1. Factor By Grouping
By L.D.

2. Problem 1
 Factor 5(x + 8) + 3x(x + 8)

3. Problem 1
 Factor 5(x + 8) + 3x(x + 8)
The first thing we do to solve this is separate the problem up.
Since the things in parenthesis on both sides are the same, we
will smash them together and combine them to make a single (x
+ 8). I will turn them red on the next slide to show we mustn’t
touch them now.

4. Problem 1
 Factor 5(x + 8) + 3x(x + 8)
 (x + 8)
The second things we must do is look at the problem without
the parenthesis problems. What we see is 5 + 3x, so we will place
that in parenthesis and let is sit next to our combined (x + 8) to
make our answer which is on the next slide.

5. Problem 1 Answer
 (5 + 3x)(x + 8)

6. Practice
 Factor
 2x (x – 5) + 3 (x -5)
 x2 (y – 3) – 8 (y – 3)

7. Practice
 Factor
 2x (x – 5) + 3 (x -5) (2x + 3)(x – 5)
 x2 (y – 3) – 8 (y – 3) (x2 – 8)(y – 3)

8. Problem 2
 Factor y3 – 4y2 + 8y - 32

9. Problem 2
 Factor y3 – 4y2 + 8y – 32
The first thing we need to do here is to get this into the proper
format. So we need to split the problem in half using the + as a
dividing line, like a wall and then we need to find the GCF of the
problems on separate sides of the wall.

10. Problem 2
 Factor y3 – 4y2 + 8y – 32
 y2(y – 4) & 8(y – 4)
Now that we have the GCFs, we will combine the problem and
make the 8 have a plus in front of it as it is positive.
y2(y – 4) + 8(y – 4)

11. Problem 2
 Factor y3 – 4y2 + 8y – 32
 y2(y – 4) & 8(y – 4)
 y2(y – 4) + 8(y – 4)
Now we create our problem by slapping the parenthesis
together and adding the y2 and 8.

12. Problem 2 Answer
 Factor y3 – 4y2 + 8y – 32
 (y2+ 8)(y – 4)

13. Practice
 5w + 10 + 3xw + 6x
 2wx + 10w + 7x + 35
 a3 + 6a – 5a2 – 30

14. Practice
 5w + 10 + 3xw + 6x (5 + 3x)(w + 2)
 2wx + 10w + 7x + 35 (2w + 7)(x + 5)
 a3 + 6a – 5a2 – 30 (a – 5)(a + 6)
The reason the last one is came out like this is becaus, when I
looked at the second half with the negative dividing line, I
decided to take -5 out of the -5a2.
a3 + 6a – 5a2 – 30
a(a2 + 6) – 5(a2 – 6)
(a – 5)(a2 + 6)

15. Bonus
 x(13 + x) – (x + 13)

16. Bonus
 x(13 + x) – (x + 13)
In this problem we will fill up the emptiness next to the dividing
line by make visible the ever invisible 1. We can then solve the
problem as we always have. We will then get (x - 1)(13 + x)