Chabot Mathematics
§5.5 Factor
Special Forms
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Review § 5.4
MTH 55
 Any QUESTIONS About
• §5.4 → Factoring TriNomials
 Any QUESTIONS About HomeWork
• §5.4 → HW-14
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
§5.5 Factoring Special Forms
 Factoring Perfect-Square Trinomials
and Differences of Squares
• Recognizing Perfect-Square Trinomials
• Factoring Perfect-Square Trinomials
• Recognizing Differences of Squares
• Factoring Differences of Squares
• Factoring SUM of Two Cubes
• Facting DIFFERENCE of Two Cubes
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Recognizing Perfect-Sq Trinoms
 A trinomial that is the square of a binomial
is called a perfect-square trinomial
A2 + 2AB + B2 = (A + B)2
A2 − 2AB + B2 = (A − B)2
 Reading the right sides first, we see that
these equations can be used to factor
perfect-square trinomials.
• A2 + 2AB + B2 = (A + B)(A + B)
• A2 − 2AB + B2 = (A − B)(A − B)
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Recognizing Perfect-Sq Trinoms
 Note that in order for the trinomial to be
the square of a binomial, it must have
the following:
1. Two terms, A2 and B2, must be squares,
such as: 9, x2, 100y2, 25w2
2. Neither A2 or B2 is being SUBTRACTED.
3. The remaining term is either
2  A  B or −2  A  B
• where A & B are the square roots of A2 & B2
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Trinom Sqs
 Determine whether each of the
following is a perfect-square trinomial.
a) x2 + 8x + 16
b) t2 − 9t − 36
c) 25x2 + 4 – 20x
 SOLUTION a) x2 + 8x + 16
1. Two terms, x2 and 16, are squares.
2. Neither x2 or 16 is being subtracted.
3. The remaining term, 8x, is 2x4,
where x and 4 are the square roots
of x2 and 16
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Trinom Sqs
 SOLUTION b) t2 – 9t – 36
1. Two terms, t2 and 36, are squares. But
2. But 36 is being subtracted so t2 – 9t –
36 is not a perfect-square trinomial
.
 SOLUTION c) 25x2 + 4 – 20x
It helps to write it in descending order.
25x2 – 20x + 4
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Trinom Sqs

1.
2.
3.
SOLUTION c) 25x2 − 20x + 4
Two terms, 25x2 and 4, are squares.
There is no minus sign before 25x2 or 4.
Twice the product of the square roots is
2  5x  2, is 20x, the opposite of the
remaining term, −20x

Thus 25x2 − 20x + 4 is a
perfect-square trinomial.
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring a Perfect-Square Trinomial
 The Two Types of Perfect-Squares
2
A
2
A
+ 2AB +
2
B
= (A +
2
B)
− 2AB +
2
B
= (A −
2
B)
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor Perf. Sqs
 Factor: a) x2 + 8x + 16
b) 25x2 − 20x + 4
 SOLUTION a)
x2 + 8x + 16 = x2 + 2  x  4 + 42 = (x + 4)2
A2 + 2 A B + B2 = (A + B)2
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor Perf. Sqs
 Factor: a) x2 + 8x + 16
b) 25x2 − 20x + 4
 SOLUTION b)
25x2 – 20x + 4 = (5x)2 – 2  5x  2 + 22 = (5x – 2)2
A2 – 2 A B + B2 = (A – B)2
Chabot College Mathematics
11
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor 16a2 – 24ab + 9b2
 SOLUTION
16a2 − 24ab + 9b2 = (4a)2 − 2(4a)(3b) + (3b)2
= (4a − 3b)2 = (4a − 3b)(4a − 3b)
 CHECK:
(4a − 3b)(4a − 3b) = 16a2 − 24ab + 9b2 
 The factorization is (4a − 3b)2.
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Expl  Factor 12a3 – 108a2 + 243a
 SOLUTION
 Always look for a common factor. This
time there is one. Factor out 3a.
12a3 − 108a2 + 243a = 3a(4a2 − 36a + 81)
= 3a[(2a)2 − 2(2a)(9) + 92]
= 3a(2a − 9)2
 The factorization is 3a(2a − 9)2
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Recognizing Differences of Squares
 An expression, like 25x2 − 36, that can
be written in the form A2 − B2 is called
a difference of squares.
 Note that for a binomial to be a
difference of squares, it must
have the following.
1. There must be two expressions, both
squares, such as: 9, x2, 100y2, 36y8
2. The terms in the binomial must have
different signs.
Chabot College Mathematics
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Difference of 2-Squares
 Diff of 2 Sqs → A2 − B2
 Note that in order for a term to be a
square, its coefficient must be a
perfect square and the power(s)
of the variable(s) must be even.
• For Example 25x4 − 36
–
–
25 = 52
The Power on x is even at 4 → x4 = (x2)2
– Also, in this case 36 = 62
Chabot College Mathematics
15
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Test Diff of 2Sqs
 Determine whether each of the
following is a difference of squares.
a) 16x2 − 25 b) 36 − y5
c) −x12 +
49
 SOLUTION a) 16x2 − 25
1. The 1st expression is a sq: 16x2 =
(4x)2
The 2nd expression is a sq: 25 = 52
2. The terms have different signs.
 Thus, 16x2 − 25 is a difference of
Chabot College Mathematics
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Test Diff of 2Sqs
SOLUTION b) 36 − y5
The expression y5 is not a square.
Thus, 36 − y5 is not a diff of squares
SOLUTION c) −x12 + 49
The expressions x12 and 49 are squares:
x12 = (x6)2 and 49 = 72
2. The terms have different signs.
 Thus, −x12 + 49 is a diff of sqs, 72 − (x6)2

1.


1.
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring Diff of 2 Squares
A2 − B2 = (A + B)(A − B)
 The Gray Area by
Square Subtraction
2
2
x y
 The Gray Area by
(LENGTH)(WIDTH)
x  y x  y 
Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor Diff of Sqs
 Factor: a) x2 − 9
b) y2 − 16w2
 SOLUTION
a) x2 − 9 = x2 – 32 = (x + 3)(x − 3)
A2 − B2 = (A + B)(A − B)
b) y2 − 16w2 = y2 − (4w)2 = (y + 4w)(y − 4w)
A2 − B2 =
Chabot College Mathematics
19
(A + B) (A − B)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor Diff of Sqs
 Factor: c) 25 − 36a12 d) 98x2 − 8x8
 SOLUTION
c) 25 − 36a12 = 52 − (6a6)2 = (5 + 6a6)(5 − 6a6)
d) 98x2 − 8x8
Always look for a common factor. This time
there is one, 2x2:
98x2 − 8x8 = 2x2(49 − 4x6)
= 2x2[(72 − (2x3)2]
= 2x2(7 + 2x3)(7 − 2x3)
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Grouping to Expose Diff of Sqs
 Sometimes a Clever Grouping will
reveal a Perfect-Sq TriNomial next to
another Squared Term
 Example Factor m2 − 4b4 + 14m + 49
 rearranging 
m2 + 14m + 49 − 4b4
 GROUPING 
Chabot College Mathematics
21
(m2 + 14m + 49) − 4b4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Grouping to Expose Diff of Sqs
 Example Factor m2 − 4b4 + 14m + 49
• Recognize m2 + 14m + 49 as Perfect
Square Trinomial → (m+7)2
• Also Recognize 4b4 as a Sq → (2b)2
(m2 + 14m + 49) − 4b4
 Perfect Sqs 
(m + 7)2 − (2b2)2
 In Diff-of-Sqs Formula: A→m+7; B→2b2
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Grouping to Expose Diff of Sqs
 Example Factor m2 − 4b4 + 14m + 49
(m + 7)2 − (2b2)2
 Diff-of-Sqs → (A − B)(A + B) 
([m+7] − 2b2)([m + 7] + 2b2)
 Simplify → ReArrange 
(−2b2 + m + 7)(2b2 + m + 7)
 The Check is Left for us to do Later
Chabot College Mathematics
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring Two Cubes
 The principle of patterns applies to the
sum and difference of two CUBES.
Those patterns
• SUM of Cubes

A  B   A  B  A  AB  B
3
3
2
2

2

• DIFFERENCE of Cubes

A  B   A  B  A  AB  B
3
Chabot College Mathematics
24
3
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
TwoCubes SIGN Significance
 Carefully note the Sum/Diff of
Two-Cubes Sign Pattern

A  B   A  B  A  AB  B
3
3
SAME Sign
2
3
SAME Sign
Chabot College Mathematics
25

2

OPP Sign

A  B   A  B  A  AB  B
3
2
2
OPP Sign
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example: Factor x3 + 64
x  64
 Factor
3
x  64
3
Recognize Pattern as Sum of CUBES
x   4
3
3
Determine Values that were CUBED
x A 4 B
Map Values to Formula
x  4x  x  4  4 
2
x  4x  4 x  16
2
Chabot College Mathematics
26
2
Substitute into Formula
Simplify and CleanUp
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example: Factor 8w3−27z3
 Factor 8w3  27 z 3
8w  27 z Recognize Pattern as Difference of CUBES
3
3
3
3
2 w  3 z  Determine CUBED Values
3
3
2w  3z 
3
3
2w  A 3z  B
Simplify by Properties of Exponents
Map Values to Formula
2w  3z 2w  2w3z   3z   Sub into Formula
2
2
2w  3z 4w  6wz  9 z  Simplify & CleanUp
2
Chabot College Mathematics
27
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example: Check 8w3−27z3
 Check 2w  3z 4w2  6wz  9 z 2  8w3  27 z 3
??
2w  3z 4w
2
 6wz  9 z
2

Use Distributive property
 8w3  12w2 z  18wz 2  12 zw 2  18 z 2 w  27 z 3

 

 8w  12w z  12w z  18wz  18wz  27 z
3
2
2
2
2
Use Comm & Assoc. properties, and Adding-to-Zero
 8w  0  0  27 z  8w  27 z 
3
Chabot College Mathematics
28
3
3
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
3
Sum & Difference Summary
 Difference of Two SQUARES
A  B   A  B  A  B 
2
2
 SUM of Two CUBES

2

2

A  B   A  B  A  AB  B
3
3
2
 Difference of Two CUBES

A  B   A  B  A  AB  B
3
3
Chabot College Mathematics
29
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring Completely
 Sometimes, a complete factorization
requires two or more steps. Factoring is
complete when no factor can be
factored further.
 Example: Factor 5x4 − 3125
• May have the Difference-of-2sqs TWICE
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring Completely
 SOLUTION
5x4 − 3125 = 5(x4 − 625)
= 5[(x2)2 − 252]
= 5(x2 − 25)(x2 + 25)
= 5(x − 5)(x + 5)(x2 + 25)
 The factorization:
5(x − 5)(x + 5)(x2 + 25)
Chabot College Mathematics
31
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring Tips
1. Always look first for a common
factor. If there is one, factor it out.
2. Be alert for perfect-square
trinomials and for binomials that
are differences of squares.
 Once recognized, they can be
factored without trial and error.
3. Always factor completely.
4. Check by multiplying.
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
WhiteBoard Work
 Problems From §5.5 Exercise Set
• 14, 22, 48, 74, 94, 110
 The SUM (Σ) &
DIFFERENCE (Δ)
of Two Cubes
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
All Done for Today
Sum of
Two
Cubes
Chabot College Mathematics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
36
5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
4
5
6
y
5
4
3
2
1
x
0
-3
-2
-1
0
1
2
3
4
5
-1
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt