# 7.6 Workbook Answers - Tequesta Trace Middle

```7-6
Choosing a Factoring Model
Extension: Factoring Polynomials with
More Than One Variable
Essential question: How can you factor polynomials with more than one
variable?
• What is the connection between x2 - 81 and
m2 - n2? They are both the difference of squares.
Standards for
Mathematical Content
2
2
Factoring by GCF
Factoring ax2 + bx + c
Factoring special products
• In the perfect square trinomial 4x 2 + 12xy + 9y 2,
what represents a and what represents b?
Math Background
Students have factored polynomials using a variety
of methods: by grouping, by using special product
patterns, by using the factors of the first and third
terms. All these polynomials involved a single
variable. Now students will factor polynomials that
involve more than one variable. They will discover
that the methods they learned for single variable
polynomials apply equally well to multi-variable
polynomials.
2x represents a; 3y represents b.
EXTRA EXAMPLE
Factor the polynomial completely. Explain each
step.
-3x2 + 36xy - 108y2
2
-3(x - 12xy + 36y2) Factor the GCF.
-3(x - 6y)(x - 6y) Perfect square trinomial
Review the methods students have learned to factor
polynomials that involve a single variable. Give
students an example of each type of polynomial
and have them do the factoring. Then give them
polynomials with two variables that can be factored
using the special product patterns. Lead them to
see that the patterns still apply. Ask what they think
this means about methods for factoring multivariable polynomials in general.
Avoid Common Errors
Students may confuse the perfect square trinomial
rule with the difference of squares rule and use
a2 - b2 = (a - b)(a - b). To help students
remember that a2 - b2 = (a - b)(a + b), have them
remember the statement “Two terms = Two signs”.
T EACH
EXPLORE
Questioning Strategies
• If the second polynomial were 3m2 - 3, how
would you factor it? Use the GCF and the
difference of squares: 3(m - 1)(m + 1).
397
Lesson 6
© Houghton Mifflin Harcourt Publishing Company
Teaching Strategies
Have groups of students develop a step-by-step
plan to factor polynomials. Ask them to make a list
of questions they can ask themselves as they work.
Such questions might include Does this polynomial
follow the pattern of Difference of Squares? Does
it follow the pattern of Perfect Squares? Is there a
GCF? What are some factors of the third coefficient?
What is the sum of these factors? Allow students to
refer to this plan, as necessary, when they factor
polynomials.
IN T RO DUC E
Chapter 7
EXAMPLE
Questioning Strategies
• How do you know which special product pattern
to use when factoring a perfect square trinomial?
Look at the first operation sign. If it is +, factor
as (a + b)(a + b). If it is -, factor as
(a - b)(a - b).
Prerequisites
1
2
• How would you factor x2y - r2s ? Use the
difference of squares rule: (xy + rs)(xy - rs)
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.
Name
Class
Notes
7-6
Date
Choosing a Factoring Method
Extension: Factoring Polynomials with More
Than One Variable
Essential question: How can you factor polynomials with more than one variable?
A-SSE.1.2
1
EXPLORE
Factoring Polynomials with Two Variables
Factor each polynomial completely. Explain each step.
2x2 - 162
2(x2 - 81)
Factor the GCF.
x(x + 9)(x - 9)
Difference of squares
3m2 - 3n2
2
2
3( m - n )
Factor the GCF.
3(m + n)(m - n)
Difference of squares
REFLECT
1a. Compare the methods of factoring the two polynomials.
The methods are the same, except the second polynomial has two variables.
1b. How would the factoring change for the polynomial 3m2 - 12n2?
After factoring the GCF, you would factor m2 - 4n2, which is still
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a difference of squares.
To factor a perfect square trinomial with more than variable you can use the same
patterns you used with perfect square trinomials in one variable.
a2 + 2ab + b2 = (a + b)(a + b)
a2 - 2ab + b2 = (a - b)(a - b)
A-SSE.1.2
2
EXAMPLE
Factoring a Perfect Square Trinomial
Factor the trinomial completely. Explain each step.
3g2 + 12gh + 12h2
3
(g2 + 4gh + 4h2)
Factor the GCF.
3(g + 2h)(g + 2h)
Perfect square trinomial pattern
397
Chapter 7
Lesson 6
REFLECT
2a. Which pattern would you use to factor 16a2 - 48ab + 36b2? Explain how you know
and then factor the trinomial.
If there is no obvious pattern shown by the trinomial, you can find the factors of the
coefficient of the third term and check their sums to find how to factor.
A-SSE.1.2
3
EXAMPLE
Factoring a Polynomial
Factor 3x2 + 21xy + 36y2 completely. Explain your steps.
2
2
3 (x + 7xy + 12y )
Factor the GCF.
Find the factors of 12 that add to 7.
Factors of 12
Sum of factors
1 and 12
13
2
and
6
8
3
and
4
7
3(x + 3y)(x + 4y)
Factor according to the sum.
REFLECT
3a. Would finding the factors and their sum be enough to factor the polynomial
2x 2 + 11xy + 12y 2? Why or why not?
No, because the x-term of the binomial has a coefficient of 2. This would affect the
middle term of the polynomial, so you can’t just find the factors and the sum.
3b. What additional conditions must you consider to factor 2x 2 + 11xy + 12y 2?
You have to consider that one factor of 12 has been doubled. You will have to find
© Houghton Mifflin Harcourt Publishing Company
© Houghton Mifflin Harcourt Publishing Company
a2 - 2ab + b2 = (a - b)(a - b); the first operation sign is -; 4(2a - 3b)(2a - 3b)
which factor by trying to divide each factor by 2, then finding the sum.
Sometimes grouping will allow you to factor a polynomial.
Chapter 7
Chapter 7
398
Lesson 6
398
Lesson 6
3
EXTRA EXAMPLE
Factor x3 + 2x2y + x + 2y completely. Explain your
steps.
EXAMPLE
Questioning Strategies
• What tells you that this polynomial is not a
“perfect square” polynomial? 36 is a perfect
• Does not have a common GCF.
• Group terms with a common factor:
(x3 + 2x2y) + (x + 2y)
square, but the coefficient of the first term, 3, is
not.
• Factor out the GCF of each group:
x2(x + 2y) + 1(x + 2y)
• Describe the relationship between the coefficient
of the second term and the coefficient of the third
term after the GCF has been factored out. The
• Factor out the GCF of the products:
(x2 + 1)(x + 2y)
coefficient of the second term is the sum of the
two factors of the coefficient of the third term.
The coefficient of the third term is the product of
these two factors.
CLOS E
Essential Question
How can you factor polynomials with more than one
variable?
EXTRA EXAMPLE
Factor m2 - 8md + 15d2 completely. Explain your
steps.
You can use the same methods you used for
polynomials of one variable: difference of squares
or perfect square trinomial patterns, sum of the
factors method, or factoring by grouping.
• Does not have a common GCF.
• Does not follow any special pattern.
• Multiply the coefficients of the first and third
terms: 1 • 15 = 15. The product is positive, so the
factors are either both positive or both negative.
Summarize
Have students write a journal entry describing
the aspect of factoring they find most difficult.
They should also note any reminders or hints they
developed to help overcome this difficulty.
• The middle term is negative, so both factors will
be negative.
• Find the negative factors of 15 and their sums.
Factors
Sum
-1 and -15
-3 and -5
-16
-8
PR ACTICE
• Factor according to the sum: (m - 3d)(m - 5d)
4
EXAMPLE
Questioning Strategies
• Suppose the polynomial were 27x3 - 27xy +
6y - 6x2. What would be the first step of the
factoring? Factor out the common GCF of 3.
Where skills are
practiced
1 EXPLORE
EXS. 3, 6, 8
2 EXAMPLE
EXS. 1, 11, 15
3 EXAMPLE
EXS. 2, 4, 5, 7, 9, 10,
12, 16
4 EXAMPLE
EXS. 13, 14
• Do you have to factor 6x2 + 12xy + 3x2 + 4y2 by
grouping? Explain. No; you can combine the like
terms 6x2 and 3x2 to get 9x2 + 12xy + 4y2 which
is a perfect square trinomial.
Chapter 7
399
Lesson 6
© Houghton Mifflin Harcourt Publishing Company
Where skills are
taught
• The sum of -3 and -5 is -8, the coefficient of
the middle term.
Notes
A-SSE.1.1b
4
EXAMPLE
Factoring a Polynomial Using Grouping
Factor 9x3 - 9xy + 2y - 2x2 completely. Explain your steps.
There is no GCF for all the terms. Since the polynomial has four terms, factor by grouping.
(9x3 - 9xy ) + (
2y - 2x2
)
Group terms that have a common
factor. The common factor of 9x3y
and -9xy is 9x . The common
factor of 2y and -2x2 is
9x
(x
2
-y
)+(
2
2
9x ( x - y ) + ( 2
)(
2
y-x
) ( -1 ) (
)
2 .
Factor out the GCF of each group.
x2 - y
)
The polynomial contains the binomial
opposites (x2 - y) and
(
( y - x2 ).
)
Write (y - x2) as (-1) x2 - y .
2
9x ( x - y ) - ( 2
(
9x - 2
)(
x2 - y
)(
x2 - y
)
Simplify.
)
Factor out
(
x2 - y ), the common
factor of the products.
REFLECT
4a. Describe the steps you would use in factoring 8x2 - 2x + 24xy - 6y.
First I would factor out 2, the GCF of all the terms. I would then follow the steps in
the Example to complete the factoring, although in this case there are no opposite
© Houghton Mifflin Harcourt Publishing Company
binomials.
PRACTICE
Choose a factoring method to factor each polynomial completely.
Explain each step.
1. x2 + 6xy + 9x2
(x + 3y)(x + 3y)
Perfect square trinomial
2. 4x2 - 4xy - 8y 2
4(x 2 - xy - 2y 2)
Factor GCF.
4(x + y)(x - 2y)
Factor according to sum.
3. x2 - 4y 2
(x + 2y)(x - 2y)
Difference of squares
4. g 2 + 3gh - 10h 2
(g - 2h)(g + 5h)
Factor according to sum.
399
Chapter 7
Lesson 6
Factor each polynomial completely.
5. 2m2 + 5mn - 3n2
6. 4x 2 - 9y 2
(2x + 3y)(2x - 3y)
7. g 2 - 7gh + 10h 2
8. 16b2 - 49c2
(g - 5h)(g - 2h)
(4b + 7c)(4b - 7c)
9. a3 - 3a2b - 4ab2
10. 6a2 + 3ab - 18b2
a(a 2 - 3ab - 4b 2)
3(2a 2 + ab - 6b 2)
a(a + b)(a - 4b)
3(a + 2b)(2a - 3b)
11. 2t 3 + 12t 2w + 18tw2
12. 6c 3 - 27c 2d + 12cd 2
2t (t 2 + 6tw + 9w 2)
3c (2c 2 - 9cd + 4d 2)
2t (t + 3w)(t + 3w)
3c (c - 4d )(2c - d )
13. x3y - 5x2y + 4x - 20
14. x4y2 + 4x2y2 - 7x2 - 28
x2y(x - 5) + 4(x - 5)
x2y2(x2 + 4) - 7(x2 + 4)
(x2y + 4)(x - 5)
(x2y2 - 7)(x2 + 4)
4
4
15. Factor x - y completely. (Hint: What special form does this polynomial
appear to follow?)
(x 2 + y 2)(x 2 - y 2)
(x 2 + y 2)(x + y)(x - y)
16. Jaime and Sam both factored the polynomial 2x 2 + 10xy + 8y 2. Which student is
correct? Explain.
Jaime
Sam
2x2 + 10xy + 8y2
2
2
2(x + 5xy + 4y )
2x2 + 10xy + 8y2
(2x + 2y)(x + 4y)
2(x + y)(x + 4y)
Jaime is correct. Sam did not completely factor the polynomial. The term (2x + 2y)
© Houghton Mifflin Harcourt Publishing Company
© Houghton Mifflin Harcourt Publishing Company
(m + 3n)(2m - n)
can be further factored to 2(x + y).
Chapter 7
Chapter 7
400
Lesson 6
400
Lesson 6
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
and apply important lesson concepts. For
additional exercises, see the Student Edition.
1. yes
2. no; 5m(m + 9)
3. no; 2p(p2 + 3)(p2 − 3)
2
4. yes
5. no; 3jk3(5k + 19)
6. no; 14(7g4 − 2g + 5)
7. 8y(3xy + 5)
8. 5r(r2 − 2s)
9. x2y(3x + y)
10. −3b(a − 2)2
11. (5t + 3)(t − 3s)(t + 3s)
12. 2(y + 4x)(y − 7x)
13. 3a(2a + 3)(a + 5)
14. xy(x − 3y)(x + 3y)
15. 12(n3 − 4)
16. 3c2(c + 4d)2
17. not factorable
2
18. 10w2(w + 4v2)(w − 2v)(w + 2v)
© Houghton Mifflin Harcourt Publishing Company
Problem Solving
1. 3(5x − 4y)(x + y)
2. 4π(2k − 1)2; 4π m2
3. 4(2x + y)(2x − y)
4. −5(x − 19)(x + 3); 525 attendees
5. D
6. G
7. C
8. J
Chapter 7
401
Lesson 6
Name
Class
Notes
7-6
Date
© Houghton Mifflin Harcourt Publishing Company
401
Chapter 7
Lesson 6
Problem Solving
© Houghton Mifflin Harcourt Publishing Company
Q
Q Q
Chapter 7
Chapter 7
402
© Houghton Mifflin Harcourt Publishing Company
Lesson 6
402
Lesson 6
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