\}[/latex]

Try It 6

Divide the rational functions [latex]f(x)=\dfrac{2x^2+5x-12}{3x^2-2x}[/latex] and [latex]g(x)=\dfrac{2x^2-x-3}{3x^2+x-2}[/latex]. State the domain of the quotient function.

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Multiplying and Dividing Rational Functions

The techniques we have learned to multiply and divide rational functions can be used to extend the number of functions being multiplied or divided. Any time we need to divide, we multiply by the reciprocal.

Example 9

Simplify: [latex]\left(\dfrac{f\cdot g}{h}\right)(x)[/latex] when [latex]f(x)=\dfrac{x^3+2x^2}{x+1}[/latex], [latex]g(x)=\dfrac{x^2-2x-3}{x}[/latex], and [latex]h(x)=\dfrac{x^2-3x}{x^2+7x+10}[/latex].

State the domain of the product/quotient function.

Solution

Blue highlights where restrictions on the domain are found.

[latex]\begin{aligned}\left(\dfrac{f\cdot g}{h}\right)(x)&=\dfrac{x^3+2x^2}{x+1}\cdot\dfrac{x^2-2x-3}{x}\div\dfrac{x^2-3x}{x^2+7x+10}\\\\&=\dfrac{x^3+2x^2}{\color{blue}{x+1}}\cdot\dfrac{x^2-2x-3}{\color{blue}{x}}\cdot \dfrac{\color{blue}{x^2+7x+10}}{\color{blue}{x^2-3x}}&&\text{Write division as multiplication of the reciprocal}\\\\&=\dfrac{x^2(x+2)}{\color{blue}{x+1}}\cdot\dfrac{(x-3)(x+1)}{\color{blue}{x}}\cdot \dfrac{\color{blue}{(x+5)(x+2)}}{\color{blue}{x(x-3)}}&&\text{Factor. Restrictions: }x\neq -1,\,0,\,3,\,-2,\,-5\\\\&=\dfrac{\cancel{x^2}(x+2)}{\cancel{(x+1)}}\cdot\dfrac{\cancel{(x-3)}\cancel{(x+1)}}{\cancel{x}}\cdot \dfrac{(x+5)(x+2)}{\cancel{x}\cancel{(x-3)}}&&\text{Cancel}\\\\&=(x+5)(x+2)^2\end{aligned}[/latex]

Domain of [latex]\left(\dfrac{f\cdot g}{h}\right)(x)=\{x\;|\;x\in\mathbb{R},\;x\neq -1,\,0,\,-2,\,-5\}[/latex]

Try It 7

Simplify: [latex]\left(\dfrac{f\cdot h}{g}\right)(x)[/latex] when [latex]f(x)=\dfrac{x^3+2x^2}{x+1}[/latex], [latex]g(x)=\dfrac{x^2-2x-3}{x}[/latex], and [latex]h(x)=\dfrac{x^2-3x}{x^2+7x+10}[/latex].

State the domain of the product/quotient function.

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Chapter 7: Rational Functions