Topics
5. Rational Functions
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United States of America
Grades 9-12

5.03 Multiplying and dividing rational expressions

Lesson
Practice

Introduction

In lesson  5.02 Rational expressions  , we compared simplifying rational expressions to simplifying rational numbers, then simplified rational expressions using factoring and polynomial division. We will connect our understanding of multiplication and division of rational numbers from Algebra 1 lesson  1.01 The real number system  to rational expressions in this lesson.

Multiplying and dividing rational expressions

Exploration

Complete the table by finding the product of each row and column:

\dfrac{3}{2}-\dfrac{1}{4}\dfrac{1}{2}-\dfrac{2}{5}\dfrac{3}{8}
\dfrac{3x}{2}
-\dfrac{1}{4x}
\dfrac{x}{2}
-\dfrac{2x}{5}
\dfrac{3}{8x}
  1. Can each product be rewritten as rational expression?
  2. What conclusions can we make about the products of rational expressions?
  3. Complete the table again, but this time find the quotient of each pair of rational expressions.
  4. What conclusions can we make about the quotients of rational expressions?

Recall that closure means a set of numbers is closed under an operation if performing that operation on numbers in the set produces a number that is also in the set. The product or quotient of two rational expressions is equivalent to a rational expression, so rational expressions are closed under multiplication and division.

To multiply two (or more) rational expressions together, we multiply the numerators to form the new numerator and multiply the denominators to form the new denominator - the same process as for multiplying fractions:\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}

To divide two rational expressions, we multiply the first rational expression by the reciprocal of the second rational expression - the same process as for dividing fractions:\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} = \frac{AD}{BC}

Since rational expressions are closed under multiplication and division, the same algebraic properties of real numbers apply to rational expressions:

A table showing the properties of real numbers. First row: commutative property of multiplication: A over B times C over D equals C over D times A over B; Second row: associative property of multiplication: left parenthesis A over B times C over D right parenthesis times E over F equals A over B times left parenthesis C over D times E over F right parenthesis; Third row: multiplicative identity: A over B times 1 equals A over B; Fourth row: multiplicative inverse: A over B times B over A equals 1; Fifth row: distributive property of multiplication: A over B left parenthesis C over D plus E over F right parenthesis equals A C over B D plus A E over B F.

Finding common factors, in particular the greatest common factor (GCF), between any of the numerators and denominators can help us use the algebraic properties to simplify rational expressions.

We need to state restrictions on the variables so we do not get an expression with 0 in the denominator, leading to an undefined expression.

Examples

Example 1

Fully simplify the expression, justifying each step. State any restrictions on the variables.\frac{21a^3 b^6}{20 c^2} \cdot \frac{16 a^2 c^3}{15 b^6}

Worked Solution
Create a strategy

We can multiply the fractions to combine numerators and denominators, identify common factors between the numerator and denominator and then use the commutative and associative properties of multiplication to group like terms together.

Any variables that lead the denominators of the rational expressions to equal zero should be excluded.

Apply the idea

The restrictions on the variables for the expression given are b \neq 0 and c \neq 0.

\displaystyle \frac{21a^3 b^6}{20 c^2} \cdot \frac{16 a^2 c^3}{15 b^6}\displaystyle =\displaystyle \dfrac{ \left( 21a^3 b^6 \right) \left( 16 a^2 c^3 \right)}{ \left (20c^2)(15b^6 \right)}Definition of multiplying rational expressions
\displaystyle =\displaystyle \dfrac{ \left(21 \cdot 16 \right) \left(a^3 \cdot a^2 \right) \left(b^6 \right) \left(c^3 \right)}{\left(20 \cdot 15 \right) \left(b^6 \right) \left (c^2 \right)}Commutative and associative properties of multiplication
\displaystyle =\displaystyle \dfrac{28 a^5 c}{25}Simplify common factors

\dfrac{28 a^5 c}{25}, b \neq 0 and c \neq 0.

Reflect and check

If we hadn't restricted the variables at the beginning, we might forget to do so at the end, because the simplified expression for this problem does not have any variables in the denominator.

When simplifying, we often don't show every step, and may combine or skip steps. For instance, we might "simplify" like terms from numerators and denominators of different fractions before multiplying the expressions together.

In that case the multiplying, using the commutative, associative and exponent properties are implied. This is what the simplifying would look like if we didn't combine steps:

The restrictions on the variables for the expression given are b \neq 0 and c \neq 0.

\displaystyle \frac{21a^3 b^6}{20 c^2} \cdot \frac{16 a^2 c^3}{15 b^6}\displaystyle =\displaystyle \dfrac{ \left( 21a^3 b^6 \right) \left( 16 a^2 c^3 \right)}{ \left (20c^2)(15b^6 \right)}Definition of multiplying rational expressions
\displaystyle =\displaystyle \dfrac{ \left(21 \cdot 16 \right) \left(a^3 \cdot a^2 \right) \left(b^6 \right) \left(c^3 \right)}{\left(20 \cdot 15 \right) \left(b^6 \right) \left (c^2 \right)}Commutative and associative properties of multiplication
\displaystyle =\displaystyle \dfrac{7 \cdot 3 \cdot 4 \cdot 4}{5 \cdot 4 \cdot 3 \cdot 5} \cdot \dfrac{a^3 a^2}{1} \cdot \dfrac{b^6}{b^6} \cdot \dfrac{c^3}{c^2}Definition of multiplying rational expressions
\displaystyle =\displaystyle \dfrac{7 \cdot 4}{5 \cdot 5} \cdot \dfrac{a^3 a^2}{1} \cdot \dfrac{b^6}{b^6} \cdot \dfrac{c^3}{c^2}Simplify coefficients
\displaystyle =\displaystyle \dfrac{7 \cdot 4}{5 \cdot 5} \cdot a^5 \cdot 1 \cdot cProduct and quotient properties of exponents
\displaystyle =\displaystyle \dfrac{28 a^5 c}{25}Definition of multiplying rational expressions

\dfrac{28 a^5 c}{25}, b \neq 0 and c \neq 0.

Example 2

Fully simplify the expression, justifying each step. State any restrictions on the variables.\frac{10 x}{y^2 z} \div \frac{3 x^2 z}{10 y}

Worked Solution
Create a strategy

We can first turn the division into a multiplication by taking the reciprocal of the second fraction. We then again want to identify common factors that will help us simplify the rational expression.

Apply the idea
\displaystyle \frac{10 x}{y^2 z} \div \frac{3 x^2 z}{10 y}\displaystyle =\displaystyle \frac{10 x}{y^2 z} \cdot \frac{10 y}{3 x^2 z}Rewrite division as multiplication
\displaystyle =\displaystyle \frac{10}{y z} \cdot \frac{10}{3 x z}Multiplicative inverse of rational expressions
\displaystyle =\displaystyle \frac{100}{3 x y z^2}Definition of multiplying rational expressions

\dfrac{100}{3 x y z^2}, x \neq 0, y \neq 0, and z \neq 0.

Reflect and check

Note that the first step was to rewrite the division as a multiplication, because we need to switch to multiplication in order to use the commutative and associative algebraic properties in the simplifying steps.

While there are some simplification steps that can be done in division form, it is much easier to make a mistake when doing so.

Example 3

Fully simplify the rational expression, justifying each step. State any restrictions on the variables.\frac{x^2 + 3x - 10}{x^3 - 8} \cdot \frac{x^2 - 9}{x^2 + 2x - 15}

Worked Solution
Create a strategy

When the numerator or denominator of a rational expression contains polynomials, we need to see if it's possible to use factoring to rewrite the polynomials in terms of multiplication. Only then can we use our multiplication properties to simplify the expression.

In particular, notice that three of the expressions are quadratic, and the other one is a difference of two cubes, so we have factoring techniques we can use on each part of the expression.

Factoring the denominators will also help us determine any restricted values:\frac{x^2 + 3x - 10}{x^3 - 8} \cdot \frac{x^2 - 9}{x^2 + 2x - 15} = \frac{\left(x + 5\right)\left(x - 2\right)}{\left(x - 2\right)\left(x^2 + 2x + 4\right)} \cdot \frac{\left(x - 3\right)\left(x + 3\right)}{\left(x + 5\right)\left(x - 3\right)}

Apply the idea

First, we will need to determine values of x for which the expression is undefined before simplifying, so if we look at the factored forms in the denominators of the multiplication problem, we can see that the binomials (x-2), (x+5), and (x-3) are in the denominator. The original expression will be undefined for x=2, x=-5, and x=3.

We also need to determine when the polynomial x^2+2x+4 is equal to zero. Since we cannot factor the polynomial, we can use the quadratic formula to determine when x^2+2x+4=0.

\displaystyle x\displaystyle =\displaystyle \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}Quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-2 \pm \sqrt{2^2-4\left(1 \right) \left(4 \right)}}{2 \left(1 \right)}Substitute a=1, b=2, and c=4
\displaystyle x\displaystyle =\displaystyle \dfrac{-2 \pm \sqrt{-12}}{2}Evaluate the multiplication

Since the discriminant is negative, we can stop evaluating because the solution will be complex. There are no real values of x that will cause the quadratic expression to be zero. Thus, there are no additional restricted values.

Second, we will use the factored forms to simplify the rational expressions:

\displaystyle \frac{x^2 + 3x - 10}{x^3 - 8} \cdot \frac{x^2 - 9}{x^2 + 2x - 15}\displaystyle =\displaystyle \frac{\left(x + 5\right)\left(x - 2\right)}{\left(x - 2\right)\left(x^2 + 2x + 4\right)} \cdot \frac{\left(x - 3\right)\left(x + 3\right)}{\left(x + 5\right)\left(x - 3\right)}Factor each expression
\displaystyle =\displaystyle \dfrac{ \left( x + 5 \right) \left(x-2 \right) \left (x-3 \right) \left(x + 3 \right)}{\left(x+5 \right) \left(x - 2 \right) \left(x-3 \right) \left(x^2 + 2x +4 \right)}Commutative and associative properties of multiplication
\displaystyle =\displaystyle \frac{1}{x^2 + 2x + 4} \cdot \frac{x + 3}{1}Simplify all common factors between numerator and denominator (multiplicative inverse: \dfrac{A}{B} \cdot \dfrac{B}{A}=1)
\displaystyle =\displaystyle \frac{x + 3}{x^2 + 2x + 4}Definition of multiplying rational expressions

Finally, we can state that the simplified form is \dfrac{x+3}{x^2 + 2x + 4}, x \neq 2, x \neq -5, and x \neq 3.

Reflect and check

We could have skipped the commutative/associative properties of multiplication step by taking the shortcut, and simplified common factors in numerator and denominator across the fraction.

Two expressions being multiplied is shown: left parenthesis x plus 5 right parenthesis left parenthesis x minus 2 right parenthesis all over left parenthesis x minus 2 right parenthesis left parenthesis x squared plus 2 x plus 4 right parenthesis times left parenthesis x minus 3 right parenthesis left parenthesis x plus 3 right parenthesis all over left parenthesis x plus 5 right parenthesis left parenthesis x minus 3 right parenthesis. Speak to your teacher for more information.

Note that this only works when all the terms are written in terms of multiplication, and one common factor needs to be in the numerator and the other in the denominator in order to create an \dfrac{A}{B} \cdot \dfrac{B}{A}=1 situation.

Idea summary

Since rational expressions are closed over multiplication and division, when multiplying rational expressions, we can use the algebraic properties of multiplication to find common factors and simplify the expressions.

Division of rational expressions can be expressed rewritten as a multiplication problem, where \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \cdot \dfrac{D}{C}.

By the definition of multiplying rational expressions, we know \dfrac{A}{B} \cdot \dfrac{C}{D}= \dfrac{AC}{BD}.

Outcomes

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.APR.D.7 (+)

Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.

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