Can each product be rewritten as a rational expression?
What conclusions can we make about the products of rational expressions?
Complete the table again, but this time find the quotient of each pair of rational expressions.
What conclusions can we make about the quotients of rational expressions?

The product or quotient of two rational expressions always results in another rational expression, even if the denominator is simply 1.

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 used when 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 used when dividing fractions:\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} = \frac{AD}{BC}

Another way to represent division is with a complex fraction. The fraction in the numerator is divided by the fraction in the denominator.

\frac{A}{B} \div \frac{C}{D} = \dfrac{\frac{A}{B}}{\frac{C}{D}}

The same algebraic properties we use with 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 will need to assume that no denominator is equal to zero to avoid undefined expressions. We can do this by stating restrictions on the variables which will relate to restrictions on the domain of an associated rational function.

Examples

Example 1

Fully simplify the expression, justifying each step. Write 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)}	Combine into a single fraction
	\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}	Rewrite using factors
	\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}	Divide out common factors in the coefficients
	\displaystyle =	\displaystyle \dfrac{7 \cdot 4}{5 \cdot 5} \cdot a^5 \cdot 1 \cdot c	Product and quotient rules of exponents
	\displaystyle =	\displaystyle \dfrac{28 a^5 c}{25}	Evaluate the multiplication

\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 as multiplication by the reciprocal
	\displaystyle =	\displaystyle \frac{10}{y z} \cdot \frac{10}{3 x z}	Multiplicative inverse
	\displaystyle =	\displaystyle \frac{100}{3 x y z^2}	Evaluate the multiplication

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

Reflect and check

Notice that whenever we mutliply or divide variables we use the laws of exponents. When we divided \dfrac{y}{y^2} we subtracted the exponents. When we multiplied z\cdot z we added the exponents.

Example 3

Fully simplify the rational expression, justifying each step. Write 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.

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.

Example 4

Rewrite the complex fraction as a simplified rational expression, assuming no denominator equals zero:

\dfrac{\dfrac{3 - u}{14}} { \dfrac{u - 3}{3u^2}}

Worked Solution

Create a strategy

To simplify this to an equivalent expression, we will multiply the first fraction by the reciprocal of the second fraction. This approach utilizes the property that dividing by a fraction is equivalent to multiplying by its reciprocal.

\displaystyle \dfrac{\dfrac{3 - u}{14}} { \dfrac{u - 3}{3u^2}}	\displaystyle =	\displaystyle \dfrac{3 - u}{14} \div \dfrac{u - 3}{3u^2}	Rewrite the complex fraction as division
	\displaystyle =	\displaystyle \dfrac{3 - u}{14} \cdot \dfrac{3u^2}{u - 3}	Multiply by the reciprocal

Apply the idea

\displaystyle \dfrac{\dfrac{3 - u}{14}} { \dfrac{u - 3}{3u^2}}	\displaystyle =	\displaystyle \frac{3 - u}{14} \cdot \frac{3u^2}{u - 3}	Replace division with multiplication by the reciprocal
	\displaystyle =	\displaystyle \dfrac{(3-u) \cdot 3u^2}{14(u - 3)}	Multiply numerators and denominators respectively
	\displaystyle =	\displaystyle \dfrac{-1(u-3) \cdot 3u^2}{14(u - 3)}	Factor out -1 from the numerator
	\displaystyle =	\displaystyle \dfrac{-3u^2}{14}	Divide out the common factor, u-3

The equivalent expression after simplifying is \dfrac{-3u^2}{14}.

Reflect and check

We can check that our simplified expression is equivalent by confirming that it simplifies to the same value as the original expression for various values of u, excluding those that would result in a zero denominator.

Original expression with u=2:

\displaystyle \frac{3 - 2}{14} \div \frac{2 - 3}{3(2)^2}	\displaystyle =	\displaystyle \frac{1}{14} \cdot \frac{3(4)}{-1}
	\displaystyle =	\displaystyle \frac{12}{-14}
	\displaystyle =	\displaystyle -\frac{6}{7}

Equivalent expression with u=2:

\displaystyle \dfrac{-3(2)^2}{14}	\displaystyle =	\displaystyle \dfrac{-3(4)}{14}
	\displaystyle =	\displaystyle \dfrac{-12}{14}
	\displaystyle =	\displaystyle -\dfrac{6}{7}

Both calculations yield the same result, confirming the expressions are equivalent.

Idea summary

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

We can use the algebraic properties of multiplication to find common factors and simplify the expressions.

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

Rational expressions can also involve complex fractions which can be simplified using the same skills used with multiplication and division: \dfrac{\dfrac{A}{B}}{\dfrac{C}{D}}= \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \cdot \dfrac{D}{C} = \dfrac{AD}{BC}

Outcomes

A2.EO.1

The student will perform operations on and simplify rational expressions.

A2.EO.1a

Add, subtract, multiply, or divide rational algebraic expressions, simplifying the result.

A2.EO.1b

Justify and determine equivalent rational algebraic expressions with monomial and binomial factors. Algebraic expressions should be limited to linear and quadratic expressions.

A2.EO.1c

Recognize a complex algebraic fraction and simplify it as a product or quotient of simple algebraic fractions.

A2.EO.1d

Represent and demonstrate equivalence of rational expressions written in different forms.

5.04 Multiply and divide rational expressions

Ideas

Multiply and divide rational expressions

Exploration

Examples

Example 1

Example 2

Example 3

Example 4

Outcomes

A2.EO.1

A2.EO.1a

A2.EO.1b

A2.EO.1c

A2.EO.1d

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