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Mathematics LibreTexts

7.2: Multiplying and Dividing Rational Expressions

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    Learning Objectives

    • Multiply rational expressions.
    • Divide rational expressions.
    • Multiply and divide rational functions.

    Multiplying Rational Expressions

    When multiplying fractions, we can multiply the numerators and denominators together and then reduce, as illustrated:

    \(\dfrac{3}{5} \cdot \dfrac{5}{9}=\dfrac{3 \cdot 5}{5 \cdot 9}=\dfrac{\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{3}}}}\color{black}{\cdot}\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{5}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{5}}}{1}}\color{black}{\cdot}\color{Cerulean}{\stackrel{\cancel{\color{black}{9}}}{3}}}\color{black}{=\dfrac{1}{3} }\)

    Multiplying rational expressions is performed in a similar manner. For example,

    \(\dfrac{y}{x} \cdot \dfrac{x}{y^{2}}=\dfrac{y \cdot x}{x \cdot y^{2}}=\dfrac{\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{y}}}}\color{black}{\cdot}\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{x}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{x}}}{1}}\color{black}{\cdot}\color{Cerulean}{\stackrel{\cancel{\color{black}{y^{2}}}}{3}}}\color{black}{=\dfrac{1}{y} }\)

    In general, given polynomials \(P\), \(Q\), \(R\), and \(S\), where \(Q≠0\) and \(S≠0\), we have

    \[\dfrac{P}{Q} \cdot \dfrac{R}{S}=\dfrac{P R}{Q S}\]

    In this section, assume that all variable expressions in the denominator are nonzero unless otherwise stated.

    Example \(\PageIndex{1}\)

    Multiply:

    \(\dfrac{12 x^{2}}{5 y^{3}} \cdot \dfrac{20 y^{4}}{6 x^{3}}\)

    Solution:

    Multiply numerators and denominators and then cancel common factors.

    \(\begin{aligned} \dfrac{12 x^{2}}{5 y^{3}} \cdot \dfrac{20 y^{4}}{6 x^{3}}&=\dfrac{240x^{2}y^{4}}{30x^{3}y^{3}}\qquad\quad\:\:\color{Cerulean}{Multiply.} \\&=\dfrac{\color{Cerulean}{\stackrel{8}{\cancel{\color{black}{240}}}}\color{Cerulean}{\stackrel{1}{ \cancel{\color{black}{x^{2}}}}}\color{Cerulean}{\stackrel{y}{ \cancel{\color{black}{y^{4}}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{30}}}{1}}\color{Cerulean}{ \stackrel{\cancel{\color{black}{x^{3}}}}{x}}\color{Cerulean}{ \stackrel{\cancel{\color{black}{y^{3}}}}{1}}} \qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{8 y}{x} \end{aligned}\)

    Answer:

    \(\dfrac{8y}{x}\)

    Example \(\PageIndex{2}\)

    Multiply:

    \(\dfrac{x-3}{x+5} \cdot \dfrac{x+5}{x+7}\)

    Solution:

    Leave the product in factored form and cancel the common factors.

    \(\begin{aligned} \dfrac{x-3}{x+5} \cdot \dfrac{x+5}{x+7} &=\dfrac{(x-3) \cdot\color{Cerulean}{\cancel{\color{black}{(x+5)}}}}{\color{Cerulean}{\cancel{\color{black}{(x+5)}}} \color{black}{\cdot(x+7)}} \\ &=\dfrac{x-3}{x+7} \end{aligned}\)

    Answer:

    \(\dfrac{x-3}{x+7}\)

    Example \(\PageIndex{3}\)

    Multiply:

    \(\dfrac{15 x^{2} y^{3}}{(2 x-1)} \cdot \dfrac{x(2 x-1)}{3 x^{2} y(x+3)}\)

    Solution:

    Leave the polynomials in the numerator and denominator factored so that we can cancel the factors. In other words, do not apply the distributive property.

    \(\begin{aligned} \dfrac{15 x^{2} y^{3}}{(2 x-1)} \cdot \dfrac{x(2 x-1)}{3 x^{2} y(x+3)}&=\dfrac{15 x^{3} y^{3}(2 x-1)}{3 x^{2} y(2 x-1)(x+3)}\qquad\qquad\color{Cerulean}{Multiply.} \\ &=\dfrac{\color{Cerulean}{\stackrel{5}{\cancel{\color{black}{15}}}}\color{Cerulean}{\stackrel{x}{\cancel{\color{black}{x^{3}}}}}\color{Cerulean}{\stackrel{y^{2}}{\cancel{\color{black}{y^{3}}}}}\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{(2x-1)}}}}}{\color{Cerulean}{\cancel{\color{black}{3}}\cancel{\color{black}{x^{2}}}\cancel{\color{black}{y}}\cancel{\color{black}{(2x-1)}}}\color{black}{(x+3)}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{5xy^{2}}{x+3} \end{aligned}\)

    Answer:

    \(\dfrac{5 x y^{2}}{x+3}\)

    Typically, rational expressions will not be given in factored form. In this case, first factor all numerators and denominators completely. Next, multiply and cancel any common factors, if there are any.

    Example \(\PageIndex{4}\)

    Multiply:

    \(\dfrac{x+5}{x-5} \cdot \dfrac{x-5}{x^{2}-25}\)

    Solution

    Factor the denominator \(x^{2}−25\) as a difference of squares. Then multiply and cancel.

    \(\begin{aligned} \dfrac{x+5}{x-5} \cdot \dfrac{x-5}{x^{2}-25} &=\dfrac{x+5}{x-5} \cdot \dfrac{x-5}{(x+5)(x-5)} \qquad\qquad\color{Cerulean}{Factor.}\\ &=\dfrac{\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{(x+5)}}}}\color{Cerulean}{\stackrel{1}{\cancel{\color{black}{(x-5)}}}}}{\color{Cerulean}{\cancel{\color{black}{(x-5)}}}\color{Cerulean}{\cancel{\color{black}{(x+5)}}}\color{black}{(x-5)}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{1}{x-5} \end{aligned}\)

    Keep in mind that 1 is always a factor; so when the entire numerator cancels out, make sure to write the factor 1.

    Answer:

    \(\dfrac{1}{x-5}\)

    Example \(\PageIndex{5}\)

    Multiply:

    \(\dfrac{x^{2}+3 x+2}{x^{2}-5 x+6} \cdot \dfrac{x^{2}-7 x+12}{x^{2}+8 x+7}\)

    Solution:

    \(\begin{aligned} \dfrac{x^{2}+3 x+2}{x^{2}-5 x+6} \cdot \dfrac{x^{2}-7 x+12}{x^{2}+8 x+7} &=\dfrac{(x+1)(x+2)}{(x-2)(x-3)} \cdot \dfrac{(x-3)(x-4)}{(x+1)(x+7)}\qquad\color{Cerulean}{Factor.} \\ &=\dfrac{\color{Cerulean}{\cancel{\color{black}{(x+1)}}}\color{black}{(x+2)}\color{Cerulean}{\cancel{\color{black}{(x-3)}}}\color{black}{(x-4)}}{\color{black}{(x-2)}\color{Cerulean}{\cancel{\color{black}{(x-3)}}\cancel{\color{black}{(x+1)}}}\color{black}{(x+7)}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{(x+2)(x-4)}{(x-2)(x+7)} \end{aligned}\)

    It is a best practice to leave the final answer in factored form.

    Answer:

    \(\dfrac{(x+2)(x-4)}{(x-2)(x+7)}\)

    Example \(\PageIndex{6}\)

    Multiply:

    \(\dfrac{-2 x^{2}+x+3}{x^{2}+2 x-8} \cdot \dfrac{3 x-6}{x^{2}+x}\)

    Solution:

    The trinomial \(−2x^{2}+x+3\) in the numerator has a negative leading coefficient. Recall that it is a best practice to first factor out a \(−1\) and then factor the resulting trinomial.

    \(\begin{aligned} \dfrac{-2 x^{2}+x+3}{x^{2}+2 x-8} \cdot \dfrac{3 x-6}{x^{2}+x} &=\dfrac{-1\left(2 x^{2}-x-3\right)}{x^{2}+2 x-8} \cdot \dfrac{3(x-2)}{x(x+1)}\qquad\quad\color{Cerulean}{Factor.} \\ &=\dfrac{-1(2 x-3)(x+1)}{(x-2)(x+4)} \cdot \dfrac{3(x-2)}{x(x+1)} \\ &=\dfrac{-1(2 x-3)\color{Cerulean}{\cancel{\color{black}{(x+1)}}} \color{black}{\cdot 3}\color{Cerulean}{\cancel{\color{black}{(x-2)}}}}{\color{Cerulean}{\cancel{\color{black}{(x-2)}}}\color{black}{(x+4) \cdot x}\color{Cerulean}{\cancel{\color{black}{(x+1)}}}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{-1(2 x-3) \cdot 3}{(x+4) \cdot x} \\ &=\dfrac{-3(2 x-3)}{x(x+4)} \quad \text { or } \quad=-\dfrac{3(2 x-3)}{x(x+4)} \end{aligned}\)

    Answer:

    \(-\dfrac{3(2 x-3)}{x(x+4)}\)

    Example \(\PageIndex{7}\)

    Multiply:

    \(\dfrac{7-x}{x^{2}+3 x} \cdot \dfrac{x^{2}+10 x+21}{x^{2}-49}\)

    Solution:

    We replace \(7−x\) with \(−1(x−7)\) so that we can cancel this factor.

    \(\begin{aligned} \dfrac{7-x}{x^{2}+3 x} \cdot \dfrac{x^{2}+10 x+21}{x^{2}-49} &=\dfrac{-1(x-7)}{x(x+3)} \cdot \dfrac{(x+3)(x+7)}{(x+7)(x-7)} \\ &=\dfrac{-1\color{Cerulean}{\cancel{\color{black}{(x-7)}}\cancel{\color{black}{(x+3)}}\cancel{\color{black}{(x+7)}}}}{x\color{Cerulean}{\cancel{\color{black}{(x+3)}}\cancel{\color{black}{(x+7)}}\cancel{\color{black}{(x-7)}}}} \\ &=\dfrac{-1}{x} \\ &=-\dfrac{1}{x} \end{aligned}\)

    Answer:

    \(-\dfrac{1}{x}\)

    Exercise \(\PageIndex{1}\)

    Multiply:

    \(\dfrac{x^{2}-64}{8-x} \cdot \dfrac{x+x^{2}}{x^{2}+9 x+8}\)

    Answer

    \(-x\) 

    Dividing Rational Expressions

    To divide two fractions, we multiply by the reciprocal of the divisor, as illustrated:

    \(\dfrac{5}{8} \div \color{OliveGreen}{ \dfrac{1}{2}}\color{black}{=}\dfrac{5}{8} \cdot \color{OliveGreen}{\dfrac{2}{1}}\color{black}{=}\dfrac{5 \cdot \color{Cerulean}{\stackrel{1}{\cancel{\color{black}{2}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{8}}}{4}} \color{black}{\cdot 1}}=\dfrac{5}{4}\)

    Dividing rational expressions is performed in a similar manner. For example,

    \(\dfrac{x}{y^{2}} \div \color{OliveGreen}{ \dfrac{1}{y}}\color{black}{=}\dfrac{x}{y^{2}} \cdot \color{OliveGreen}{\dfrac{y}{1}}\color{black}{=}\dfrac{x \cdot \color{Cerulean}{\stackrel{1}{\cancel{\color{black}{y}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{y^{2}}}}{y}} \color{black}{\cdot 1}}=\dfrac{x}{y}\)

    In general, given polynomials P, Q, R, and S, where \(Q≠0\), \(R≠0\), and \(S≠0\), we have

    \[\dfrac{P}{Q} \div \dfrac{R}{S}=\dfrac{P}{Q} \cdot \dfrac{S}{R}=\dfrac{P S}{Q R}\]

    Example \(\PageIndex{8}\)

    Divide:

    \(\dfrac{8 x^{5} y}{25 z^{6}} \div \dfrac{20 x y^{4}}{15 z^{3}}\)

    Solution:

    First, multiply by the reciprocal of the divisor and then cancel.

    \(\begin{aligned} \dfrac{8 x^{5} y}{25 z^{6}} \div \color{Cerulean}{\dfrac{20 x y^{4}}{15 z^{3}}} &\color{black}{=}\dfrac{8 x^{5} y}{25 z^{6}} \cdot\color{Cerulean}{ \dfrac{15 z^{3}}{20 x y^{4}}}\qquad\color{Cerulean}{}Multiply\:by\:the\:reciprocal\:of\:the\:divisor. \\ &=\dfrac{120 x^{5} y z^{3}}{500 x y^{4} z^{6}}\\ &=\dfrac{\color{Cerulean}{\stackrel{6}{\cancel{\color{black}{120}}}\stackrel{x^{4}}{\cancel{\color{black}{x^{5}}}}\cancel{\color{black}{y}}\cancel{\color{black}{z^{3}}}}}{\color{Cerulean}{\stackrel{\cancel{\color{black}{500}}}{25}\cancel{\color{black}{x}}\stackrel{\cancel{\color{black}{y^{4}}}}{y^{3}}\stackrel{\cancel{\color{black}{z^{6}}}}{z^{3}}}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{6x^{4}}{25y^{3}z^{3}} \end{aligned}\)

    Answer:

    \(\dfrac{6 x^{4}}{25 y^{3} z^{3}}\)

    Example \(\PageIndex{9}\)

    Divide:

    \(\dfrac{x+2}{x^{2}-4} \div \dfrac{x+3}{x-2}\)

    Solution:

    After multiplying by the reciprocal of the divisor, factor and cancel.

    \(\begin{aligned} \dfrac{x+2}{x^{2}-4} \div \color{Cerulean}{\dfrac{x+3}{x-2}} &=\dfrac{x+2}{x^{2}-4} \cdot \color{Cerulean}{\dfrac{x-2}{x+3}}\qquad\quad\:\qquad\qquad\color{Cerulean}{Multiply\:by\:the\:reciprocal\:of\:the\:divisor.} \\ &=\dfrac{(x+2)}{(x+2)(x-2)} \cdot \dfrac{(x-2)}{(x+3)}\qquad\quad\color{Cerulean}{Factor.} \\ &=\dfrac{\color{Cerulean}{\cancel{\color{black}{(x+2)}}\cancel{\color{black}{(x-2)}}}}{\color{Cerulean}{\cancel{\color{black}{(x+2)}}\cancel{\color{black}{(x-2)}}}\color{black}{(x+3)}}\qquad\quad\color{Cerulean}{Cancel.} \\ &=\dfrac{1}{x+3} \end{aligned}\)

    Answer:

    \(\dfrac{1}{x+3}\)

    Example \(\PageIndex{10}\)

    Divide:

    \(\dfrac{x^{2}-6 x-16}{x^{2}+4 x-21} \div \dfrac{x^{2}+9 x+14}{x^{2}-8 x+15}\)

    Solution:

    Begin by multiplying by the reciprocal of the divisor. After doing so, factor and cancel.

    \(\begin{aligned} \dfrac{x^{2}-6 x-16}{x^{2}+4 x-21} \div \color{Cerulean}{\dfrac{x^{2}+9 x+14}{x^{2}-8 x+15}} &\color{black}{=}\dfrac{x^{2}-6 x-16}{x^{2}+4 x-21} \cdot \color{Cerulean}{\dfrac{x^{2}-8 x+15}{x^{2}+9 x+14}} \\ &=\dfrac{(x+2)(x-8)}{(x-3)(x+7)} \cdot \dfrac{(x-3)(x-5)}{(x+2)(x+7)} \\ &=\dfrac{\color{Cerulean}{\cancel{\color{black}{(x+2)}}\color{black}{(x+8)}\color{Cerulean}{\cancel{\color{black}{(x-3)}}}}\color{black}{(x-5)}}{\color{Cerulean}{\cancel{\color{black}{(x-3)}}\color{black}{(x+7)}\color{Cerulean}{\cancel{\color{black}{(x+2)}}}}\color{black}{(x+7)}} \\ &=\dfrac{(x-8)(x-5)}{(x+7)^{2}} \end{aligned}\)

    Answer:

    \(\dfrac{(x-8)(x-5)}{(x+7)^{2}}\)

    Example \(\PageIndex{11}\)

    Divide:

    \(\dfrac{9-4 x^{2}}{x+2} \div(2 x-3)\)

    Solution:

    Just as we do with fractions, think of the divisor \((2x−3)\) as an algebraic fraction over 1.

    \(\begin{aligned} \dfrac{9-4 x^{2}}{x+2} \div(2 x-3) &=\dfrac{9-4 x^{2}}{x+2} \div \dfrac{2 x-3}{1} \\ &=\dfrac{(3+2 x)(3-2 x)}{x+2} \cdot \dfrac{1}{2 x-3} \qquad\color{Cerulean}{Replace\:(3-2x)=(-1)\cdot(2x-3)}\\ &=\dfrac{(3+2 x) \cdot(-1)\color{Cerulean}{\cancel{\color{black}{(2 x-3)}}}}{(x+2)\color{Cerulean}{\cancel{\color{black}{(2 x-3)}}}} \qquad\:\quad\color{Cerulean}{Cancel.} \\ &=-\dfrac{3+2 x}{x+2} \text { or }=-\dfrac{2 x+3}{x+2} \end{aligned}\)

    Answer:

    \(-\dfrac{2 x+3}{x+2}\)

    Exercise \(\PageIndex{2}\)

    Divide:

    \(\dfrac{4 x^{2}+7 x-2}{25 x^{2}} \div \dfrac{1-4 x}{100 x^{4}}\)

    Answer

    \(-4 x^{3}-8 x^{2}\)

    Multiplying and Dividing Rational Functions

    The product and quotient of two rational functions can be simplified using the techniques described in this section. The restrictions to the domain of a product consist of the restrictions of each function.

    Example \(\PageIndex{12}\)

    Calculate \((f⋅g)(x)\) and determine the restrictions to the domain.

    \(f(x)=\dfrac{16 x^{2}-1}{5 x}, g(x)=\dfrac{1}{1-4 x}\)

    Solution:

    In this case, the domain of \(f(x)\) consists of all real numbers except 0, and the domain of \(g(x)\) consists of all real numbers except \(\dfrac{1}{4}\).

    Therefore, the domain of the product consists of all real numbers except 0 and \(\dfrac{1}{4}\). Multiply the functions and then simplify the result.

    \(\begin{aligned}(f \cdot g)(x) &=f(x) \cdot g(x) \\ &=\dfrac{16 x^{2}-1}{5 x} \cdot \dfrac{1}{1-4 x} \\ &=\dfrac{(4 x+1)(4 x-1)}{5 x} \cdot \dfrac{1}{1-4 x} \\ &=\dfrac{(4 x+1)(4 x-1)}{5 x \cdot(-1)(4 x-1)} \\ &=\dfrac{4 x+1}{-5 x} \\ &=-\dfrac{4 x+1}{5 x} \end{aligned}\)

    Answer:

    \((f \cdot g)(x)=-\dfrac{4 x+1}{5 x}\), where \(x\neq 0, \dfrac{1}{4}\)

    The restrictions to the domain of a quotient will consist of the restrictions of each function as well as the restrictions on the reciprocal of the divisor.

    Example \(\PageIndex{13}\)

    Calculate \((f/g)(x)\) and determine the restrictions.

    \(f(x)=\dfrac{x^{2}-64}{x^{2}-11 x+24}, g(x)=\dfrac{x+8}{x-3}\)

    Solution:

    \(\begin{aligned}(f / g)(x) &=f(x) \div g(x) \\ &=\dfrac{x^{2}-64}{x^{2}-11 x+24} \div \dfrac{x+8}{x-3} \\ &=\dfrac{x^{2}-64}{x^{2}-11 x+24} \cdot \dfrac{x-3}{x+8}\qquad\quad\:\:\color{Cerulean}{Multiply\:by\:the\:reciprocal\:of\:the\:divisor.} \\ &=\dfrac{(x+8)(x-8)}{(x-3)(x-8)} \cdot\dfrac{(x-3)}{(x+8)}\qquad\:\:\color{Cerulean}{Factor.} \\ &=\dfrac{\color{Cerulean}{\cancel{\color{black}{(x+8)}}\cancel{\color{black}{(x-8)}}\cancel{\color{black}{(x-3)}}}}{\color{Cerulean}{\cancel{\color{black}{(x-3)}}\cancel{\color{black}{(x-8)}}\cancel{\color{black}{(x+8)}}}}\qquad\color{Cerulean}{Cancel.} \\ &=\dfrac{1}{1} \\ &=1 \end{aligned}\)

    In this case, the domain of \(f(x)\) consists of all real numbers except 3 and 8, and the domain of \(g(x)\) consists all real numbers except 3. In addition, the reciprocal of \(g(x)\) has a restriction of −8. Therefore, the domain of this quotient consists of all real numbers except 3, 8, and −8.

    Answer:

    \((f / g)(x) = 1\), where \(x\neq 3, 8, -8\)

    Key Takeaways

    • After multiplying rational expressions, factor both the numerator and denominator and then cancel common factors. Make note of the restrictions to the domain. The values that give a value of 0 in the denominator are the restrictions.
    • To divide rational expressions, multiply by the reciprocal of the divisor.
    • The restrictions to the domain of a product consist of the restrictions to the domain of each factor.
    • The restrictions to the domain of a quotient consist of the restrictions to the domain of each rational expression as well as the restrictions on the reciprocal of the divisor.

    Exercise \(\PageIndex{2}\) Multiplying Rational Expressions

    Multiply. (Assume all denominators are nonzero.)

    1. \(\dfrac{2 x}{3} \cdot \dfrac{9}{4 x^{2}}\)
    2. \(-\dfrac{5 x}{3 y} \cdot \dfrac{y^{2}}{25 x}\)
    3. \(\dfrac{5 x^{2}}{2 y} \cdot \dfrac{4 y^{2}}{15 x^{3}}\)
    4. \(\dfrac{16 a^{4}}{7 b^{2}} \cdot \dfrac{49 b^{3}}{2 a^{3}}\) 
    5. \(\dfrac{x-6}{12 x^{3}} \cdot \dfrac{24 x^{2}}{x-6}\)
    6. \(\dfrac{x+10}{2x−1}\cdot\dfrac{x−2}{x+10}\) 
    7. \(\dfrac{(y-1)^{2}}{y+1} \cdot \dfrac{1}{y-1}\)
    8. \(\dfrac{y^{2}-9}{y+3} \cdot \dfrac{2 y-3}{y-3}\) 
    9. \(\dfrac{2 a-5}{a-5} \cdot \dfrac{2 a+5}{4 a^{2}-25}\)
    10. \(\dfrac{2 a^{2}-9 a+4}{a^{2}-16} \cdot\left(a^{2}+4 a\right)\)
    11. \(\dfrac{2 x^{2}+3 x-2}{(2 x-1)^{2}} \cdot \dfrac{2 x}{x+2}\)
    12. \(\dfrac{9x^{2}+19x+2}{4−x^{2}}\cdot\dfrac{x^{2}−4x+4}{9x^{2}−8x−1}\) 
    13. \(\dfrac{x^{2}+8x+16}{16−x^{2}}\cdot\dfrac{x^{2}−3x−4}{x^{2}+5x+4}\) 
    14. \(\dfrac{x^{2}−x−2}{x^{2}+8x+7}\cdot\dfrac{x^{2}+2x−15}{x^{2}−5x+6}\)
    15. \(\dfrac{x+1}{x−3}\cdot\dfrac{3−x}{x+5}\) 
    16. \(\dfrac{2x−1}{x−1}\cdot\dfrac{x+6}{1−2x}\) 
    17. \(\dfrac{9+x}{3x+1}\cdot\dfrac{3}{x+9}\) 
    18. \(\dfrac{1}{2+5x}\cdot\dfrac{5x+2}{5x}\)
    19. \(\dfrac{100-y^{2}}{y-10} \cdot \dfrac{25 y^{2}}{y+10}\)
    20. \(\dfrac{3 y^{3}}{6 y-5} \cdot \dfrac{36 y^{2}-25}{5+6 y}\)
    21. \(\dfrac{3 a^{2}+14 a-5}{a^{2}+1} \cdot \dfrac{3 a+1}{1-9 a^{2}}\)
    22. \(\dfrac{4a^{2}−16a}{4a−1}\cdot\dfrac{1−16a^{2}}{4a^{2}−15a−4}\) 
    23. \(\dfrac{x+9}{-x^{2}+14 x-45} \cdot\left(x^{2}-81\right) \)
    24. \(\dfrac{1}{2+5 x} \cdot\left(25 x^{2}+20 x+4\right)\)
    25. \(\dfrac{x^{2}+x−6}{3x^{2}+15x+18}\cdot\dfrac{2x^{2}−8}{x^{2}−4x+4}\) 
    26. \(\dfrac{5x^{2}−4x−1}{5x^{2}−6x+1}\cdot\dfrac{25x^{2}−10x+1}{3−75x^{2}}\)
    Answer

    1. \(\dfrac{3}{2x}\)

    3. \(\dfrac{2y}{3x}\)

    5. \(\dfrac{2}{x}\)

    7. \(\dfrac{y−1}{y+1}\)

    9. \(\dfrac{1}{a−5}\)

    11. \(\dfrac{2x}{2x−1}\)

    13. \(−1\) 

    15. \(−\dfrac{x+1}{x+5}\) 

    17. \(\dfrac{3}{3x+1}\)

    19. \(−25y^{2}\)

    21. \(-\dfrac{a+5}{a^{2}+1}\)

    23. \(-\dfrac{(x+9)^{2}}{x-5}\)

    25. \(\dfrac{2}{3}\)

    Exercise \(\PageIndex{3}\) Dividing Rational Expressions

    Divide. (Assume all denominators are nonzero.)

    1. \(\dfrac{5 x}{8} \div \dfrac{15 x^{2}}{4}\)
    2. \(\dfrac{3}{8 y} \div \dfrac{15}{2 y^{2}}\)
    3. \(\dfrac{\dfrac{5 x^{9}}{3 y^{3}}}{\dfrac{25 x^{10}}{9 y^{5}}}\)
    4. \(\dfrac{\dfrac{12 x^{4} y^{2}}{21 z^{5}}}{\dfrac{6 x^{3} y^{2}}{7 z^{3}}}\)
    5. \(\dfrac{(x-4)^{2}}{30 x^{4}} \div \dfrac{x-4}{15 x}\)
    6. \(\dfrac{5 y^{4}}{10(3 y-5)^{2}} \div \dfrac{10 y^{5}}{2(3 y-5)^{3}}\)
    7. \(\dfrac{x^{2}-9}{5 x} \div(x-3)\)
    8. \(\dfrac{y^{2}-64}{8 y} \div(8+y)\)
    9. \(\dfrac{(a-8)^{2}}{2 a^{2}+10 a} \div \dfrac{a-8}{a}\)
    10. \(\dfrac{2}{4 a^{2} b^{3}(a-2 b)} \div 12 a b(a-2 b)^{5}\)
    11. \(\dfrac{x^{2}+7 x+10}{x^{2}+4 x+4} \div \dfrac{1}{x^{2}-4}\)
    12. \(\dfrac{2 x^{2}-x-1}{2 x^{2}-3 x+1} \div \dfrac{1}{4 x^{2}-1}\)
    13. \(\dfrac{y+1}{y^{2}-3 y} \div \dfrac{y^{2}-1}{y^{2}-6 y+9}\)
    14. \(\dfrac{9-a^{2}}{a^{2}-8 a+15} \div \dfrac{2 a^{2}-10 a}{a^{2}-10 a+25}\)
    15. \(\dfrac{a^{2}-3 a-18}{2 a^{2}-11 a-6} \div \dfrac{a^{2}+a-6}{2 a^{2}-a-1}\)
    16. \(\dfrac{y^{2}-7 y+10}{y^{2}+5 y-14} \div \dfrac{2 y^{2}-9 y-5}{y^{2}+14 y+49}\)
    17. \(\dfrac{6 y^{2}+y-1}{4 y^{2}+4 y+1} \div \dfrac{3 y^{2}+2 y-1}{2 y^{2}-7 y-4}\)
    18. \(\dfrac{x^{2}−7x−18}{x^{2}+8x+12}\div\dfrac{x^{2}−81}{x^{2}+12x+36}\) 
    19. \(\dfrac{4a^{2}−b^{2}}{b+2a}\div (b−2a)^{2}\) 
    20. \(\dfrac{x^{2}−y^{2}}{y+x}\div (y−x)^{2}\) 
    21. \(\dfrac{5 y^{2}(y-3)}{4 x^{3}} \div \dfrac{25 y(3-y)}{2 x^{2}}\)
    22. \(\dfrac{15 x^{3}}{3(y+7)} \div \dfrac{25 x^{6}}{9(7+y)^{2}} \)
    23. \(\dfrac{3 x+4}{x-8} \div \dfrac{7 x}{8-x}\)
    24. \(\dfrac{3x−2}{2x+1}\div \dfrac{2−3x}{3x}\) 
    25. \(\dfrac{(7 x-1)^{2}}{4 x+1} \div \dfrac{28 x^{2}-11 x+1}{1-4 x}\)
    26. \(\dfrac{4 x}{(x+2)^{2}} \div \dfrac{2-x}{x^{2}-4}\)
    27. \(\dfrac{a^{2}-b^{2}}{a} \div(b-a)^{2}\)
    28. \(\dfrac{(a−2b)^{2}}{2b}\div (2b^{2}+ab−a^{2})\)
    29. \(\dfrac{x^{2}−6x+9}{x^{2}+7x+12}\div \dfrac{9−x^{2}}{x^{2}+8x+16}\) 
    30. \(\dfrac{2x^{2}−9x−5}{25−x^{2}}\div \dfrac{1−4x+4x^{2}}{−2x^{2}−9x+5}\) 
    31. \(\dfrac{3x^{2}−16x+5}{100−4x^{2}}\div\dfrac{ 9x^{2}−6x+1}{3x^{2}+14x−5}\) 
    32. \(\dfrac{10x^{2}−25x−15}{x^{2}−6x+9}\div\dfrac{9−x^{2}}{x^{2}+6x+9}\)
    Answer

    1. \(\dfrac{1}{6x}\) 

    3. \(\dfrac{3y^{2}}{5x}\)

    5. \(\dfrac{x−4}{2x3}\)

    7. \(\dfrac{x+3}{5x}\)

    9. \(\dfrac{a−8}{2(a+5)}\) 

    11. \((x+5)(x−2)\)

    13. \(\dfrac{y−3}{y(y−1)}\)

    15. \(\dfrac{a−1}{a−2}\)

    17. \(\dfrac{y−4}{y+1}\)

    19. \(\dfrac{1}{2a−b}\)

    21. \(−\dfrac{y}{10x}\)

    23. \(−\dfrac{3x+4}{7x}\)

    25. \(−\dfrac{7x−1}{4x+1}\)

    27. \(\dfrac{a+b}{a(a-b)}\)

    29. \(−\dfrac{(x−3)(x+4)}{(x+3)^{2}}\) 

    31. \(−\dfrac{1}{4}\)

    Exercise \(\PageIndex{4}\) Dividing Rational Expressions

    Recall that multiplication and division are to be performed in the order they appear from left to right. Simplify the following.

    1. \(\dfrac{1}{x^{2}} \cdot \dfrac{x-1}{x+3} \div \dfrac{x-1}{x^{3}}\)
    2. \(\dfrac{x−7}{x+9}\cdot\dfrac{1}{x^{3}}\div\dfrac{x−7}{x}\) 
    3. \(\dfrac{x+1}{x−2}\div\dfrac{x}{x−5}\cdot\dfrac{x^{2}}{x+1}\) 
    4. \(\dfrac{x+4}{2x+5}\div \dfrac{x−3}{2x+5}\cdot\dfrac{x+4}{x−3}\) 
    5. \(\dfrac{2x−1}{x+1}\div\dfrac{x−4}{x^{2}+1}\cdot\dfrac{x−4}{2x−1}\) 
    6. \(\dfrac{4x^{2}−1}{3x+2}\div\dfrac{2x−1}{x+5}\cdot\dfrac{3x+2}{2x+1}\)
    Answer

    1. \(\dfrac{x}{x+3}\)

    3. \(\dfrac{x(x−5)}{x−2}\) 

    5. \(\dfrac{x^{2}+1}{x+1}\)

    Exercise \(\PageIndex{5}\) Multiplying and Dividing Rational Functions

    Calculate \((f⋅g)(x)\) and determine the restrictions to the domain.

    1. \(f(x)=\dfrac{1}{x}\) and \(g(x)=\dfrac{1}{x−1}\)
    2. \(f(x)=\dfrac{x+1}{x−1}\) and \(g(x)=x^{2}−1\)
    3. \(f(x)=\dfrac{3x+2}{x+2}\) and \(g(x)=\dfrac{x^{2}−4}{(3x+2)^{2}}\) 
    4. \(f(x)=\dfrac{(1−3x)}{2x−6}\) and \(g(x)=\dfrac{(x−6)^{2}}{9x^{2}−1}\) 
    5. \(f(x)=\dfrac{25x^{2}−1}{x^{2}+6x+9}\) and \(g(x)=\dfrac{x^{2}−9}{5x+1}\) 
    6. \(f(x)=\dfrac{x^{2}−49}{2x^{2}+13x−7}\) and \(g(x)=\dfrac{4x^{2}−4x+1}{7−x}\)
    Answer

    1. \((f⋅g)(x)=\dfrac{1}{x(x−1)} ; x≠0, 1 \)

    3. \((f⋅g)(x)=\dfrac{x−2}{3x+2}; x≠−2, −\dfrac{2}{3}\) 

    5. \((f⋅g)(x)=\dfrac{(x−3)}{(5x−1)x+3}; x≠−3, −\dfrac{1}{5}\)

    Exercise \(\PageIndex{6}\) Multiplying and Dividing Rational Functions

    Calculate \((f/g)(x)\) and state the restrictions.

    1. \(f(x)=\dfrac{1}{x}\) and \(g(x)=\dfrac{x−2}{x−1}\) 
    2. \(f(x)=\dfrac{(5x+3)^{2}}{x^{2}}\) and \(g(x)=\dfrac{5x+3}{6−x}\) 
    3. \(f(x)=\dfrac{5−x}{(x−8)^{2}}\) and \(g(x)=\dfrac{x^{2}−2}{5x−8}\) 
    4. \(f(x)=\dfrac{x^{2}−2x−1}{5x^{2}−3x−10}\) and \(g(x)=\dfrac{2x^{2}−5x−3}{x^{2}−7x+12}\)
    5. \(f(x)=\dfrac{3x^{2}+11x−4}{9x^{2}−6x+1}\) and \(g(x)=\dfrac{x^{2}−2x+1}{3x^{2}−4x+1}\) 
    6. \(f(x)=\dfrac{36−x^{2}}{x^{2}+12x+36}\) and \(g(x)=\dfrac{x^{2}−12x+3}{6x^{2}+4x−12}\)
    Answer

    1. \((f/g)(x)=\dfrac{x−1}{x(x−2)}; x≠0, 1, 2\) 

    3. \((f/g)(x)=−\dfrac{1}{(x−8)(x+5)}; x≠±5, 8\) 

    5. \((f/g)(x)=\dfrac{(x+4)}{(x−1)}; x≠\dfrac{1}{3}, 1\)

    Exercise \(\PageIndex{7}\) Discussion Board Topics

    1. In the history of fractions, who is credited for the first use of the fraction bar?
    2. How did the ancient Egyptians use fractions?
    3. Explain why \(x=7\) is a restriction to \(\dfrac{1}{x}\div\dfrac{x−7}{x−2}\).
    Answer

    1. Answer may vary

    3. Answer may vary