FACTORING
RULES
*GCF( Greatest Common Factor) – First Rule
4 TERMS
Grouping
3 TERMS
Perfect Square Trinomial
a2  2ab  b 2  (a  b)2
a2  2ab  b 2  (a  b)2
AC Method with Grouping
2 TERMS
Difference Of Two Squares
a2  b 2  (a  b)(a  b)
Sum or Difference Of Two Cubes
a3  b3  (a  b)(a2  ab  b 2 )
a3  b3  (a  b)(a2  ab  b 2 )
2
 GCF 
Greatest Common Factor
First Rule to Always Check
1)
y 3  3y

y  y2  3 y

y y2  3


2) 8a 3  16a 2

8a 2  a  8a 2  2

8a 2  a  2 
3
3) ab  a  2ax
2
 ab  aa  2ax 
a  b  a  2x 
4) 12 p3  16 p 2t  48t 3
 4  3p  4  4p t  4  12t 
4  3 p  4 p t  12t 
3
3
2
2
3
3
4
5) 4a  2b  4c  2d
 2  2a  2b  2  2c  2d 
2  2a  b  2c  d 
1
1 2
2
6)   R h   r h
3
3
 1
 1  2
2
 h R   hr 
 3 
 3


1
  h R2  r 2
3

5
7)
x  x  4  3  x  4
 x  4  x  3 
8)
2y  y  1  7  y  1
 y  1 2y  7 
9)
a  b   a  b 
 a  b  a  b    a  b 
 a  b   a  b   1
 a  b  a  b  1
2
6
4 TERMS - Grouping
1) 2 x  2 y  ax  ay
[ 2 x  2 y  ax  ay ]
Group 1
GCF
2
Group 2
GCF
a

2  x  y   a  x  y 

GCF
 x y
 x  y  2  a 
7
2) 2 x z  4 x z  32 xz  64 z
3
2
2 z[ x  2 x 16 x  32]
3
Group 1
GCF
x2
2
Group 2
GCF
16
2

2 z  x  x  2   16  x  2  

GCF
 x  2
2 z  x  2   x  16 
2
8
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.
9
Factoring Trinomials
by
Using The
AC Method
With
Grouping
Factor the trinomial completely.
6 y  14 y  40 y
4
3
2
The first rule of factoring is to factor
out the Greatest Common Factor
(GCF).
[2 y  3 y  2 y  7 y  2 y  20]
2
2
2
2
11
Stop! Check that you have factored
the (GCF) correctly by distributing it
back through the remaining
polynomial to obtain the original
trinomial.
2 y [3 y  7 y  20]
2
2
[2 y  3 y  2 y  7 y  2 y  20]
2
2
2
2
6 y  14 y  40 y
4
3
2
12
6 y  14 y  40 y
4
3
2
After factoring out the (GCF), the remaining
polynomial is of the form
ax  bx  c
2
2 y [3 y  7 y  20]
2
2
To factor ax  bx  c , we must find two integers
whose product is ac and whose sum is b.
2
To factor 3 y  7 y  20 , we must find two integers
whose product is -60 and whose sum is 7.
2
13
2 y 2 [3 y 2  7 y  20]
Key number   60
FACTORS OF
 60
SUM OF FACTORS OF  60
1(60)  60
2(30)  60
3(20)  60
4(15)  60
5(12)  60
1  (60)  59
2  ( 30)  28
3  ( 20)  17
4  ( 15)  11
5  (12)  7
12(5)  60
12   5  7
14
2 y [3 y  7 y  20]
2
2
ac =  60
b=7
12   5  7
12(5)  60
Replace b = 7 in our original expression with
b = 12 + (-5).
2 y [3 y 7 y  20]
2
2
2 y [3 y 12 y  5y  20]
2
2
15
2 y 2 [3 y 2  7 y  20]
FINISH FACTORING BY GROUPING
2 y [3 y  12 y 5 y  20]
2
2
Group 1
GCF
3y
Group 2
GCF
5
2 y [3 y ( y  4) 5( y  4)]
2
GCF
3y
GCF
5
16
6 y  14 y  40 y
4
3
2
2 y [3 y  7 y  20]
2
2
2 y [3 y  12 y 5 y  20]
2
2
Group 1
GCF
3y
Group 2
GCF
5
2 y [3 y ( y  4)  5( y  4)]
2
GCF
(y  4)
FACTORED COMPLETELY
2 y ( y  4)(3 y  5)
2
17
Practice Problems
1)
12a 2  4a  16
2) 6a 2  29ab  28b 2
3)
8 x 2  30 x  18
4)
3h 2  10h  8
5) 10m 2  7 mn  12n 2
6)
6 y 2  3 y  18
18
1)
12a 2  4a  16
GCF
KEY #
FACTORS OF
SUM OF FACTORS OF
19
2) 6a 2  29ab  28b 2
GCF
KEY #
FACTORS OF
SUM OF FACTORS OF
20
3) 8 x 2  30 x  18
GCF
KEY #
FACTORS OF
SUM OF FACTORS OF
21
4)
3h  10h  8
2
GCF
KEY #
FACTORS OF
SUM OF FACTORS OF
22
5) 10m 2  7mn  12n 2
GCF
KEY #
FACTORS OF
SUM OF FACTORS OF
23
6)
6 y  3 y  18
2
GCF
KEY #
FACTORS OF
SUM OF FACTORS OF
24
Answers To Practice Problems
1)
4(3a  4)( a  1)
2) (3a  4b)(2a  7b)
3)
2(4 x  3)( x  3)
4)
1(3h  2)( h  4)
5)
(5m  4n)(2m  3n)
6)
3(2 y  3)( y  2)
25
Perfect Square Trinomials
a  2ab  b   a  b 
2
2
a  2ab  b   a  b 
2
1)
2
2
2
25m 2  70mn  49n 2
a  2ab  b   a  b 
2
2
2




 5m   2  5m  7n    7n 
 b 
 a 
 5m  7n 
2
2
2
26
2)
4 x  20 x  25 x
4
3
2
x 2  4 x 2  20 x  25 
a  2ab  b   a  b 
2
2
2
2
2






x 2  2 x   2  2 x  5    5  
 b  
 a 
x
2
 2x  5 
2
27
2 TERMS
1) Difference of Two
Squares
2) Sum and Difference of
Two Cubes
28
Difference of Two Squares
a  b  a  b a  b 
2
2
1) x 2  9
x
2
  3    x  3  x  3 
2
2) 2 p 2  200
2  p 2  100 
2
2

2  p   10    2  p  10  p  10 


29
3) x  81
4
 x   9   x
x
2
2
2
2
2


 9   x  3  x  3 
9 x 9
2
54
4) 6t 
25
 2 9 
6 t  
25 

2
 2  3 2 
 3  3 
6  t       6  t   t  
 5  
 5  5 

30
Sum and Difference of Two Cubes

 a  b  a
a  b   a  b  a  ab  b
2
a b
2
3
3
3
3
2
2
 ab  b


31
1)  x  125
3

 x  125
3

  x    5  


3
3

a  b   a  b  a  ab  b
3
3

2
  x  5  x  5 x  25
2
2


32
2) 16r  128rs
3


4
16r r  8s
3
3
16r  r    2s  


3
3

a  b   a  b  a  ab  b
3
3

2
16r  r  2s  r  2rs  4s
2
2
2


33
3)  x  216
3

 x  216
3

  x    6  


3
3

a  b   a  b  a  ab  b
3
3

2
  x  6  x  6 x  36
2
2


34
4) 64m x  8n x
3
3

8 x 8m  n
3
3

8 x  2m    n  


3
3

a  b   a  b  a  ab  b
3
3

2
8 x  2m  n  4m  2mn  n
2
2
2


35
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 realworld applications.
36
Solving Quadratic Equations
A quadratic equation is an equation
that can be written in standard
form
ax  bx  c  0
2
where a, b, and c represent real
numbers, and
a0
37
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 ab  0 then a=0 or b=0
38
Solve Each Equation
1)
 x  3  x  2   0
x  3  0 and x  2  0
x 3
x2
2)
7a  3a  10   0
7a  0 and 3a  10  0
10
a0
a
3
39
3)
9a  a  3   3a  25
9a 2  27a  3a  25
9a 2  30a  25  0
 3a  5  3a  5   0
3a  5  0
5
a
3
40
4)
 n  8  n  3   30
n 2  3n  8n  24  30
n 2  5n  24  30
n 2  5n  6  0
 n  2  n  3   0
n  2  0 and n  3  0
n  2
n  3
41
5)
x 3  3 x 2  2x  0
x  x 2  3x  2  0
x  x  1 x  2   0
x  0, x  1  0, and x  2  0
x  0,  1,  2
6)
6n 3  6n  0
6n  n 2  1  0
6n  n  1 n  1  0
6n  0, n  1  0, n  1  0
n  0,  1, 1
42
REAL-WORLD
APPLICATIONS
USING
QUADRATIC
EQUATIONS
43
44
45
The height h in feet reached by a dolphin t seconds after
2
breaking the surface of the water is given by h  16t  32t
How long will it take the dolphin to jump out of the water
and touch the trainer’s hand?
46
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.
s  16t 2  32t  48
47