1. Multiplying & Dividing
Rational Expressions
By L.D.

2. 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

3. 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 30
x =

4. Problem 1
3g4 2g2
4g3 5g
x

5. Problem 1
3g4 2g2
4g3 5g
The first thing to do is to multiply them. This can be done
to get
6g6
20g4
x

6. Problem 1
6g6
20g4
Now this can be simplified to get
3g2
10

7. 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.

8. Problem 2
2c4 6c
5c2 3c3
x

9. Problem 2
2c4 6c
5c2 3c3
The first thing to do here is to simplify the separate
fractions.
2c2 2c
5 c3
x
x

10. Problem 2
2c2 2c
5 c3
Now to multiply!
2c2 2c 4c3
5 c3 5c3
x
x
=

11. Problem 2
4c3
5c3
Even now, the answer can be simplified to just simple 4/5.

12. 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.

13. Problem 3
u2 – 5u + 4 2u + 2
3u2 – 12u u2 + 6u – 7
x

14. 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

15. 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

16. 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

17. 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

18. 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 3
x = = 4

19. Problem 4
x2 + 5x – 24 x2 - 9
x2 + 9x + 8 6x – 18
÷

20. 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

21. 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

22. Problem 4
(x + 8)( x -3) 6(x – 3)
(x + 8)(x + 1) (x + 3)(x – 3)
Now to cancel out.
x

23. 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.