Chabot Mathematics
§5.3 GCF
Grouping
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 § 5.2
MTH 55
 Any QUESTIONS About
• §5.2 → PolyNomial Multiplication
 Any QUESTIONS About HomeWork
• §5.2 → HW-17
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
PolyNomial Factoring Defined
 To factor a polynomial is to find
an equivalent expression that is
a product. An equivalent
expression of this type is called a
factorization of the polynomial
• Factoring Breaks an algebraic
expression into its simplest pieces
–“Simplest”  Smallest Powers
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factoring Monomials
 Find three factorizations of 24x3.
 SOLUTION
a) 24x3 = (6  4)(x  x2)
= 6x  4x2
b) 24x3 = (6  4)(x2  x)
= 6x2  4x
c) 24x3 = ((−6)(−4))x3
= (−6)(−4x3)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Greatest Common Factor (GCF)
 Find the prime factorization of 105 & 60
• Use Factor-Tree
60
105
5
2
 21
3

7
 30
 15
2
3
Chabot College Mathematics
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
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  GCF
 Thus
60  2  2  3  5
105  3  5  7
 Recognize the Factors that both
numbers have in COMMON
 The GREATEST Common Factor is the
PRODUCT of all the COMMON Factors
 In This Case the GCF:
60  15  4
GCF  3 5  15
Chabot College Mathematics
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105  15  7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Examples  GCF
 Find the GCF for Monomials:
14p4q and 35pq3
 The Prime Factorizations
• 14p4q = 2  7  p  p  p  p  q
• 35pq3 = 5  7  p  q  q  q
 Thus the GCF = 7  p  q = 7pq
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Examples  GCF
 Find the GCF for Three Monomials:
15x2
30xy2
57x3y
 The Prime Factorizations
• 15x2 = 3  5  x  x
• 30xy2 = 2  3  5  x  y  y
• 57x3y = 3  19  x  x  x  y
ID the
Common
Factors
 Thus the GCF = 3  x = 3x
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring When Terms Have a
Common Factor
 To factor a polynomial with two or more terms
of the form ab + ac, we use the distributive law
with the sides of the equation switched:
ab + ac = a(b + c).
 Multiply
Factor
 4x(x2 + 3x − 4)
4x3 + 12x2 − 16x
 = 4xx2 + 4x3x − 4x4
 = 4x3 + 12x2 − 16x
Chabot College Mathematics
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= 4xx2 + 4x3x − 4x4
= 4x(x2 + 3x − 4)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor by Distributive
 Factor: 9a − 21
 SOLUTION
 The prime factorization of 9a is 33a
 The prime factorization of 21 is 37
 The largest common factor is 3.
 9a − 21 = 33a − 37 (UNdist the 3)
= 3(3a − 7)

 Chk: 3(3a − 7) = 3  3a − 3  7 = 9a − 21
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Factor by Distributive
 Factor: 28x6 + 32x3.
 SOLUTION
 The prime factorization of 28x6 is

227xxxxxx
 The prime factorization of 32x3 is

22222xxx
 The largest common factor is
2  2  x  x  x or 4x3.
 28x6 + 32x3 = (4x3  7x ) + (4x3  8)
= 4x3(7x3 + 8)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factor 12x5 − 21x4 + 24x3
 The prime factorization of 12x5 is
223xxxxx=3xxx22xx
 The prime factorization of 21x4 is
37xxxx =3xxx7x
 The prime factorization of 24x3 is
2223xxx =3xxx222
 The largest common factor is 3  x  x  x or 3x3.
 12x5 – 21x4 + 24x3 = 3x3  4x2 – 3x3  7x + 3x3  8
= 3x3(4x2 – 7x + 8)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Distributive factoring
 Factor: 9a3b4 + 18a2b3
 SOLUTION
9a 3 b 4  3  3  a 2  a  b3  b
 The Prime
Factorizations: 18a 2 b3  2  3  3  a 2  b3
 The Greatest Common Factor is 9a2b3
 Distributing OUT the GCF Produces
the factorization:
9a3b4 + 18a2b3 = 9a2b3(ab + 2)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Distributive factoring
 Factor: −4xy + 8xw − 12x
 SOLUTION
 The Expanded Factorizations
• −4xy = −4x  y
• +8xw = − 2  −4x  w
• − 12x = 3  −4x
 Thus the Factored expression:
−4xy + 8xw − 12x = −4x(y − 2w + 3)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring Out a Negative GCF
 When the coefficient of the term of greatest
degree is negative, it is sometimes preferable
to factor out the −1 that is understood along
with the GCF
• e.g. Factor Out the GCF for
Factor out
only the 3 .
Or factor
out the –3
Chabot College Mathematics
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 
 
 3w  3w  5
Both are
Correct
 3w4  9w3  15  3 w4   3  3w3   35
3
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
PolyNomial Factoring Tips
 Factor out the Greatest Common Factor
(GCF), if one exists.
 The GCF multiplies a polynomial with
the same number of terms as the
original polynomial.
 Factoring can always be checked
by multiplying.
• Multiplication should yield the
original polynomial.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factoring by GROUPING
 Sometimes algebraic expressions
contain a common factor with
two or more terms.
 Example: Factor x2(x + 2) + 3(x + 2)
 SOLUTION: The binomial (x + 2) is a
factor of BOTH x2(x + 2) & 3(x + 2).
 Thus, (x + 2) is a common factor; so
x2(x + 2) + 3(x + 2) = (x + 2)x2 + (x + 2)3
= (x + 2)(x2 + 3)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Grouping Game Plan
 If a polynomial can be split into
groups of terms and the groups
share a common factor, then the
original polynomial can be factored.
 This method, known as
factoring by grouping, can be
tried on any polynomial with
four or more terms
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Examples  Grouping
 Factor by grouping.
a) 3x3 + 9x2 + x + 3
b) 9x4 + 6x − 27x3 − 18
 Solution
a) 3x3 + 9x2 + x + 3 = (3x3 + 9x2) + (x + 3)
= 3x2(x + 3) + 1(x + 3)
= (x + 3)(3x2 + 1)
Don’t Forget the “1”
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Examples  Grouping
 Factor by grouping.
a) 3x3 + 9x2 + x + 3
b) 9x4 + 6x − 27x3 − 18
 Solution
b) 9x4 + 6x − 27x3 − 18
= (9x4 + 6x) + (−27x3 − 18)
= 3x(3x3 + 2) + (−9)(3x3 + 2)
= (3x3 + 2)(3x − 9)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Example  Grouping
 Factor: y5 + 5y3 + 3y2 + 15
 SOLUTION
y5 + 5y3 + 3y2 + 15
= (y5 + 5y 3) + (3y 2 + 15)
=
y 3 (y 2
+ 5) +
3(y 2
= (y 2 + 5) (y 3 + 3)
Chabot College Mathematics
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+ 5)
Grouping
Factoring each
binomial
Factoring out the
common factor
(a BiNomial)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factor 4ab + 2ac + 8xb + 4xc
 Try grouping terms which have
something in common. Often, this can
be done in more than one way.
 For example
4 ab  2ac  8 xb  4 xc
Grp-1
or
Grp-2
( 4 ab  2ac )  ( 8 xb  4 xc ) ( 4 ab  8 xb )  ( 2ac  4 xc )
a’s & x’s Grouping
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b’s & c’s Grouping
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factor 4ab + 2ac + 8xb + 4xc
 Next, find the greatest common factor
for the polynomial in each set of
parentheses.
Grouping Set-1
Grouping Set-2
( 4 ab  2ac )  ( 8 xb  4 xc )
( 4 ab  8 xb )  ( 2ac  4 xc )
 The GCF for
(4ab + 2ac) is 2a
 The GCF for
(4ab + 8xb) is 4b
 The GCF for
(8xb + 4xc) is 4x
 The GCF for
(2ac + 4xc) is 2c
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factor 4ab + 2ac + 8xb + 4xc
 Write each of the polynomials in
parentheses as the product of the GCF
and the remaining polynomial
( 4 ab  2ac )  ( 8 xb  4 xc )
( 4 ab  8 xb )  ( 2ac  4 xc )
 2a ( 2b  c )  4 x ( 2b  c )  4 b ( a  2x )  2c ( a  2x )
 Apply the distributive property to any
common factors
( 2a  4 x )( 2b  c )
Chabot College Mathematics
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( 4 b  2c )( a  2x )
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factor 4ab + 2ac + 8xb + 4xc
 Examine the Factorizations
( 4 ab  2ac )  ( 8 xb  4 xc )
2( a  2x )( 2b  c )
( 4 ab  8 xb )  ( 2ac  4 xc )
2( 2b  c )( a  2x )
 Notice that it did not matter how the
terms were originally grouped, the
factored forms of the polynomials are
IDENTICAL
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
WhiteBoard Work
 Problems From §5.3 Exercise Set
• 22, 32, 52, 56, 68, 84

Factor by
Grouping
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
All Done for Today
Factoring
4-Term
Polynomials
Chabot College Mathematics
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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
28
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
 
 
 3w  3w  5
 3w4  9w3  15  3 w4   3  3w3   3 5
3
Chabot College Mathematics
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3
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
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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
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-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
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-1
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt
Factor 4ab + 2ac + 8xb + 4xc
 Divide each polynomial in parentheses
by the GCF
( 4 ab  2ac )  ( 8 xb  4 xc )
( 4 ab  8 xb )  ( 2ac  4 xc )
4 ab  2ac 4 ab 2ac


2a
2a
2a
4 ab  8 xb 4 ab 8 xb


4b
4b
4b
 a  2x
 2b  c
8 xb  4 xc 8 xb 4 xc


4x
4x
4x
 2b  c
Chabot College Mathematics
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2ac  4 xc 2ac 4 xc


2c
2c
2c
 a  2x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt