factoring rules 2 *gcf( greatest common factor) – first rule 4 terms grouping 3 terms perfect...
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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 2a 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
GCFGreatest 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
22) 8 16
8 2
2
8
8
a a a
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
23) 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 16)
3
3
1
3R 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
29)
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 GroupGCF GCF
GCFx 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 GroupGCxF GCF
GCFx
x z x z xz z
z x 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 26 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 yy2 22 2[ 7 02 2 23 2 ]y yyy y
4 3 26 14 40y y y
13
2ax 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 26 14 40y y y
22 3 20[ ]2 7yy y 2ax 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 y 5 0]2y
60
22 3 20[ ]2 7yy y
6012 12( 5) 5 7
2 22 [3y y2 22 [3y y
16
FINISH FACTORING BY GROUPING
22 3 20[ ]2 7yy y
2
3
2
Group 1 Group 2G
5CF GCF
2 012 5 ]3[ 2
y
y yy y
2
3GCF C
5G F
[ ( 4) ( 45 )]2 3
y
yy y y
1722 ( 4)(3 5)y y y
FACTORED COMPLETELY
4 3 26 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 2G
5CF 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 #
21) 12 4 16 a a
20
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
2 22) 6 29 28 a ab b
21
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
23) 8 30 18x x
22
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
24) 3 10 8 h h
23
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
2 25) 10 7 12 m mn n
24
SUM OF FACTORS OF
GCF
FACTORS OF
KEY #
26) 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
22 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
22 2
22 2
2
2
a ab b a b
a ab b a b
27
4 3 2
2
2 2
22
2
22
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 2a 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 22 2 2
2
3) 81
9 9 9
9 3 3
x
x x x
x x x
2
2
22
544) 6
259
625
3 3 36 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
m x n x
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
100
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-WORLDAPPLICATIONS
USINGQUADRATICEQUATIONS
44
45
46
The height h in feet reached by a dolphin t seconds afterbreaking the surface of the water is given by hHow long will it take the dolphin to jump out of the water and touch the trainer’s hand?
216 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.
216 32 48s t t
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