If you thought that adding and subtracting rational expressions was difficult, you are in for a nice surprise.  Multiplying and dividing rational expressions is far easier.  In fact, learning both multiplying and dividing rational expressions boils down to learning only how to multiply them.  Division only involves one extra step.

                Consider how we multiply regular fractions.  We multiply straight across the top, straight across the bottom, then we simplify.  East enough, right?  For instance consider the problem

\(\Large \frac{3}{4} \times \Large \frac{5}{7}\)

To multiply these two fractions, we simply multiply across the top, and across the bottom, like so

\(\Large \frac{3}{4} \times \Large \frac{5}{7} = \Large \frac{{3 \cdot 5}}{{4 \cdot 7}} = \Large \frac{{15}}{{28}}\)

The fraction \(15/28\) is already in simplest form, so we’re done.  The procedure for multiplying rational expressions is exactly the same.  Only the expressions are slightly more complex.  Alright, let’s look at an example:

Multiply

\(\Large \frac{{{p^2} + 4p - 5}}{{p + 2}} \times \Large \frac{{p + 2}}{{4{p^2} - 4{p^3}}}\)

To solve this problem, we simply multiply across the top, and multiply across the bottom, then simplify.

\(\Large \frac{{{p^2} + 4p - 5}}{{p + 2}} \times \Large \frac{{p + 2}}{{4{p^2} - 4{p^3}}} = \Large \frac{{\left( {{p^2} + 4p - 5} \right)\left( {p + 2} \right)}}{{\left( {p + 2} \right)\left( {4{p^2} - 4{p^3}} \right)}}\)

\( = \Large \frac{{{p^2} + 4p - 5}}{{4{p^2} - 4{p^3}}}\), by cancelling \(p + 2\)

\( = \Large \frac{{\left( {p + 5} \right)\left( {p - 1} \right)}}{{4{p^2}\left( {1 - p} \right)}}\), by factoring both numerator and denominator

\( =  - \Large \frac{{p + 5}}{{4{p^2}}}\), since \(\Large \frac{{p - 1}}{{1 - p}} =  - 1\)

So then, we have

\(\Large \frac{{{p^2} + 4p - 5}}{{p + 2}} \times \Large \frac{{p + 2}}{{4{p^2} - 4{p^3}}} =  - \Large \frac{{p + 5}}{{4{p^2}}}\)

So how does division of rational expressions work?  Well, consider how division of fractions works.  Since dividing fractions is akin to multiplying the first fraction by the inverse of the second, we use the simple technique of flipping the second fraction and then multiplying them!  Easy as pie, right?

                For instance, consider the problem

\(\Large \frac{3}{8} \div \Large \frac{2}{3}\)

Here dividing by \(2/3\) is the same thing as multiplying by \(3/2\) (since \(2/3\) and \(3/2\) are inverses of each other).  So we simply flip the second fraction and multiply straight across.

\(\Large \frac{3}{8} \div \Large \frac{2}{3} = \Large \frac{3}{8} \times \Large \frac{3}{2} = \Large \frac{{3 \cdot 3}}{{8 \cdot 2}} = \Large \frac{9}{{16}}\)

We use the same procedure to divide rational expressions.  Consider the following problem.

Divide

\(\Large \frac{{4x}}{{{x^2} - 9x + 8}} \div \Large \frac{{x - 3}}{{ - {x^2} + 9x - 8}}\)

Simply flip the second rational expression, and then multiply.

\(\Large \frac{{4x}}{{{x^2} - 9x + 8}} \div \Large \frac{{x - 3}}{{ - {x^2} + 9x - 8}} = \Large \frac{{4x}}{{{x^2} - 9x + 8}} \times \Large \frac{{ - {x^2} + 9x + 8}}{{x - 3}}\)

\( = \Large \frac{{\left( {4x} \right)\left( { - {x^2} + 9x + 8} \right)}}{{\left( {{x^2} - 9x + 8} \right)\left( {x - 3} \right)}}\)

\( = \Large \frac{{ - \left( {4x} \right)\left( {{x^2} - 9x - 8} \right)}}{{\left( {{x^2} - 9x + 8} \right)\left( {x - 3} \right)}}\), by factoring-1 from \( - {x^2} + 9x + 8\)

\( = \Large \frac{{ - 4x}}{{x - 3}}\), by cancelling \({x^2} - 9x - 8\)

Then our final result is

\(\Large \frac{{4x}}{{{x^2} - 9x + 8}} \div \Large \frac{{x - 3}}{{ - {x^2} + 9x - 8}} = \Large \frac{{ - 4x}}{{x - 3}}\)

Below you can download some free math worksheets and practice.