5.3: Divide Rational Expressions
- Page ID
- 83138
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Dividing rational expressions is very much the same as fraction division in arithmetic. The first step is to change the division to multiplication and take the reciprocal of the second fraction. Two examples are shown below. Compare the similarities!
\(\begin{array} &&\text{Divide rational numbers} && \text{Divide rational expressions} \\ &\dfrac{3}{4} ÷ \dfrac{9}{20} && \dfrac{x - 6}{x^2 - 16} ÷ \dfrac{4x-24}{x^2 + x - 12} \\ &\dfrac{3}{4} \cdot \dfrac{20}{9} &\textcolor{green}{\text{Multiply by the reciprocal}} & \dfrac{x - 6}{x^2 - 16} \cdot \dfrac{x^2 + x - 12}{4x-24} \\ &=\dfrac{3}{4} \cdot \dfrac{4 \cdot 5}{3 \cdot 3} &\textcolor{green}{\text{Factor and cancel}} & \dfrac{x - 6}{(x-4)(x+4)} \cdot \dfrac{(x+4)(x-3)}{4(x-6)} \\ &\dfrac{5}{3} &\textcolor{green}{\text{What's left?}} & \dfrac{x-3}{4(x-4)} \\ &\text{This answer is in simplest form.} &\textcolor{green}{\text{Simplify answer}}& \dfrac{x-3}{4x-16} \end{array}\)
Complex Fractions
There are two dueling notations for dividing rational expressions. The above notation uses the symbol \(÷\) for division. The competing notation is called a complex fraction, which maintains fraction notation. A horizontal fraction bar indicates division.
A complex fraction is a fraction in which either the numerator is a fraction, or the denominator is a fraction, or both. To simplify complex fractions, translate the main fraction bar to division.
\(\begin{array} &\dfrac{\frac{a}{b}}{\frac{c}{d}} &= \dfrac{a}{b} ÷ \dfrac{c}{d} &\text{Step \(1\): Translate to \(÷\) notation.} \\ &= \dfrac{a}{b} \cdot \dfrac{d}{c} &\text{Step \(2\): Multiply by the reciprocal.} \\ &= \dfrac{ad}{bc} &\text{Step \(3\): Multiply and simplify.} \end{array}\)
Simplify \(\dfrac{\frac{8u^3}{35w^4}}{\frac{24u}{5w^3}}\)
Solution
Do you see the main fraction bar? \(\dfrac{\frac{8u^3}{35w^4}}{\frac{24u}{5w^3}}\) \(\textcolor{blue}{\longleftarrow \text{Change fraction bar to } ÷} \)
\( \dfrac{8u^3}{35w^4} ÷\dfrac{24u}{5w^3} = \dfrac{\cancel{8u^3}^{\textcolor{red}{u^2}}}{\cancel{35w^4}^{\textcolor{red}{7w}}} \cdot \dfrac{\cancel{5w^3}^{\textcolor{red}{1}}}{\cancel{24u}^{\textcolor{red}{3}}} = \dfrac{u^2}{21w} \)
Simplify \(\dfrac{\frac{2}{h}}{10h}\)
Solution
Let’s convert \(10h\) to fraction form: \(\dfrac{\frac{2}{h}}{\frac{10h}{1}}\) \(\textcolor{blue}{\longleftarrow \text{Change fraction bar to } ÷} \)
\( \dfrac{2}{h} ÷\dfrac{10h}{1} = \dfrac{\cancel{2}^{\textcolor{red}{1}}}{h} \cdot \dfrac{1}{\cancel{10h}^{\textcolor{red}{5h}}} = \dfrac{1}{5h^2} \)
Simplify \(\dfrac{\frac{t^2-1}{t+2}}{\frac{5-5t}{t^2+3t+2}}\)
Solution
The given expression is a complex fraction.
\(\begin{array}&\dfrac{\frac{t^2-1}{t+2}}{\frac{5-5t}{t^2+3t+2}} &= \dfrac{t^2-1}{t+2} ÷ \dfrac{5-5t}{t^2+3t+2} &\text{Change the main fraction bar to \(÷\)} \\ &= \dfrac{t^2-1}{t+2} \cdot \dfrac{t^2+3t+2}{5-5t} &\text{Multiply by the reciprocal.} \\ &= \dfrac{(t+1)(t-1)}{(t+2)} \cdot \dfrac{(t+2)(t+1)}{5(1-t)} = \dfrac{(t+1) (\cancel{t-1})}{(\cancel{t+2})} \cdot \dfrac{(\cancel{t+2})(t+1)}{5(\cancel{1-t})^{\textcolor{red}{-1}}} &\text{Factor and cancel. Opposites, too!} \\ &= −\dfrac{(t+1)^2}{5} & \end{array}\)
In our final example (below), we divide three rational expressions. As a student, allow yourself to apply your knowledge of fractions. The process is the same for rational expressions. Before doing example 4, you might try to simplify the following: \(\dfrac{1}{2} ÷ \dfrac{3}{4} ÷ \dfrac{3}{5}\).
Order of operations requires us to move from left to right, taking two fractions at a time.
\(\left( \dfrac{1}{2} ÷ \dfrac{3}{4} \right) ÷ \dfrac{3}{5} = \left( \dfrac{1}{2} \cdot \dfrac{4}{3} \right) ÷ \dfrac{3}{5} = \left( \dfrac{2}{3} \right) ÷ \dfrac{3}{5} = \dfrac{2}{3} \cdot \dfrac{5}{3} = \dfrac{10}{9}\)
Simplify \(\dfrac{2y}{24-6y} ÷ \dfrac{y-2}{y^2 -3y-4} ÷ \dfrac{y^2+y}{3}\)
Solution
\( \begin{array} &\left( \dfrac{2y}{24-6y} ÷ \dfrac{y-2}{y^2 -3y-4} \right) ÷ \dfrac{y^2+y}{3} &= \left( \dfrac{2}{6} \cdot \dfrac{y}{\cancel{4-y}^{\textcolor{red}{-1}}} \dfrac{(\cancel{y-4})(y+1)}{y-2} \right) ÷ \dfrac{y^2+y}{3} \\ &= \left( \dfrac{1}{\cancel{3}} \cdot \dfrac{\cancel{y}}{-1} \cdot \dfrac{\cancel{y+1}}{y-2} \right) \cdot \dfrac{\cancel{3}}{\cancel{y} (\cancel{y+1})} \\ &= -\dfrac{1}{y-2} \\ &= \dfrac{1}{2-y} \end{array} \)
Try It! (Exercises)
For #1-3, divide the rational numbers without using a calculator. Give answer in reduced fraction form.
- \(\dfrac{7^{10}}{4^5} ÷ \dfrac{7^{12}}{4^6}\)
- \(\dfrac{13^8 \cdot 5^{10}}{6^7} ÷ \dfrac{13^8 \cdot 5^8}{3 \cdot 6^6 \cdot 5}\)
- \(2^9 ÷ \dfrac{4 \cdot 2^6}{8^2} ÷ \dfrac{2^3}{6}\)
For #4-20, divide and simplify.
- \(\dfrac{24u^2}{9w^8} ÷ \dfrac{12u}{45w^{10}}\)
- \(\dfrac{\frac{36a^2b^2}{25c^3}}{\frac{72ab}{5c^5}}\)
- \(\dfrac{60x^6y^{10}}{11z^5} ÷ \dfrac{9x^8}{44z^2} ÷ \dfrac{8y^7}{x}\)
- \(\dfrac{10p(p−q)^4}{3q(q−7)^2} ÷ \dfrac{2p^2(p−q)^5}{15q(q−7)^4}\)
- \(\dfrac{40v(v−2)^{12}}{33(5v−6)^9} ÷ \dfrac{24v^3(v−12)^{10}}{11(5v−6)^{10}}\)
- \(\dfrac{\frac{d−5}{20d}}{\frac{(d−5)^3}{25d^2}}\)
- \(\dfrac{\frac{4r^2−9}{r}}{2r+3}\)
- \(\dfrac{x^2+3x−18}{x^2−2x−3} ÷ \dfrac{x^2+12x+36}{x^2−6x−7}\)
- \(\dfrac{4m^3+3m^2}{8m^2} ÷ \dfrac{4m^3+7m^2+3m}{4m}\)
- \(\dfrac{\frac{12t^3}{t^2−4}}{\frac{8t^2}{4t−16}} \)
- \(\dfrac{5n+6}{\frac{5n^2−4n−12}{4−n^2}}\)
- \(\dfrac{6p^2+6p-72}{3−p} ÷ (3p + 12)\)
- \(\dfrac{y^2+2y}{4y−32} ÷ \dfrac{y+2}{12y−24} ÷ \dfrac{6y−12}{5y}\)
- \(\dfrac{w^2−25}{w^2−2w−35} ÷ \dfrac{4−4w}{w^4+2w^3} ÷ \dfrac{w^3−5w^2}{w^2−8w+7}\)
- \(\dfrac{b^4−1}{b^2−5b−6} ÷ \dfrac{b^4+b^2}{b^2−12b+36}\)
- \(\dfrac{\frac{8a^3−1}{2a^2+7a−4}}{\frac{16a}{4a+16}}\)
- \(\dfrac{\frac{64c^3+27}{16c^3}}{\frac{16c^2−9}{32c^6}}\)