1. No category

7-2
Factoring by GCF
Going Deeper
Essential question: How can you factor polynomials completely by grouping?
2
Standards for
Mathematical Content
Questioning Strategies
• Are (x + 5) and (5 + x) binomial opposites?
Explain. No; (x + 5) and (5 + x) are equal by
A-SSE.1.1 Interpret expressions that represent a
quantity in terms of its context.*
A-SSE.1.1b Interpret complicated expressions
by viewing one or more of their parts as a single
entity.*
A-SSE.1.2 Use the structure of an expression to
identify ways to rewrite it.
the commutative property. To be opposites, the
binomials would have to be (x - 5) and
(5 - x); that is, they would have to have the
same variables and numbers in the reverse order
with a subtraction sign between them.
EXTRA EXAMPLE
Use grouping and binomial opposites to factor
2x3 + 6x2 - 3 - x. (2x2 - 1)(x + 3)
Math Background
Factoring by grouping is a factoring method that
can be used with some polynomials that have four
terms. This method can be used when the terms of
the polynomial can be grouped so that each group
has the same greatest common factor (GCF).
CLOS E
Essential Question
How can you factor polynomials completely by
grouping?
IN T RO DUC E
Summarize
Have students write a journal entry in which they
describe how they factored one of the polynomials
in the Practice exercises. They should include how
analyzing the terms indicated that grouping was an
appropriate method to use and a justification for
each step that they took in their work.
T EACH
EXAMPLE
Questioning Strategies
• A student factored 2x4 + 6x3 - 8x2 - 24 as
2x(x2 - 4)(x + 3). Is this correct? Explain.
No; (x2 - 4) can be factored further as
(x - 2)(x + 2). The correct factorization is
2x(x - 2)(x + 2)(x + 3).
PR ACTICE
Where skills are
taught
EXTRA EXAMPLE
Use grouping to factor -3x4 - 6x3 + 3x2 + 6x.
-3x(x + 2)(x - 1)(x + 1)
Chapter 7
377
Where skills are
practiced
1 EXAMPLE
EXS. 1–2
2 EXAMPLE
EXS. 3–4
Lesson 2
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First, determine whether all the terms have a
common GCF and, if so, factor that out. Then,
group the terms that have a common factor and
factor out the GCF of each group. Use that GCF
as a common factor to rewrite the polynomial in
factored form. As a final step, check that none of
the binomial factors can be factored further.
Review how to use the GCF to factor an expression
with a common binomial factor, such as
3(x + 1) - 2x(x + 1). Once students recognize that
the common factor in both terms is the binomial
(x + 1), remind them that a binomial can be
factored out of a term just as a monomial can. Tell
students that they can factor some polynomials
with four terms by grouping the terms in pairs and
factoring out common monomials and binomials.
1
EXAMPLE
Name
Class
Notes
7-2
Date
Factoring by GCF
Going Deeper
Essential question: How can you factor polynomials completely by grouping?
When a polynomial has four terms, you may be able to make two groups of terms
and factor out the GCF from each group. Before doing this, however, you should
check if there is a GCF for all the terms that can be factored out first.
A-SSE.1.2
1
EXAMPLE
Factoring out a GCF and Grouping
Factor -2x4 - 2x3 - 6x2 - 6x.
-2x ( x3 + x2 + 3x + 3
)
Factor out -2x , the GCF of all the
terms.
( -2x )[ ( x3 + x2 ) + ( 3x + 3 ) ]
Group terms that have a common
factor. The common factor of x3
and x2 is
x2
of 3x and 3 is
( -2x ) [
( x + 1 )+
2
x
3
( x + 1 )]
( -2x )( x + 1 ) ( x2 + 3 )
. The common factor
3
.
Factor out the GCF of each group.
Factor out ( x + 1 ) , the common
factor of the products inside the
brackets.
© Houghton Mifflin Harcourt Publishing Company
REFLECT
1a. Why did the operation signs change when -2x was factored out of each term?
Subtracting terms with positive coefficients is the same as adding terms
with negative coefficients. The polynomial could be written as
-2x4 + (-2x3) + (-6x2) + (-6x). When -2x is factored out, the coefficients
become positive again, but the operation signs remain as +.
1b. Does performing grouping before factoring out -2x change the factorization?
Explain.
No; if grouping is done first, the steps would be:
(-2x4 - 2x3) + (-6x2 - 6x)
-2x3(x + 1) + (-6x)(x + 1)
-2x3(x + 1) - 6x(x + 1)
(-2x3 - 6x)(x + 1)
-2x(x2 + 3)(x + 1)
377
Chapter 7
Lesson 2
Binomials have opposites. For example, (3 - x) and -(3 - x) are opposites. The expression
-(3 - x) can be rewritten as (-3 + x) or (x - 3). So, (3 - x) and (x - 3) are opposites.
2
A-SSE.1.1b
EXAMPLE
Factoring with Binomial Opposites
( 4x3 - 8x2 ) + ( 6 - 3x )
4x2
( x - 2 )+
3
( 2-x )
Group terms in the order in which
they appear.
The GCF of the terms in the first group is
4x2 . The GCF of the terms in the second
3 . Factor out the GCF of each group.
group is
4x2
( x - 2 )+
3
( x - 2 )-
3
( -1 )( x - 2 )
The polynomial contains the binomial
( x-2 )
Simplify.
opposites ( x - 2 ) and ( 2 - x ).
Write (2 - x) as (-1)(x - 2 ).
4x2
( x - 2 )( 4x2 - 3 )
Factor out ( x - 2 ) , the common factor
of the products.
REFLECT
2a. How could you have rearranged the terms of the polynomial so that you would not
have encountered binomial opposites when factoring? Show how you would have
factored the polynomial if you had first rearranged the terms.
Rearrange the terms so that subtracting 3x comes before adding 6.
4x3 - 8x2 - 3x + 6
© Houghton Mifflin Harcourt Publishing Company
© Houghton Mifflin Harcourt Publishing Company
Factor 4x3 - 8x2 + 6 - 3x.
(4x3 - 8x2) - (3x - 6)
4x2(x - 2) - 3(x - 2)
(x - 2)(4x2 - 3)
PRACTICE
Factor each polynomial by grouping.
1. -3a3 - 9a2 - 6a - 18
2. -2c3 + 4c2 - 8c + 16
-3(a2 + 2)(a + 3)
3
2
3. 5s - 5s + 2 - 2s
-2(c2 + 4)(c - 2)
4. 4z3 - 16z2 + 40 - 10z
(5s2 - 2)(s - 1)
Chapter 7
Chapter 7
2(2z2 - 5)(z - 4)
378
Lesson 2
378
Lesson 2
ADD I T I O N A L P R AC T I C E
AND PRO BL E M S O LV I N G
Assign these pages to help your students practice
and apply important lesson concepts. For
additional exercises, see the Student Edition.
Answers
Additional Practice
1. c(8c + 7)
2. 3n2(n + 4)
3. 3x(5x4 - 6)
4. 4(−2s4 + 5t3 - 7)
5. 6n(n5 + 3n3 − 4)
2
6. 5m2(−m - m + 1)
7. 16t(-t + 2)
8. 3x and 4x + 1
9. (m + 5)(3m + 4)
10. (b − 3)(16b + 1)
11. (x + 4)(2x2 + 3)
12. (4n + 3)(n2 + 1)
13. (5d − 3)(2d + 7)
14. (4n − 5)(3n2 − 2)
15. (b − 3)(5b3 − 1)
16. (t 2 − 2)(t − 5)
Problem Solving
© Houghton Mifflin Harcourt Publishing Company
1. 4x ft; (x + 1) ft
2. −3(x2 + 9x − 275)
3. 4(3x + 7); 31 feet
4. (5x + 4) m; (x2 − 2) m
5. C
6. G
7. C
8. F
Chapter 7
379
Lesson 2
Name
Class
Date
Notes
7-2
© Houghton Mifflin Harcourt Publishing Company
Additional Practice
379
Chapter 7
Lesson 2
© Houghton Mifflin Harcourt Publishing Company
© Houghton Mifflin Harcourt Publishing Company
Problem Solving
Chapter 7
Chapter 7
380
Lesson 2
380
Lesson 2