« Simplifying Rational Expressions

Adding and Subtracting Rational Expressions »

Multiplication

Example 1

Multiply and then simplify the product

$\frac{{2x + 4}}{x} \cdot \frac{3}{{6x + 12}}$

Solution

Example 2

Multiply the following rational expressions:

$\frac{{{x^2} + 6x + 9}}{{{x^2} - 9}} \cdot \frac{{3x - 9}}{{{x^2} + 2x - 3}}$

Solution

1: Factor all numerators and denominators:

$\frac{{{x^2} + 6x + 9}}{{{x^2} - 9}} \cdot \frac{{3x - 9}}{{{x^2} + 2x - 3}} = \frac{{(x + 3)(x + 3)}}{{(x - 3)(x + 3)}} \cdot \frac{{3(x - 3)}}{{(x + 3)(x - 1)}}$

2: Cancel all common factors:

$\frac{{{x^2} + 6x + 9}}{{{x^2} - 9}} \cdot \frac{{3x - 9}}{{{x^2} + 2x - 3}} = \frac{{\cancel{{(x + 3)}}\cancel{{(x + 3)}}}}{{\cancel{{(x - 3)}}\cancel{{(x + 3)}}}} \cdot \frac{{3\cancel{{(x - 3)}}}}{{\cancel{{(x + 3)}}(x - 1)}}$

3: Multiply the denominators and numerators:

$\frac{{{x^2} + 6x + 9}}{{{x^2} - 9}} \cdot \frac{{3x - 9}}{{{x^2} + 2x - 3}} = \frac{{\cancel{{(x + 3)}}\cancel{{(x + 3)}}}}{{\cancel{{(x - 3)}}\cancel{{(x + 3)}}}} \cdot \frac{{3\cancel{{(x - 3)}}}}{{\cancel{{(x + 3)}}(x - 1)}} = \frac{1}{1} \cdot \frac{3}{{x - 1}} = \frac{3}{{x - 1}}$

Division of rational expressions

When we divide rational functions we multiply by the reciprocal.

Example 3:

Perform the indicated operations:

$\frac{{2{x^2} + x - 6}}{{{x^2} - 2x - 8}}:\frac{{2{x^2} - x - 3}}{{{x^2} - 3x - 4}}$

Solution 3:

$$\frac{{2{x^2} + x - 6}}{{{x^2} - 2x - 8}}:\frac{{2{x^2} - x - 3}}{{{x^2} + 3x - 4}} =$$ $$= \frac{{2{x^2} + x - 6}}{{{x^2} - 2x - 8}} \cdot \frac{{{x^2} - 3x - 4}}{{2{x^2} - x - 3}} =$$ $$= \frac{{2\left( {x - \frac{3}{2}} \right)(x + 2)}}{{(x + 2)(x - 4)}} \cdot \frac{{(x - 4)(x + 1)}}{{2\left( {x - \frac{3}{2}} \right)(x + 1)}} =$$ $$= \frac{{\bcancel{{(2x - 3)}}\cancel{{(x + 2)}}}}{{\cancel{{(x + 2)}}\bcancel{{(x - 4)}}}} \cdot \frac{{\bcancel{{(x - 4)}}\cancel{{(x + 1)}}}}{{\bcancel{{(2x - 3)}}\cancel{{(x + 1)}}}} = 1$$

Example 4:

Perform the indicated operations:

$\frac{{\frac{{x + 4}}{{2x - 6}}}}{{\frac{{3x + 12}}{{4x - 12}}}}$

Solution 4:

$$\frac{{\frac{{x + 4}}{{2x - 6}}}}{{\frac{{3x + 12}}{{4x - 12}}}} = \frac{{x + 4}}{{2x - 6}} \cdot \frac{{4x - 12}}{{3x + 12}} =$$ $$= \frac{{\cancel{{x + 4}}}}{{2\cancel{{(x - 3)}}}} \cdot \frac{{4\cancel{{(x - 3)}}}}{{3\cancel{{(x + 4)}}}} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$$