# Multiplication of Rational Expressions

## Key Questions

• It is.

In fact, the multiplication of two fraction is given by the multiplication of numerators and denominators, separately. So, since the multiplication of real numbers is commutative, the multiplication of fraction (which you can see as a multiplication of real number in two separate levels) is also commutative.

For example, if you have two rational expression of variable $x$, such as $\setminus \frac{f \left(x\right)}{g \left(x\right)}$ and $\setminus \frac{h \left(x\right)}{k \left(x\right)}$, their product will be $\setminus \frac{f \left(x\right) \setminus \cdot h \left(x\right)}{g \left(x\right) \setminus \cdot k \left(x\right)}$. Since both $f \left(x\right) \setminus \cdot h \left(x\right)$ and $g \left(x\right) \setminus \cdot k \left(x\right)$ are commutative, you get that $\setminus \frac{f \left(x\right) \setminus \cdot h \left(x\right)}{g \left(x\right) \setminus \cdot k \left(x\right)} = \setminus \frac{h \left(x\right) \setminus \cdot f \left(x\right)}{k \left(x\right) \setminus \cdot h \left(x\right)}$, and this last term is of course $\setminus \frac{h \left(x\right)}{k \left(x\right)} \setminus \cdot \setminus \frac{f \left(x\right)}{h \left(x\right)}$

• It's very easy: you only need to multiply numerators and denominators separately! So, the general formula is $\setminus \frac{a}{b} \setminus \cdot \setminus \frac{c}{d} = \setminus \frac{a c}{b d}$.

For example, if you want to compute $\setminus \frac{2}{3} \setminus \cdot \setminus \frac{4}{9}$, you simply get \frac{2 \cdot 4}{3\cdot 9} = \frac{8}{27