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# Products and Quotients of Rational Expressions

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Practice Products and Quotients of Rational Expressions
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Rational Expression Multiplication and Division
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How can you use your knowledge of multiplying fractions to multiply the following rational expressions?

$\frac{10y+20}{5y-15} \cdot \frac{y-3}{y^2+10y+16}$

### Guidance

Rational expressions are examples of fractions, so you multiply and divide rational expressions in the same ways that you multiply and divide fractions. As with fractions, after multiplying or dividing you will want to simplify your result. You will also want to state any restrictions that would cause the denominator of either original rational expression to be equal to zero. To multiply:

$\frac{x^2+2x}{x+3}\cdot \frac{x^2+4x+3}{x}$

First, factor all expressions that can be factored:

$\frac{x(x+2)}{x+3}\cdot \frac{(x+3)(x+1)}{x}$

Next, multiply the numerators and multiply the denominators to create one big rational expression. Leave in factored form:

$\frac{x(x+2)(x+3)(x+1)}{x(x+3)}$

Simplify:

$\frac{\cancel{x}(x+2)\cancel{(x+3)}(x+1)}{\cancel{x}\cancel{(x+3)}}$

$=(x+2)(x+1)$

$=x^2+3x+2$

Finally, state the restrictions based on the original rational expressions:

$x\ne -3$ and $x\ne 0$ .

To divide rational expressions, recall that dividing one fraction by another is the same as multiplying the first fraction by the reciprocal of the second fraction. For example, $\frac{x^2+2x}{x+3}\div \frac{x}{x^2+4x+3}$ is equivalent to, and can be rewritten as, $\frac{x^2+2x}{x+3}\cdot \frac{x^2+4x+3}{x}$ , which can then be solved using the same steps as above.

#### Example A

Multiply the following rational expressions and state the restrictions.

$\frac{4x-8}{x^2-7x+10} \cdot \frac{x^2-3x-10}{x^2-4}$

Solution: Begin by factoring the numerator and denominator of each expression:

$\frac{4(x-2)}{(x-5)(x-2)}\cdot \frac{(x-5)(x+2)}{(x-2)(x+2)}$

Next, multiply the numerators and the denominators to create one rational expression:

$\frac{4(x-2)(x-5)(x+2)}{(x-5)(x-2)(x-2)(x+2)}$

Simplify by removing common factors from the numerator and denominator that divide to make 1:

$\frac{4\cancel{(x-2)}\cancel{(x-5)}\cancel{(x+2)}}{\cancel{(x-5)}\cancel{(x-2)}\cancel{(x+2)}(x-2)}$

The final answer is: $\frac{4}{(x-2)}$ with restrictions: $x\ne 5$ , $x\ne 2$ , and $x\ne -2$ .

#### Example B

Divide the following rational expressions and state the restrictions.

$\frac{m^2-4}{m^2+9m+14} \div \frac{3m^2-6m}{m^2-49}$

Solution: To divide rational expressions, multiply by the reciprocal of the divisor. Then, follow the process for multiplying rational expressions.

$\frac{m^2-4}{m^2+9m+14} {\color{red}\cdot \frac{m^2-49}{3m^2-6m}}$

Begin by factoring the numerator and denominator of each expression.

$\frac{(m-2)(m+2)}{(m+7)(m+2)}\cdot \frac{(m-7)(m+7)}{3m(m-2)}$

Next, multiply the numerators and the denominators to create one rational expression:

$\frac{(m-2)(m+2)(m-7)(m+7)}{3m(m+7)(m+2)(m-2)}$

Simplify by removing common factors from the numerator and denominator that divide to make 1:

$\frac{\cancel{(m+2)}\cancel{(m-2)}\cancel{(m+7)}(m-7)}{3m \cancel{(m-2)} \cancel{(m+7)} \cancel{(m+2)}}$

The final answer is: $\frac{(m-7)}{3m}$ with restrictions: $x\ne -7$ , $x\ne 2$ , $x\ne 0$ , $x\ne 7$ and $x\ne -2$ . Note that when dividing rational expressions, for restrictions you must consider ALL factors that ever appear in a denominator. This means that both the numerator and denominator of the second rational expression must be considered for restrictions.

#### Example C

Simplify the following rational expressions and state the restrictions.

$\frac{12x^2+13x-35}{5x^2-21x+18} \div \frac{3x^2+16x+21}{5x^2+9x-18}$

Solution: To divide rational expressions, multiply by the reciprocal of the divisor. Then, follow the process for multiplying rational expressions.

$\frac{12x^2+13x-35}{5x^2-21x+18} {\color{red}\times \frac{5x^2+9x-18}{3x^2+16x+21}}$

Begin by factoring the numerator and denominator of each expression.

$\frac{(4x-5)(3x+7)}{(5x-6)(x-3)} \times \frac{(5x-6)(x+3)}{(3x+7)(x+3)}$

Next, multiply the numerators and the denominators to create one rational expression:

$\frac{(4x-5)(3x+7)(5x-6)(x+3)}{(5x-6)(x-3)(3x+7)(x+3)}$

Simplify by removing common factors from the numerator and denominator that divide to make 1:

$\frac{(4x-5)\cancel{(3x+7)}\cancel{(5x-6)}\cancel{(x+3)}}{\cancel{(5x-6)}(x-3)\cancel{(3x+7)}\cancel{(x+3)}}$

The final answer is: $\frac{(4x-5)}{(x-3)}$ with restrictions: $x\ne \frac{6}{5}$ , $x\ne 3$ , $x\ne -\frac{7}{3}$ , and $x\ne -3$ .

#### Concept Problem Revisited

$\frac{10y+20}{5y-15} \cdot \frac{y-3}{y^2+10y+16}$

Begin by factoring the numerator and denominator of each expression.

$\frac{10(y+2)}{5(y-3)} \cdot \frac{y-3}{(y+8)(y+2)}$

Express the factored rational expressions as a single rational expression.

$\frac{10(y+2)(y-3)}{5(y-3)(y+8)(y+2)}$

Cancel the common factors that exist in the numerator and the denominator.

$\frac{\overset{{\color{red}2}}{\cancel{10}}\cancel{(y+2)}\cancel{(y-3)}}{\cancel{5}\cancel{(y-3)}(y+8)\cancel{(y+2)}}$

The result of cancelling the common factors is the answer. Don't forget to include the restrictions.

$\boxed{\frac{2}{y+8}; y \ne 3; y \ne -8; y \ne -2}$

### Vocabulary

Rational Expression
A rational expression is an algebraic expression that can be written in the form $\frac{a(x)}{b(x)}$ where $b \ne 0$ .
Restriction
Any value of the variable in a rational expression that would result in a zero denominator is called a restriction on the denominator.

### Guided Practice

Multiply or divide each of the following and state the restrictions.

1. $\frac{x+7}{x^2-5x-36} \div \frac{x^2-2x-63}{x+4} \cdot \frac{x^2-15x+54}{x^2-36}$

2. $\frac{y^2-25}{y^2-6y} \cdot \frac{y^2-12y+36}{y^2+2y-15} \div \frac{y^2-11y+30}{y^2+4y-21}$

3. $\frac{2x^2+7x-4}{6x^2+x-2} \cdot \frac{15x^2+7x-2}{5x^2+19x-4}$

1. Write the term after the division sign as a reciprocal and multiply.

$\frac{x+7}{x^2-5x-36} {\color{red}\cdot \frac{x+4}{x^2-2x-63}} \cdot \frac{x^2-15x+54}{x^2-36}$

Factor the numerator and denominator of each expression.

$\frac{x+7}{(x-9)(x+4)} \cdot \frac{x+4}{(x-9)(x+7)} \cdot \frac{(x-9)(x-6)}{(x+6)(x-6)}$

Express the factored rational expressions as a single rational expression.

$\frac{(x+7)(x+4)(x-9)(x-6)}{(x-9)(x+4)(x-9)(x+7)(x+6)(x-6)}$

Cancel the common factors that exist in the numerator and the denominator.

$\frac{\overbrace{\cancel{(x+7)}\cancel{(x+4)}\cancel{(x-9)}\cancel{(x-6)}}^{{\color{red}1}}}{\cancel{(x-9)}\cancel{(x+4)}(x-9)\cancel{(x+7)}(x+6)\cancel{(x-6)}}$

The result of cancelling the common factors is the answer.

$\boxed{=\frac{1}{(x-9)(x-6)};x \ne 9;x \ne -4;x \ne -7;x \ne -6;x \ne 6;}$

2. Write the term after the division sign as a reciprocal and multiply.

$\frac{y^2-25}{y^2-6y} \cdot \frac{y^2-12y+36}{y^2+2y-15} {\color{red}\cdot \frac{y^2+4y-21}{y^2-11y+30}}$

Factor the numerator and denominator of each expression.

$\frac{(y+5)(y-5)}{y(y-6)} \cdot \frac{(y-6)(y-6)}{(y+5)(y-3)}\cdot \frac{(y+7)(y-3)}{(y-6)(y-5)}$

Express the factored rational expressions as a single rational expression.

$\frac{(y+5)(y-5)(y-6)(y-6)(y+7)(y-3)}{y(y-6)(y+5)(y-3)(y-6)(y-5)}$

Cancel the common factors that exist in the numerator and the denominator.

$\frac{\cancel{(y+5)}\cancel{(y-5)}\cancel{(y-6)}\cancel{(y-6)}(y+7)\cancel{(y-3)}}{y\cancel{(y-6)}\cancel{(y+5)}\cancel{(y-3)}\cancel{(y-6)}\cancel{(y-5)}}$

The result of cancelling the common factors is the answer.

$\boxed{=\frac{y+7}{y}; y \ne 0;y \ne 6;y \ne -5;y \ne 3;y \ne 5;y \ne -7}$

3. Factor the numerator and denominator of each expression.

$\frac{(2x-1)(x+4)}{(2x-1)(3x+2)} \cdot \frac{(5x-1)(3x+2)}{(5x-1)(x+4)}$

Express the factored rational expressions as a single rational expression.

$\frac{(2x-1)(x+4)(5x-1)(3x+2)}{(2x-1)(3x+2)(5x-1)(x+4)}$

Cancel the common factors that exist in the numerator and the denominator.

$\frac{\overbrace{\cancel{(2x-1)}\cancel{(x+4)}\cancel{(5x-1)}\cancel{(3x+2)}}^{{\color{red}1}}}{\cancel{(2x-1)}\cancel{(3x+2)}\cancel{(5x-1)}\cancel{(x+4)}}$

The result of cancelling the common factors is the answer.

$\boxed{=1; x \ne \frac{1}{2}; x \ne -\frac{2}{3}; x \ne \frac{1}{5}; x \ne -4}$

### Practice

Multiply or divide each of the following and state the restrictions for each.

1. $\frac{3x+9}{6x} \cdot \frac{x^2}{x^2-9}$
2. $\frac{c^2+5c+6}{c-1} \cdot \frac{c^2-1}{c+3}$
3. $\frac{a^2+3a}{3a-9} \cdot \frac{a^2-a-6}{2a^2+6a}$
4. $\frac{y-3}{y+3} \cdot \frac{y^2-9}{y+3} \cdot \frac{y^2+6y+9}{y^2-6y+9}$
5. $\frac{m^2-4m-5}{m^2-5m} \cdot \frac{m^2-6m+5}{m^2-1} \cdot \frac{m}{m-5}$
6. $\frac{x^2-x-20}{x^2-25} \div \frac{3x+12}{x+5}$
7. $\frac{d^2-9}{3-3d} \div \frac{d^2+5d+6}{d^2+3d-4}$
8. $\frac{4x^2-20x}{3x+6} \div \frac{x-5}{x^2-x-6}$
9. $\frac{4n^2-9}{2n^3+2n^2-4n} \div \frac{2n^2-n-3}{3n^2-6n+3}$
10. $\frac{e^2+10e+21}{2e^2+7e-15} \div \frac{e^2+8e+15}{e^2+10e+25}$
11. $\frac{x^2+2x-15}{x^2-6x+8} \cdot \frac{x^2+2x-8}{x^2-6x+9} \cdot \frac{x^2-7x+12}{x^2-x-30}$
12. $\frac{2x^2+5x-3}{4x^2-12x+5} \div \frac{3x^2+13x+12}{6x^2-7x-20}$
13. $\frac{5m^2-20}{m^2+14m+33} \cdot \frac{m^2+10m-11}{m^2-8m+12} \cdot \frac{m^2-3m-18}{m^2+m-2}$
14. $\frac{2y^2+5y-12}{y^2+9y+14} \div \frac{6y^2-7y-3}{3y^2+25y+8} \cdot \frac{y^2+3y-28}{y^2-16}$
15. $\frac{x^2-49}{x^2+3x-88} \cdot \frac{x^2+6x-55}{x^2-11x+28} \div \frac{x^2+2x-35}{x^2-12x+32}$