Multiplying And Dividing Rational Expressions

In this video, we are going to multiply and divide rational expressions.

We can start by either directly performs the operation or reduce across.

For example: $\frac{15}{4}\times\frac{32}{20}$

First, reduce the numbers that are across from each other $\frac{15}{4}\times\frac{32}{20}$ $\frac{3}{1}\times\frac{8}{4}$

Further reduce, which can be done vertically with the 8 and the 4 $\frac{3}{1}\times\frac{2}{1}$

Multiply $\frac{6}{1}$

And we have $6$

With variables, it is the same concept $\frac{3x+15}{x^2+7x+12}\times\frac{x^2-9}{x+5}$

Factor the expressions if necessary $\frac{3(x+5)}{(x+4)(x+3)}\times\frac{(x+3)(x+3)}{(x+5)}$

Reduce diagonally $\frac{3}{(x+4)}\times\frac{(x+3)}{1}$

Multiple across $\frac{3(x+3)}{x+4}$

Division is similar to multiplication.
Remember to Keep, Change, Flip

For example: $\frac{x+8}{3x}\div\frac{4x+32}{15x^4}$

Keep the first fraction, change the operation from division to multiplication, and flip the second fraction $\frac{x+8}{3x}\times\frac{15x^4}{4x+32}$

Factor if necessary $\frac{x+8}{3x}\times\frac{15x^4}{4(x+8)}$

Reduce expressions $\frac{1}{1}\times\frac{5x^3}{4}$

Multiply across $\frac{5x^3}{4}$

Video-Lesson Transcript

Let’s go over how to multiply and divide rational expressions.

For example: $\dfrac{3x^2}{4} \times \dfrac{12y}{x^3}$

Just like normal fractions, we can multiply straight across or cross-cancel first.

I advise you to do cross-cancel first.

Just a review, $\dfrac{15}{4} \times \dfrac{32}{20}$

We’ll cross-cancel by finding the greatest common factor of $4$ and $8$ – which is $4$.

Then identify what number goes into $15$ and $20$. So, $5$ does. $= \dfrac{3}{1} \times \dfrac{8}{4}$

Then, we can reduce the second expression and we’ll have $= \dfrac{3}{1} \times \dfrac{2}{1}$

Let’s multiply across $= \dfrac{6}{1}$ $= 6$

Let’s do the same thing with our example. $\dfrac{3x^2}{4} \times \dfrac{12y}{x^3}$

Let’s see what we can reduce. $4$ and $12$ can be reduced.

Then let’s subtract the exponents of $x$. $= \dfrac{3}{1} \times \dfrac{3y}{x^1}$

Now, we can multiply across. $= \dfrac{9y}{x}$ Let’s have a more complex example. $\dfrac{3x + 15}{x^2 + 7x + 12} \times \dfrac{x^2 - 9}{x + 5}$

Here, we can’t cross-cancel because the terms are being added and subtracted.

Because in cross-cancel, things are being multiplied.

An example of this is: $\dfrac{(3) (4)}{(5)} \times \dfrac{(10) (7)}{(6)}$

This can be cross-cancelled because everything is multiplied.

What if terms aren’t multiplied?

We have to factor them out first. So we can come up with multiplication.

Only then can we cross-cancel. $\dfrac{3 (x + 5)}{(x + 4) (x + 3)} \times \dfrac{(x + 3) (x - 3)}{(x + 5)}$

Now that we factored this, let’s see what we can reduce.

We can cancel $x + 5$ and $x +3$

So, our final answer is $\dfrac{3 (x - 3)}{x + 4}$

Now, let’s look at division.

Before we go into the lesson, let’s have a quick look at the division of fractions. $\dfrac{12}{5} \div \dfrac{2}{9}$

In order to solve this, we do keep-change-flip.

Keep the first one. Then change to multiplication. And lastly, flip the second expression. $\dfrac{12}{5} \times \dfrac{9}{2}$

Then follow the same rule in multiplication.

Cross-cancel if you can. $= \dfrac{6}{5} \times \dfrac{9}{1}$ $= \dfrac{54}{5}$

Now, let’s apply the same rules to this problem. $\dfrac{x + 8}{3x} \div \dfrac{4x + 32}{15x^4}$

So let’s do keep-change-flip.

Keep the first one, change to multiplication and flip the last one. $= \dfrac{x + 8}{3x} \times \dfrac{15x^4}{4x + 32}$

Let’s try to factor this before we reduce. $= \dfrac{x + 8}{3x} \times \dfrac{15x^4}{4 (x + 8)}$

Now, let’s reduce. $x +8$ cancelled. Then $3$ and $15$. Then we’ll subtract the exponent of the $x$-es.

To get the final answer of $= \dfrac{5x^3}{4}$