Multiplying rational expressions
As is the case with ordinary fractions, we multiply two rational expressions by simply multiplying their numerators and denominators:
$\dfrac{\color{#026fc1}{P}}{\color{#026fc1}{Q}} \times \dfrac{\color{#c1026f}{R}}{\color{#c1026f}{S}} = \dfrac{\color{#026fc1}{P}\color{#c1026f}{R}}{\color{#026fc1}{Q}\color{#c1026f}{S}} \qquad $ \t
Note
Before actually calculating the products on the top and bottom, you should first simplify by factoring and cancelling as above if possible (see the examples that follow).
#[Examples][Ejemplos]#
\\ $\dfrac{\color{#026fc1}{x+1}}{\color{#026fc1}{x}} \times \dfrac{\color{#c1026f}{x-1}}{\color{#c1026f}{2x+1}} $ \t ${}= \dfrac{\color{#026fc1}{(x+1)}\color{#c1026f}{(x-1)}}{\color{#026fc1}{x}\color{#c1026f}{(2x+1)}}$ \t
\\ \t ${}= \dfrac{x^2-1}{2x^2+1}$ \t
\\ \t
\\ $\color{#026fc1}{2} \times \dfrac{\color{#c1026f}{4x}}{\color{#c1026f}{x-1}} $ \t ${}= \dfrac{\color{#026fc1}{2}}{\color{#026fc1}{1}} \times \dfrac{\color{#c1026f}{4x}}{\color{#c1026f}{x-1}} $ \t
\\ \t ${}= \dfrac{\color{#026fc1}{2}\color{#c1026f}{(4x)}}{\color{#026fc1}{(1)}\color{#c1026f}{(x-1)}}$ \t
\\ \t ${}= \dfrac{8x}{x-1}$ \t
\\ \t
\\ $\dfrac{\color{#026fc1}{4x}}{\color{#026fc1}{x-1}} \times \color{#c1026f}{(x^2+1)} $ \t ${}= \dfrac{\color{#026fc1}{4x}}{\color{#026fc1}{x-1}} \times \dfrac{\color{#c1026f}{x^2+1}}{\color{#c1026f}{1}} $ \t
\\ \t ${}= \dfrac{\color{#026fc1}{4x}\color{#c1026f}{(x^2+1)}}{\color{#026fc1}{(x-1)}\color{#c1026f}{(1)}}$ \t
\\ \t ${}= \dfrac{4x^3+4x}{x-1}$ \t
\\ \t
\\ $\dfrac{\color{#026fc1}{x-4}}{\color{#026fc1}{6x}} \times \dfrac{\color{#c1026f}{4x^3}}{\color{#c1026f}{2x+1}} $ \t ${}= \dfrac{\color{#026fc1}{(x-4)}\color{#c1026f}{4x^3}}{\color{#026fc1}{6x}\color{#c1026f}{(2x+1)}}$ \t
\\ \t ${}=\dfrac{(x-4)2x}{3(2x+1)}$ \t $\color{#6968d0}{2x}$.
\\ \t ${}= \dfrac{2x^2-8x}{6x+3}$ \t
\\ \t
\\ $\dfrac{\color{#026fc1}{x-1}}{\color{#026fc1}{x}} \times \dfrac{\color{#c1026f}{2x^2-x}}{\color{#c1026f}{x^2-2x-1}} $ \t ${}= \dfrac{\color{#026fc1}{(x-1)}\color{#c1026f}{(2x^2-x)}}{\color{#026fc1}{x}\color{#c1026f}{(x^2-2x-1)}}$ \t
\\ \t ${}=\dfrac{(x-1)(x)(2x-1)}{x(x-1)(x-1)}$ \t
\\ \t ${}=\dfrac{2x-1}{x-1}$ \t $\color{#6968d0}{x}$ $\color{#6968d0}{(x-1)}$.