Chabot Mathematics
§5.6 Factoring
Strategies
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Review § 5.5
MTH 55
 Any QUESTIONS About
• §5.5 → Factoring: TriNomials, Special
Forms
 Any QUESTIONS About HomeWork
• §5.5 → HW-19
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
To Factor a Polynomial
A. Always look for a common factor first.
If there is one, factor out the Greatest
Common Factor (GCF). Be sure to
include it in your final answer.
B. Then look at the number of terms
•
TWO Terms: If you have a Difference of
SQUARES, factor accordingly:
A2 − B2 = (A − B)(A + B)
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
To Factor a Polynomial
• TWO Terms: If you have a SUM of
CUBES, factor accordingly:
A3 + B3 = (A + B)(A2 − AB + B2)
• TWO Terms: If you have a DIFFERENCE
of CUBES, factor accordingly:
A3 − B3 = (A − B)(A2 + AB + B2)
• THREE Terms: If the trinomial is a
perfect-square trinomial, factor accordingly:
A2 + 2AB + B2 = (A + B)2
or
A2 – 2AB + B2 = (A – B)2.
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
To Factor a Polynomial
•
THREE Terms: If it is not a perfectsquare trinomial, try using FOIL Guessing
•
FOUR Terms: Try factoring by
grouping
C. Always factor completely: When a
factor can itself be factored, be sure to
factor it. Remember that some
polynomials, like A2 + B2, are PRIME
D. CHECK by Multiplying the Factors
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Choosing the Right Method
 Example: Factor
5t4 − 3125
 SOLUTION
A. Look for a common factor:
5t4 − 3125 = 5(t4 − 625).
B. The factor t4 − 625 is a diff of squares:
(t2)2 − 252. We factor it, being careful
to rewrite the 5 from step (A):
5(t4 − 625) = 5(t2 − 25)(t2 + 25)
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Example  Factor 5t4 − 3125
C. Since t2 − 25 is not prime, we continue
factoring:
5(t2 − 25)(t2 + 25) =
5(t − 5)(t + 5)(t2 + 25)
SUM of squares with
no common factor.
It canNOT be factored!
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Example  Factor 5t4 − 3125
D. Check:
5(t − 5)(t + 5)(t2 + 25)
= 5(t2 − 25)(t2 + 25)
= 5(t4 − 625)
= 5t4 − 3125

The factorization is VERIFIED as
5(t − 5)(t + 5)(t2 + 25)
Chabot College Mathematics
8
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  2x3 + 14x2 + 3x + 21
 SOLUTION
A. We look for a common factor.
There is none.
B. Because there are four terms,
try factoring by grouping:
2x3 + 14x2 + 3x + 21
= (2x3 + 14x2) + (3x + 21)
= 2x2 (x + 7) + 3 (x + 7)
= (x + 7)(2x2 + 3)
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  2x3 + 14x2 + 3x + 21
C. Nothing can be factored further, so we
have factored completely.
D. Check by Forward FOIL:
(x + 7)(2x2 + 3) = 2x3 + 3x + 14x2 + 21
= 2x3 + 14x2 + 3x + 21

Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  −x5 − 2x4 + 24x3
 SOLUTION
A. We note that there is a
common factor, −x3:
−x5 − 2x4 + 24x3 = −x3(x2 + 2x − 24)
B. The factor x2 + 2x − 24 is not a
perfect-square trinomial. We factor it
by FOIL trial and error:
−x5 − 2x4 + 34x3 = −x3(x2 + 2x − 24)
= −x3(x − 4)(x + 6)
Chabot College Mathematics
11
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  −x5 − 2x4 + 24x3
C. Nothing can be factored further, so we
have factored completely
D. Check: −x3(x − 4)(x + 6)
= −x3(x2 + 2x − 24)
= −x5 − 2x4 + 24x3

Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  x2 − 18x + 81
 SOLUTION
A. Look for a common factor.
There is none.
B. This polynomial is a perfect-square
trinomial. Factor accordingly:
x2 − 18x + 81 = x2 − 2  9  x + 92
= (x − 9)(x − 9)
= (x − 9)2
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  x2 − 18x + 81
C. Nothing can be factored further, so we
have factored completely.
D. Check:
(x − 9)(x − 9) = x2 − 18x + 81.

Chabot College Mathematics
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  12x2y3 + 20x3y4 + 4x2y5
 SOLUTION
A. We first factor out the largest common
factor, 4x2y3:
4x2y3(3 + 5xy + y2)
B. The constant term in 3 + 5xy + y2 is not
a square, so we do not have a perfectsquare trinomial. It cannot be factored
using grouping or trial and error. The
Trinomial term cannot be factored.
Chabot College Mathematics
15
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  12x2y3 + 20x3y4 + 4x2y5
C. Nothing can be factored further,
so we have factored completely
D. Check:
4x2y3(3 + 5xy + y2)
= 12x2y3 + 20x3y4 + 4x2y5
Chabot College Mathematics
16

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  ab + ac + wb + wc
 SOLUTION
A. We look for a common factor.
There is none.
B. There are four terms. We try factoring
by grouping:
ab + ac + wb + wc = a(b + c) + w(b + c)
= (b + c)(a + w)
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  ab + ac + wb + wc
C. Nothing can be factored further, so we
have factored completely.
D. Check by FOIL Multiplication:
(b + c)(a + w) = ba + bw + ca + cw
= ab + ac + wb + wc

Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  36x2 + 36xy + 9y2
 SOLUTION
A. Look for common factor. The GCF is 9,
but Let’s hold off for now
B. There are three terms. Note that the
first term and the last term are squares:
36x2 = (6x)2 and 9y2 = (3y)2.
 We see that twice the product of 6x and
3y is the middle term, 2  6x  3y = 36xy,
so the trinomial is a perfect square.
Chabot College Mathematics
19
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  36x2 + 36xy + 9y2
B. To Factor the Trinomial Square, we
write the binomial squared:
36x2 + 36xy + 9y2 = (6x + 3y)2
= (6x+3y)(6x+3y) = 3(2x + y)3(2x + y)
= 3∙3(2x + y)(2x + y) = 9(2x + y)2
C. Cannot Factor Further.
D. Check: 9(2x + y)2 = 9(2x + y)(2x + y)
= 9(4x2 + 2xy + 2yx + y2)
= 36x2 + 36xy + 9y2 
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  a8 − 16b4
 SOLUTION
A. We look for a common factor.
There is none.
B. There are two terms.
Since a8 = (a4)2 and 16b4 = (4b2)2, we
see that we have a difference of
squares  (a4)2 − (4b2)2
Thus, a8 − 16b4 = (a4 + 4b2)(a4 − 4b2)
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Factor  a8 − 16b4
C. The factor (a4 − 4b2) is itself a
difference of squares. Thus,
(a4 − 4b2) = (a2 − 2b)(a2 + 2b)
D. Check:
(a4 + 4b2)(a2 − 2b)(a2 + 2b)
= (a4 + 4b2)(a4 − 4b2)
= a8 − 16b4 
Chabot College Mathematics
22
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Example  Factor: 4x2 – 14x + 12
 SOLUTION
 Look for a common factor  Find “2”:
4x2 – 14x + 12 = 2(2x2 – 7x + 6)
 The other factor has three terms. The
trinomial is not a square. Try to
FOIL factor using trial and error
4x2 – 14x + 12 = 2(2x – 3)(x – 2)
 Cannot Factor Further; Check Later
Chabot College Mathematics
23
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
Example  18y9 – 27y8
 SOLUTION
 Look for a common factor – Find 9y8
18y9 – 27y8 = 9y8(2y – 3)
 The other factor has two terms but is
not a difference of squares and not a
sum or difference of cubes
 No factor with more than one term
can be factored further
 Check: 9y8(2y − 3) = 18y9 − 27y8 
Chabot College Mathematics
24
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
WhiteBoard Work
 Problems From §5.6 Exercise Set
• 28, 36, 62, 68, 78, 82, 86

Find &
Factor
the
Trinomial
Squares
Chabot College Mathematics
25
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt
All Done for Today
Factoring
difference
of
2 Squares
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.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
27
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.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
28
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-25_sec_5-6_Factoring_Strategy.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
29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-25_sec_5-6_Factoring_Strategy.ppt