Multiplying & Dividing Rational Expressions

By L.D.

Table of Contents

Slide 3: Mini Lesson: Multiplying Fractions

Slide 4: Problem 1, 3g4/4g3 x 2g2/5g

Slide 7: Mini Lesson (Simplifying)

Slide 8: Problem 2, 2c4/5c2 x 6c /3c3

Slide 12: Mini Lesson: Simplifying Trick

Slide 13: Problem 3, u2 – 5u + 4/3u2 – 12u x 2u + 2/u2 + 6u – 7

Slide 18: Mini Lesson: Dividing Fractions

Slide 19: Problem 4, x2 + 5x – 24/x2 + 9x + 8 ÷ x2 – 9/6x - 18

Mini Lesson: Multiplying Fractions

Do you remember how to multiply fractions?

Try to multiply 4/6 and 9/5.

We just multiplied straight across, 4 x 9 and 6 x 5.

Now it is time to simplify. Find the GCF of 36 and 30. Since it is 6, each number will be individually divided by 6 to make the final answer be 6/5.

4 9 36

6 5 30x =

Problem 1

3g4 2g2

4g3 5gx

Problem 1

3g4 2g2

4g3 5g

The first thing to do is to multiply them. This can be done to get

6g6

20g4

x

Problem 1

6g6

20g4

Now this can be simplified to get

3g2

10

Mini Lesson (Simplifying)

When presented with two fractions to multiply, like 12/24 and 11/66, there is a way to avoid multiplying out such large numbers.

Here’s how you do it, you simplify them and then multiply them.

By doing this you avoid all the trouble of multiplying big numbers and having to simplify them later.

So instead of having

12/24 x 11/66 = 132/1584 (now simplify)

There is instead

1/2 x 1/6 = 1/12

Another thing that can be simplified is variables, if you have t15/t20. It can be simplified to t3/t4.

Problem 2

2c4 6c

5c2 3c3x

Problem 2

2c4 6c

5c2 3c3

The first thing to do here is to simplify the separate fractions.

2c2 2c

5 c3

x

x

Problem 2

2c2 2c

5 c3

Now to multiply!

2c2 2c 4c3

5 c3 5c3x

x

=

Problem 2

4c3

5c3

Even now, the answer can be simplified to just simple 4/5.

Mini Lesson: Simplifying Trick

Say you need to simplify a problem like

5(x -3) 7(11-x)

2 7(x – 3)

A special way to make it easier is that if you have the same thing in the two problems (the placement needs to be that one of those “same things” has to be in a denominator of one problem and the other has to be in the numerator) you can cancel it out.

Problem 3

u2 – 5u + 4 2u + 2

3u2 – 12u u2 + 6u – 7x

Problem 3

u2 – 5u + 4 2u + 2

3u2 – 12u u2 + 6u – 7

The first thing that needs to happen to the problem here is that it must be completely factored out on the different numerators and denominators like below.

(x – 4)(x – 1) 2(x + 1)

3x(x – 4) (x + 7)(x – 1)

x

x

Problem 3

(x – 4)(x – 1) 2(x + 1)

3x(x – 4) (x + 7)(x – 1)

The last mini lesson taught that it was possible to cancel out certain special parts of the problem.

(x – 4) 2(x + 1)

3x(x – 4) (x + 7)

x

x

Problem 3

(x – 4) 2(x + 1)

3x(x – 4) (x + 7)

Now we will do the normal canceling out.

1 2(x + 1)

3x (x + 7)

x

x

Problem 3

1 2(x + 1)

3x (x + 7)

Now we will multiply these and you will get your final answer!

2(x + 1)

3x(x + 7)

x

Mini Lesson: Dividing Fractions

For those who have forgotten how to divide fractions, here it is!

For an example I will divide 2/3 and 1/6.To do this I will simply flip the thing that my main number (2/3) is being divided into and multiply them.

2 6 12

3 1 3x = = 4

Problem 4

x2 + 5x – 24 x2 - 9

x2 + 9x + 8 6x – 18÷

Problem 4

x2 + 5x – 24 x2 - 9

x2 + 9x + 8 6x – 18

What to do in this situation is to first flip it and make it into a multiplication problem.

x2 + 5x – 24 6x – 18

x2 + 9x + 8 x2 - 9

÷

x

Problem 4

x2 + 5x – 24 6x – 18

x2 + 9x + 8 x2 – 9

The next thing to do is to factor the problem.

(x + 8)( x -3) 6(x – 3)

(x + 8)(x + 1) (x + 3)(x – 3)

x

x

Problem 4

(x + 8)( x -3) 6(x – 3)

(x + 8)(x + 1) (x + 3)(x – 3)

Now to cancel out.

x

Problem 4

6(x – 3)

(x + 1)(x + 3)

That’s the final answer, I could go farther, but I would rather not since I like the way it looks now.

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