## Section14.5Multiplication and Division of Rational Expressions

##### Multiplication of Rational Expressions.

As suggested in the last section, multiplying rational expressions entails writing the numerators of the expression over the product of the denominators of the expression. We then simplify the result. Several examples follow.

###### Example14.5.1.

Multiply and simplify the result. Make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{x+3}{x-2} \cdot \frac{x^2-4}{x^2+6x+9} \end{equation*}
Solution
\begin{align*} \frac{x+3}{x-2} \cdot \frac{x^2-4}{x^2+6x+9}\amp=\frac{(x+3)(x^2-4)}{(x-2)(x^2+6x+9)}\\ \amp=\frac{(x+3)(x+2)(x-2)}{(x-2)(x+3)(x+3)}\\ \amp=\frac{x+3}{x+3} \cdot \frac{x-2}{x-2} \cdot \frac{x+2}{x+3}\\ \amp=\frac{x+2}{x+3}, x \neq 2 \end{align*}

Note that we did not need to state the restriction $x \neq -3$ because a factor of $(x+3)$ is still present in the simplified result.

###### Example14.5.2.

Multiply and simplify the result. Make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{10x-35}{5x^2+25x} \cdot \frac{x^2+8x+15}{2x-7} \end{equation*}
Solution
\begin{align*} \frac{10x-35}{5x^2+25x} \cdot \frac{x^2+8x+15}{2x-7}\amp=\frac{(10x-35)(x^2+8x+15)}{(5x^2+25x)(2x-7)}\\ \amp=\frac{5(2x-7)(x+3)(x+5)}{5x(x+5)(2x-7)}\\ \amp=\frac{5}{5} \cdot \frac{2x-7}{2x-7} \cdot \frac{x+5}{x+5} \cdot \frac{x+3}{x}\\ \amp=\frac{x+3}{x}, x \neq \frac{7}{2}, x \neq -5 \end{align*}
###### Example14.5.3.

Multiply and simplify the result. Make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{8}{2x+22} \cdot (x^2+22x+121) \end{equation*}
Solution
\begin{align*} \frac{8}{2x+22} \cdot (x^2+22x+121)\amp=\frac{8}{2x+22} \cdot \frac{x^2+22x+121}{1}\\ \amp=\frac{8(x^2+22x+121)}{(2x+22) \cdot 1}\\ \amp=\frac{8(x+11)(x+11)}{2(x+11)}\\ \amp=\frac{x+11}{x+11} \cdot \frac{8}{2} \cdot \frac{x+11}{1}\\ \amp=4(x+11), x \neq -11 \end{align*}
##### Division of Rational Expressions.

Recall that when we divide one fraction by another, we rewrite the quotient as a product after reciprocating the divisor. That is:

\begin{equation*} \frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \cdot \frac{d}{c}\text{.} \end{equation*}

When working with rational expressions that include division, we always first rewrite the quotient as a product and then simplify the result. When we convert the quotient to a product, we need to state any domain restrictions in the original divisor that are lost by reciprocating the divisor.

###### Example14.5.4.

Perform the division and simplify the result. Make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{3}{x-10} \div \frac{9}{x-5} \end{equation*}
Solution
\begin{align*} \frac{3}{x-10} \div \frac{9}{x-5}\amp=\frac{3}{x-10} \cdot \frac{x-5}{9},x \neq 5\\ \amp=\frac{3(x-5)}{(x-10) \cdot 9}, x \neq 5\\ \amp=\frac{x-5}{3(x-10)},x \neq 5 \end{align*}

When presented with a quotient in which one of the expressions is a rational expression but the other expression is a polynomial, it can be useful to write the polynomial as a rational expression with a denominator of $1\text{.}$ Two examples follow.

###### Example14.5.5.

Perform the division and simplify the result. Make sure that you state any necessary domain restrictions.

\begin{equation*} (x^2+8x+12) \div \frac{x^2+4x-12}{x^2-36} \end{equation*}
Solution
\begin{align*} \amp(x^2+8x+12) \div \frac{x^2+4x-12}{x^2-36}\\ \amp \phantom{={}} \phantom{={}} =\frac{x^2+8x+12}{1} \div \frac{x^2+4x-12}{x^2-36}\\ \amp \phantom{={}} \phantom{={}} =\frac{x^2+8x+12}{1} \cdot \frac{x^2-36}{x^2+4x-12}, x \neq 6, x \neq -6\\ \amp \phantom{={}} \phantom{={}} =\frac{(x^2+8x+12)(x^2-36)}{1 \cdot (x^2+4x-12)}, x \neq 6, x \neq -6\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+6)(x+2)(x+6)(x-6)}{(x+6)(x-2)},x \neq 6\\ \amp \phantom{={}} \phantom{={}} =\frac{x+6}{x+6} \cdot \frac{(x+2)(x+6)(x-6)}{x-2}, x \neq 6\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+2)(x+6)(x-6)}{x-2}, x \neq 6, x \neq -6 \end{align*}
###### Example14.5.6.

Perform the division and simplify the result. Make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{x^2-18x+72}{9x^2-16} \div (9x^2-16x) \end{equation*}
Solution
\begin{align*} \frac{x^2-18x+72}{9x^2-16} \div (9x^2-16x)\amp=\frac{x^2-18x+72}{9x^2-16} \div \frac{9x^2-16x}{1}\\ \amp=\frac{x^2-18x+72}{9x^2-16} \cdot \frac{1}{9x^2-16x}\\ \amp=\frac{(x^2-18x+72) \cdot 1}{(9x^2-16)(9x^2-16x)}\\ \amp=\frac{(x-12)(x-6)}{(3x+4)(3x-4) \cdot x(9x-16)}\\ \amp=\frac{(x-12)(x-6)}{x(3x+4)(3x-4)(9x-16)} \end{align*}

When there is repeated division between rational expression, every divisor ends up being reciprocated when the expression is rewritten as a product. The is a result of the fact that operations are performed left-to-right. An example follows.

###### Example14.5.7.

Perform the division and simplify the result. Make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{x-3}{x-7} \div \frac{x+8}{x-12} \div \frac{x+8}{x-7} \end{equation*}
Solution
\begin{align*} \amp\frac{x-3}{x-7} \div \frac{x+8}{x-12} \div \frac{x+8}{x-7}\\ \amp \phantom{={}} \phantom{={}} =\left(\frac{x-3}{x-7} \div \frac{x+8}{x-12}\right) \div \frac{x+8}{x-7}\\ \amp \phantom{={}} \phantom{={}} =\left(\frac{x-3}{x-7} \cdot \frac{x-12}{x+8}\right) \div \frac{x+8}{x-7}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{(x-3)(x-12)}{(x-7)(x+8)} \div \frac{x+8}{x-7}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{(x-3)(x-12)}{(x-7)(x+8)} \cdot \frac{x-7}{x+8}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{(x-3)(x-12)(x-7)}{(x-7)(x+8)(x+8)}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{x-7}{x-7} \cdot \frac{(x-3)(x-12)}{(x+8)(x+8)}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{(x-3)(x-12)}{(x+8)^2}, x \neq 12, x \neq 7 \end{align*}

### ExercisesExercises

Multiply or divide as indicated. Simplify each result and make sure that you state any necessary domain restrictions.

###### 1.

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

Solution

\begin{aligned}[t] \frac{x-4}{2x+3} \cdot \frac{x-1}{x-4}\amp=\frac{(x-4)(x-1)}{(2x+3)(x-4)}\\ \amp=\frac{x-4}{x-4} \cdot \frac{x-1}{2x+3}\\ \amp=\frac{x-1}{2x+3}, x \neq 4 \end{aligned}

###### 2.

$\frac{12x+18}{x^2-1} \cdot \frac{x+1}{6x+9}$

Solution

\begin{aligned}[t] \frac{12x+18}{x^2-1} \cdot \frac{x+1}{6x+9}\amp=\frac{(12x+18)(x+1)}{(x^2-1)(6x+9)}\\ \amp=\frac{6(2x+3)(x+1)}{(x-1)(x+1) \cdot 3(2x+3)}\\ \amp=\frac{x+1}{x+1} \cdot \frac{2x+3}{2x+3} \cdot \frac{6}{3(x-1)}\\ \amp=\frac{2}{x-1}, x \neq -1, x \neq -\frac{3}{2} \end{aligned}

###### 3.

$\frac{4x^2-20x+24}{x^2-25} \cdot \frac{5x+25}{x^2-9}$

Solution

\begin{aligned}[t] \frac{4x^2-20x+24}{x^2-25} \cdot \frac{5x+25}{x^2-9}\amp=\frac{(4x^2-20x+24)(5x+25)}{(x^2-25)(x^2-9)}\\ \amp=\frac{4(x^2-5x+6) \cdot 5(x+5)}{(x-5)(x+5)(x-3)(x+3)}\\ \amp=\frac{20(x-3)(x-2)(x+5)}{(x-5)(x+5)(x-3)(x+3)}\\ \amp=\frac{x-3}{x-3} \cdot \frac{x+5}{x+5} \cdot \frac{20(x-2)}{(x-5)(x+3)}\\ \amp=\frac{20(x-2)}{(x-5)(x+3)}, x \neq 3, x \neq -5 \end{aligned}

###### 4.

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

Solution

\begin{aligned}[t] (x^2+4x+4) \cdot \frac{x+2}{x^2-4}\amp=\frac{x^2+4x+4}{1} \cdot \frac{x+2}{x^2-4}\\ \amp=\frac{(x^2+4x+4)(x+2)}{x^2-4}\\ \amp=\frac{(x+2)(x+2)(x+2)}{(x+2)(x-2)}\\ \amp=\frac{x+2}{x+2} \cdot \frac{(x+2)(x+2)}{x-2}\\ \amp=\frac{(x+2)^2}{x-2}, x \neq -2 \end{aligned}

###### 5.

$\frac{x^2-9}{x+1} \div \frac{x+3}{x^2-1}$

Solution

\begin{aligned}[t] \amp\frac{x^2-9}{x+1} \div \frac{x+3}{x^2-1}\\ \amp \phantom{={}} \phantom{={}} =\frac{x^2-9}{x+1} \cdot \frac{x^2-1}{x+3}, x \neq 1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x^2-9)(x^2-1)}{(x+1)(x+3)}, x \neq 1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+3)(x-3)(x+1)(x-1)}{(x+1)(x+3)}, x \neq 1\\ \amp \phantom{={}} \phantom{={}} =\frac{x+3}{x+3} \cdot \frac{x+1}{x+1} \cdot \frac{(x-3)(x-1)}{1}, x \neq 1\\ \amp \phantom{={}} \phantom{={}} =(x-3)(x-1), x \neq 1, x \neq -1, x \neq -3 \end{aligned}

###### 6.

$\frac{x+5}{x^2-6x-7} \cdot \frac{x^2+1}{x^2+11x+30} \div \frac{x-7}{x+1}$

Solution

\begin{aligned}[t] \amp\frac{x+5}{x^2-6x-7} \cdot \frac{x^2+1}{x^2+11x+30} \div \frac{x-7}{x+1}\\ \amp \phantom{={}} \phantom{={}} =\frac{x+5}{x^2-6x-7} \cdot \frac{x^2+1}{x^2+11x+30} \cdot \frac{x+1}{x-7}, x \neq -1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+5)(x^2+1)(x+1)}{(x^2-6x+7)(x^2+11x+30)(x-7)}, x \neq -1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+5)(x^2+1)(x+1)}{(x-7)(x+1)(x+5)(x+6)(x-7)}, x \neq -1\\ \amp \phantom{={}} \phantom{={}} =\frac{x+5}{x+5} \cdot \frac{x+1}{x+1} \cdot \frac{x^2+1}{(x+6)(x-7^2)}, x \neq -1\\ \amp \phantom{={}} \phantom{={}} =\frac{x^2+1}{(x+6)(x-7)^2}, x \neq -1, x \neq -5 \end{aligned}

###### 7.

$\frac{x^2+2x-15}{4x+20} \div (x-3)$

Solution

\begin{aligned}[t] \frac{x^2+2x-15}{4x+20} \div (x-3)\amp=\frac{x^2+2x-15}{4x+20} \div \frac{x-3}{1}\\ \amp=\frac{x^2+2x-15}{4x+20} \cdot \frac{1}{x-3}\\ \amp=\frac{x^2+2x-15}{(4x+20)(x-3)}\\ \amp=\frac{(x+5)(x-3)}{4(x+5)(x-3)}\\ \amp=\frac{x+5}{x+5} \cdot \frac{x-3}{x-3} \cdot \frac{1}{4}\\ \amp=\frac{1}{4}, x \neq -5, x \neq 3 \end{aligned}

###### 8.

$\frac{2x}{x-2} \div \frac {x+2}{x} \div \frac{7x}{x^2-4}$

Solution

\begin{aligned}[t] \amp\frac{2x}{x-2} \div \frac{x+2}{x} \div \frac{7x}{x^2-4}\\ \amp \phantom{={}} \phantom{={}} =\frac{2x}{x-2} \cdot \frac {x}{x+2} \cdot \frac{x^2-4}{7x}\\ \amp \phantom{={}} \phantom{={}} =\frac{2x \cdot x \cdot (x^2-4)}{(x-2) \cdot (x+2) \cdot 7x}\\ \amp \phantom{={}} \phantom{={}} =\frac{2x^2(x+2)(x-2)}{7x(x-2)(x+2)}\\ \amp \phantom{={}} \phantom{={}} =\frac{x}{x} \cdot \frac{x+2}{x+2} \cdot \frac{x-2}{x-2} \cdot \frac{2x}{7}\\ \amp \phantom{={}} \phantom{={}} =\frac{2x}{7}, x \neq 0, x \neq 2, x \neq -2 \end{aligned}