# Dividing Rational Expressions

Because rational expressions are just fractions with variables, the process of dividing rational expressions is exactly the same as the process of dividing fractions. **You just find the reciprocal of the second rational expression (flip it, so the the denominator becomes the numerator and the numerator becomes the denominator), and then multiply the rational expressions (numerator times numerator and denominator times denominator).** It's really that easy!

Just like with fractions, you can cross-cancel before you multiply. So, if you have polynomials or polynomial expressions, factor them completely and cross-cancel any terms that you find in both a numerator and a denominator. Remember you can cancel out terms attached with multiplication but if numbers or variable are attached by addition or subtraction they have to stay together.

*Example*:

$$\require{cancel}\eqalign{ \dfrac{x^2+x-6}{x^2}\div\dfrac{(x+3)^2}{x}=&\\ \dfrac{x^2+x-6}{x^2}\times\dfrac{x}{(x+3)^2}=&\qquad&&\text{Replace the divisor expression with its reciprocal and multiply}\\ \dfrac{\overset{(x-2)(x+3)}{\bcancel{x^2+x-6}}}{x^2}\times\dfrac{x}{\underset{(x+3)(x+3)}{\bcancel{(x+3)^2}}}=& \quad&&\text{Factor completely}\\ \dfrac{\overset{(x-2)\bcancel{(x+3)}}{\bcancel{x^2+x-6}}}{\underset{x}{\bcancel{x^2}}}\times\dfrac{\bcancel{x}}{\underset{\bcancel{(x+3)}(x+3)}{\bcancel{(x+3)^2}}}=& \quad&&\text{Cross cancel}\\ \dfrac{\overset{(x-2)\bcancel{(x+3)}}{\bcancel{x^2+x-6}}}{\underset{x}{\bcancel{x^2}}}\times\dfrac{\bcancel{x}}{\underset{\bcancel{(x+3)}(x+3)}{\bcancel{(x+3)^2}}}=&\dfrac{x-2}{x(x+3)}&&\text{Multiply numerator times numerator and denominator times denominator}}$$

Dividing rational expressions is easy! Just find the reciprocal of the second (divisor) fraction and then multiply. To make your life easiest, make sure you factor and cross-cancel as you go!

#### Practice Problems:

## Divide Rational Expressions

Simplify the rational expressions:

- $\dfrac{x}{x^2}\div \dfrac{x}{x^2}$
- $\dfrac{x}{3x}\div \dfrac{x}{2x}$
- $\dfrac{x+1}{x^2}\div \dfrac{x+1}{7x}$
- $\dfrac{9x}{3x^2+6x+9}\div \dfrac{x^2}{3x^2-27}$
- $\dfrac{2x-2}{7x^2+14x}\div \dfrac{2x}{x+2}$
- $\dfrac{x^2-5x-24}{x^2+2x+1}\div \dfrac{x^2-8x}{x+1}$
- $\dfrac{9x}{x^2+4x-21}\div \dfrac{3x^2+9x}{x^2}$
- $\dfrac{x^3-3x^2-10x}{x^2+5x-36}\div \dfrac{10x}{x^2+7x-18}$

#### Answer Key: