Complete Factoring Rules for Grade 8.ppt

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About This Presentation

Complete Factoring Rules for Grade 8.ppt


Slide Content

FACTORING
RULES

2
*GCF( Greatest Common Factor) – First Rule
4 TERMS
Grouping

3 TERMS
Perfect Square Trinomial


AC Method with Grouping
2 TERMS
Difference Of Two Squares



Sum or Difference Of Two Cubes

2 2 2
2 2 2
a 2ab b (a b)
a 2ab b (a b)
   
   
2 2
a b (a b)(a b)   
3 3 2 2
3 3 2 2
a b (a b)(a ab b )
a b (a b)(a ab b )
    
    

3
GCF
Greatest Common Factor
First Rule to Always Check
 
 
 
3
2
2
1) 3
3
3
y y
y
y
y y
y

  

 

3
2 2
2
2
2) 8 16
8 2
2
8
8
aaa
a
aa
a
  

4
 
 
3 2 3
3 2 3
3 2 3
4) 12 16 48
3 4 14 4 2
34 4 12
4
p p t t
p p t t
p p t t
  
  



 
 
 
2
3) 2
2
2
a
ab a ax
b aa ax
b a xa
 
 
 

5
 
 
2 2 2 2
2
5) 4 2 4 2
2 2
2 2
a b c d
a b c d
a b c d
  
    
  
 
2 2
2 2
2 2
1 1
3 3
1 1
6)
3
3
1
3
R h r h
R r
R
h h
h r


 

 
 
 
 



  
 
 


6


7) 3
3
4 4
4
x xx
xx

 

 
8) 2 7
2
1 1
1 7
y y
yy
y 






 
2
9)
1
1
a b a b
a
a b a b
a b
a b
a b
b
a b
  
 
  







7
4 TERMS - Grouping



2
1 2

1
2
)2 2
[2
2
2
]

  
  
   
 
 


Group Group
GCF GCF
GCF
x y
a
x y ax ay
x y x y
x y x y
x
a
y
a
a
a

8


 
2
3 2
3
16
2
2
1 2

2
2
2)2 4 32 64
2[ 2 16 32]
2 2 2
2 2
16
16

  
  
   
 
 


Group Group
GC
x
F GCF
GCF
x
xz xz xz z
zx x x
z x
x
x
z x
x

9
3 TERMS
1) Perfect Square
Trinomials
2) AC Method With
Grouping
We will explore factoring trinomials using
the ac method with grouping next and
come back to Perfect Square Trinomials
later.

Factoring Trinomials
by
Using The
AC Method
With
Grouping

11
4 3 2
6 14 40y y y 
2 22 2
[ 7 02 2 23 2 ]y yyy y    
The first rule of factoring is to factor
out the Greatest Common Factor
(GCF).
Factor the trinomial completely.

12
Stop! Check that you have factored
the (GCF) correctly by distributing it
back through the remaining
polynomial to obtain the original
trinomial.
2 2
[3 20]2 7 y yy
2 22 2
[ 7 02 2 23 2]y yyy y    
4 3 2
6 14 40y y y 

13
2
ax cbx 
To factor , we must find two integers
whose product is -60 and whose sum is 7.
To factor , we must find two integers
whose product is ac and whose sum is b.
After factoring out the (GCF), the remaining
polynomial is of the form
4 3 2
6 14 40y y y 
22
3 20[ ]2 7yy y
2
ax cbx
2
73 20y y

14
Key number 60
6012 12( 5) 5 7   
FACTORS OF 60 SUM OF FACTORS OF 60
22
3 20[ ]2 7yy y
1( 60) 60 1 ( 60) 59
2( 30) 60 2 ( 30) 28
3( 20) 60 3 ( 20) 17
4( 15) 60 4 ( 15) 11
5( 12) 60 5 ( 12) 7
    
    
    
    
    

15
ac = b = 7
Replace b = 7 in our original expression with
b = 12 + (-5).
7 0]2y
12y5 0]2y
60
22
3 20[ ]2 7yy y
6012 12( 5) 5 7   
2 2
2[3yy
2 2
2[3yy

16
FINISH FACTORING BY GROUPING
22
3 20[ ]2 7yy y
2
3
2
Group 1 Group 2
G
5
CF GCF

2 012 5 ]3[ 2
y
y yy y



2
3
GCF C
5
G F

[ ( 4) ( 45 )]2 3
y
yy y y

 


17
2
2 ( 4)(3 5)y y y 
FACTORED COMPLETELY
4 3 2
6 14 40y y y 
22
3 20[ ]2 7yy y
GC
2
F
( 4)
2[ ( 4) (5 ]3 4)
y
y yy y

 

2
3
2
Group 1 Group 2
G
5
CF GCF

2 012 5 ]3[ 2
y
y yy y




18
Practice Problems
2
2 2
2
2
2 2
2
1) 12 4 16
2) 6 29 28
3) 8 30 18
4) 3 10 8
5) 10 7 12
6) 6 3 18
 
 
 
  
 
  
a a
a ab b
x x
h h
m mn n
y y

19
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
2
1) 12 4 16 a a

20
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
2 2
2) 6 29 28 a ab b

21
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
2
3) 8 30 18x x 

22
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
2
4) 3 10 8  h h

23
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
2 2
5) 10 7 12 m mn n

24
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
2
6) 6 3 18y y  

25
Answers To Practice Problems
1) 4(3 4)( 1)
2) (3 4 )(2 7 )
3) 2(4 3)( 3)
4) 1(3 2)( 4)
5) (5 4 )(2 3 )
6) 3(2 3)( 2)
 
 
 
  
 
  
a a
a b a b
x x
h h
m n m n
y y

26
Perfect Square Trinomials

 
 
2 2
2
2 2
2 2
2
1) 25 70 49
5 2 5 7 7
5 7
2
ba
m mn n
m
a ab
m n n
m
b a
n
b
 
   
 
   
 





 


2
2 2
2
2 2
2
2
a ab b a b
a ab b a b
   
   

27

 
 
4 3 2
2
2 2
22
2
2
2
2 2
2) 4 20 25
4 20 25
2 2 2 5
2
5
2 5
a b
a ab b a b
x x x
x x x
x x x
x x
 
  
 
 
  
     
  
 
 
 

 

28
2 TERMS
1) Difference of Two
Squares
2) Sum and Difference of
Two Cubes

29
Difference of Two Squares

2 2
a b a b a b   

2
2 2
1) 9
3 3 3
x
x x x

   
   
2
2
2 2
2) 2 200
2 100
2 10 2 10 10
p
p
p p p

 
 
    
 

30
  
 
4
2 2
2 2 2
2
3) 81
9 9 9
9 3 3
x
x x x
x x x

   
  

2
2
2
2
54
4) 6
25
9
6
25
3 3 3
6 6
5 5 5
t
t
t t t

 

 
 
 
    
        
      

31
Sum and Difference of Two Cubes
 
 
3 3 2 2
3 3 2 2
a b a b a ab b
a b a b a ab b
    
    

32
 

 
 
3
3
3
3
3 2
3
2
2
1) 125
125
5
5 5 25
x
x
x
x x
a b a b a ab b
x

 
 
  
 
  
 

 

33
 

 
  
4 3
3 3
3 3
3 3 2
2 2
2
2) 16 128
16 8
16 2
16 2 2 4
a b a b
r rs
r
a ab
r s
r r s
r r s r rs s
b


 
 
 
 

  

34
 

 
 
3
3
3
3
3 2
3
2
2
3) 216
216
6
6 6 36
x
x
x
x x
a b a b a ab b
x

 
 
  
 
  
 

 

35
 

 
  
3 3 2
3 3
3 3
3 3
2
2
2
4) 64 8
8 8
8 2
8 2 4 2
a b a b a ab
mx nx
x m n
x m n
x m n m n
b
m n


 
 






 

36
What purpose does factoring
serve?
Factoring is an algebraic process which allows
us to solve quadratic equations pertaining to
real-world applications, such as remodeling a
kitchen or building a skyscraper.
We will cover the concept of solving quadratic
equations and then investigate some real-
world applications.

37
Solving Quadratic Equations
A quadratic equation is an equation
that can be written in standard form
where a, b, and c represent real
numbers, and
2
0ax bx c  
0a

38
We will solve some quadratic equations
using factoring and the
Zero-Factor Property.
When the product of two real numbers is 0,
at least one of them is 0.
If a and b represent real numbers, and
if then a=0 or b=00ab

39
Solve Each Equation
1) 3 2 0
3 0 and 2 0
3 2
x x
x x
x x
  
   
 
 2) 7 3 10 0
7 0 and 3 10 0
10
0
3
a a
a a
a a
  
   
 

40

  
2
2
3) 9 3 3 25
9 27 3 25
9 30 25 0
3 5 3 5 0
3 5 0
5
3
a a a
a a a
a a
a a
a
a
  
  
  
  
 

41


2
2
2
4) 8 3 30
3 8 24 30
5 24 30
5 6 0
2 3 0
2 0 and 3 0
2 3
n n
n n n
n n
n n
n n
n n
n n
  
   
  
  
  
   
 

42
 

3 2
2
5) 3 2 0
3 2 0
1 2 0
0, 1 0, and 2 0
0, 1, 2
x x x
x x x
x x x
x x x
x
  
  
  
    
  
 

3
2
6) 6 6 0
6 1 0
6 1 1 0
6 0, 1 0, 1 0
0, 1, 1
n n
n n
n n n
n n n
n
 
 
  
    
 

43
REAL-WORLD
APPLICATIONS
USING
QUADRATIC
EQUATIONS

44

45

46
The height h in feet reached by a dolphin t seconds after
breaking the surface of the water is given by h
How long will it take the dolphin to jump out of the water
and touch the trainer’s hand?
2
16 32t t 

47
From the top of the building a ball is thrown straight up with
an initial velocity of 32 feet per second. The equation below
gives the height s of the ball t seconds after thrown. Find the
maximum height reached by the ball and the time it takes for
the ball to hit the ground.
2
16 32 48s t t  
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