Multiplication and division of rational expressions

When multiplying rational expressions, first simplify each expression if possible, then multiply the numerators and the denominators. Simplify again if possible.

That is, $\dfrac{p}{q}\cdot \dfrac{r}{s}=\dfrac{pr}{qs}$

To divide rational expressions, take the reciprocal of the second expression (the divisor) and multiply it by the first.

That is, $\dfrac{p}{q}\div \dfrac{r}{s}=\dfrac{p}{q}\cdot \dfrac{s}{r}$

Example 1: Simplify, $\dfrac{4x+20}{3}\cdot \dfrac{2x}{x+5}$.

Solution:

First, simplify each expression by factoring, if possible.

$\dfrac{4x+20}{3}\cdot \dfrac{2x}{x+5}=\dfrac{4(x+5)}{3}\cdot \dfrac{2x}{x+5}$.

Multiply the numerators,

$\hphantom{00000}=\dfrac{4(x+5)\:2x}{3(x+5)}$.

Canceling $x+5$ and simplifying,

$\hphantom{00000}=\boxed{\dfrac{8x}{3}}$

Example 2: Divide, $\dfrac{2x^2-5x-3}{3x^2+14x-5}\div \dfrac{4x-12}{x+5}$

Solution:

Take the reciprocal of the second expression (the divisor) and change the division to multiplication,

$\dfrac{2x^2-5x-3}{3x^2+14x-5}\div \dfrac{4x-12}{x+5}=\dfrac{2x^2-5x-3}{3x^2+14x-5}\cdot \dfrac{x+5}{4x-12}$.

Now, factor where possible:

$2x^2-5x-3=(2x+1)(x-3)$

$3x^2+14x-5=(3x-1)(x+5)$

$4x-12=4(x-3)$

Substituting these into the expression,

$\dfrac{2x^2-5x-3}{3x^2+14x-5}\div \dfrac{4x-12}{x+5}=\dfrac{(2x+1)(x-3)}{(3x-1)(x+5)}\cdot \dfrac{x+5}{4(x-3)}$

Canceling the common factors $x-3$ and $x+5$, we get

$\hphantom{00000}=\boxed{\dfrac{2x+1}{4(3x-1)}}$

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