6-6
Special Products of Binomials
Going Deeper
Essential question: How can you find special products of binomials?
TE ACH
Standards for
Mathematical Content
1
A-SSE.1.2 Use the structure of an expression to
identify ways to rewrite it.
A-APR.1.1 … multiply polynomials.
EXAMPLE
Questioning Strategies
• When squaring binomials, how many terms do
you expect in the answer? three terms
Prerequisites
• What do you notice about the product of the
inside terms and the product of the outside terms
when squaring binomials? They are the same.
Why? They result from multiplying the x-term in
FOIL method
Math Background
Students have used the FOIL method to multiply
binomials. However, for special products of
binomials (sum and difference product, square
of a binomial, cube of a binomial), there are rules
that can be used to find products. These rules are
especially useful in terms of saving time and effort
when multiplying special products, especially the
cube of a binomial.
one binomial by the constant in the other. Since
the x-terms and the constants are the same, the
products are the same.
• What do you notice about the product of the
inner terms and the product of the outer terms
with a sum and difference product? The product
of the inner terms and the product of the outer
terms are opposites, and their sum is zero.
IN T RO DUC E
• In a sum and difference product, what can you
say about the sign of the last term? The sign of
Give students some products of binomials and
have them review multiplying them using FOIL.
As the last two examples, give students a square
of a binomial and a sum and difference product.
Ask students to observe any patterns they see
while multiplying out these products using FOIL.
Have students offer conjectures about whether any
patterns they see can be converted into a special
product rule.
the last term is always negative since you are
multiplying a positive constant and a negative
constant.
A. (4y + 5)2 16y2 + 40y + 25
B. (4y - 5)2 16y2 - 40y + 25
C. (4y + 5)(4y - 5) 16y2 - 25
Avoid Common Errors
When students see a square of a binomial such
as (2x + 1)2, they are likely to square only the two
terms, losing the x-term of the trinomial. To avoid
this error, have students rewrite the expression to
show the multiplication:
(2x + 1)2 = (2x + 1)(2x + 1).
Chapter 6
357
Lesson 6
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EXTRA EXAMPLE
Multiply using FOIL.
Name
Class
Notes
6-6
Date
Special Products of Binomials
Going Deeper
Essential question: How can you find special products of binomials?
The special products (ax + b)2, (ax - b)2, and (ax + b)(ax - b) can all be
found using the FOIL method. The products (ax + b)2 and (ax - b)2 are
called squares of binomials and the product (ax + b)(ax - b) is called the
sum and difference product.
A-SSE.1.2
1
EXAMPLE
A
Multiplying Special Cases
Multiply (2x + 5)2 using FOIL.
2
(2x + 5)2 = (2x + 5)(2x + 5) = 4x + 10x + 10x + 25
2
= 4x + 20x + 25
B
2
Multiply (2x - 5) using FOIL.
2
(2x - 5)2 = (2x - 5)(2x - 5) = 4x - 10x - 10x + 25
2
= 4x - 20x + 25
C
Multiply (2x - 5)(2x + 5) using FOIL.
2
(2x - 5)(2x + 5) = 4x + 10x - 10x - 25
2
= 4x - 25
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REFLECT
1a. In the final answer of Part A, which two terms of the trinomial are perfect squares?
How can you use the coefficients 2 and 5 to produce the coefficient of x in the product?
Generalize these results to write a rule for the product (ax + b)2 in terms of a, b, and x.
4x2 and 25; 20x = 2(2)(5)x; (ax)2 + 2abx + b2
1b. In the final answer of Part B, which two terms of the trinomial are perfect squares?
How can you use the coefficients 2 and 5 to produce the coefficient of x in the product?
Generalize these results to write a rule for the product (ax – b)2 in terms of a, b, and x.
4x2 and 25; -20x = -2(2)(5)x; (ax)2 - 2abx + b2
1c. In the final answer of Part C, which two terms of the trinomial are perfect squares?
What is the coefficient of the x-term and how was it created? Generalize these
results to write a rule for the product (ax - b)(ax + b) in terms of a, b, and x.
4x2 and 25; There is no x-term; the coefficient is 0 because
10x + (-10x) = 0; (ax)2 - b2
1d. In Part C, suppose the product had been (2x + 5)(2x - 5). Would the answer have
been different? Explain.
No; multiplication is commutative, so (2x - 5)(2x + 5) = (2x + 5)(2x - 5) = 4x2 - 25.
357
Chapter 6
Lesson 6
The squares of binomials and the sum and difference product occur so frequently that it is
helpful to recognize their patterns and develop rules for them. These rules, along with the
rules for the cubes of binomials, are summarized in the table.
Special Product Rules
(a + b)(a - b) = a2 - b2
(a + b)2 = a2 + 2ab + b2
Square of a Binomial
(a - b)2 = a2 - 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
Cube of a Binomial
2
(a - b)3 = a3 - 3a2b + 3ab2 - b3
A-SSE.1.2
EXAMPLE
Justifying and Applying a Special Product Rule
Justify the sum and difference rule. Then use it to find the product
(4x2 + 15)(4x2 - 15).
A
Justify the rule.
(a + b)(a - b) = a · a + a(-b) +
ba
+ b(-b)
= a2
ba
2
+ -b
=
B
a2
- ab
+
-
b2
Distribute a and then b.
Multiply monomials.
Combine like terms.
Find the product (4x2 + 15)(4x2 - 15).
2
(4x2 + 15)(4x2 - 15) = ( 4x )2 - ( 15 )2
4
= 16x - 225
Sum and difference rule
Simplify.
REFLECT
2a. Error Analysis A student was asked to find the square of 7x + 3. The student
quickly wrote (7x + 3)2 = 49x2 + 9. Identify the student’s error and provide the
correct answer.
The student forgot the middle term, 2ab, in the rule for the square of a binomial.
The correct answer is 49x2 + 42x + 9.
2b. Show how to justify the rule for the cube of a binomial, (a + b)3.
(a + b)3 = (a + b)(a + b)(a + b)
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Sum and Difference
= (a + b)(a2 + 2ab + b2)
= a3+ 2a2b + ab2 + ba2 + 2ab2 + b3
= a3 + 3a2b + 3ab2 + b3
Chapter 6
Chapter 6
358
Lesson 6
358
Lesson 6
2
CLOS E
EXAMPLE
Questioning Strategies
• In part A, how could the FOIL method help you?
Essential Question
How can you find special products of binomials?
It would help you find the four terms in the
product (before like terms are combined).
You can use the distributive property, the FOIL
method, or the special product rules to find special
products of binomials.
• In part B, what would the product be if the
binomials being multiplied were 4x + 15 and
4x - 15? 16x2 - 225
Summarize
Have students make a table for multiplying
special products of binomials. The table should
include each method (distributive property,
FOIL, special product rules), an example of
each method, and a list of advantages and
disadvantages for each method.
Avoid Common Errors
When using the rules for special products, students
often forget to apply exponents to the coefficients
of terms in the binomial. Suggest that students
first write the coefficient and variable within
parentheses, with the exponent applied to both,
and then simplify.
PR ACTICE
EXTRA EXAMPLE
Find the product (2x2 + 7)(2x2 - 7). 4x4 - 49
Where skills are
taught
Highlighting
the Standards
1 EXAMPLE
EXS. 1–12
2 EXAMPLE
EXS. 13–19
Exercises 20–26: Students apply the sum and
difference rule to multiply integers by mental math.
2 EXAMPLE
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and its Reflect questions offer
an opportunity to address Mathematical
Practice Standard 3 (Construct viable
arguments and critique the reasoning of
others). Students use what they have learned
about multiplying polynomials to justify
the special product rule for the sum and
difference. Then, students identify an error
made by another student when applying
the special product rule for the square of a
binomial.
Chapter 6
Where skills are
practiced
359
Lesson 6
Notes
PRACTICE
Find each product.
1. (x + 4)2
2. (x - 1)(x + 1)
x2 + 8x + 16
x2 - 1
3. (3x - 8)2
4. (6x + 1)2
36x2 + 12x + 1
9x2 - 48x + 64
5. (9x - 7)(9x + 7)
6. (4x - 5)(4x + 5)
81x2 - 49
16x2 - 25
7. (2x + 9)(2x - 9)
8. (10x + 3)(10x - 3)
4x2 - 81
100x2 - 9
2
2
9. (8x + 7)
10. (5x - 4)
25x2 - 40x + 16
64x2 + 112x + 49
12. (1 - 5x)2
11. (4 + 3x)(4 - 3x)
16 - 9x2
1 - 10x + 25x2
13. Justify the rule for the square of a binomial, (a + b)2 = a2 + 2ab + b2. Then use it to
expand (2x3 + 6y)2.
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(a + b)2 = (a + b)(a + b) = a · a + a · b + b · a + b · b = a2 + ab + ba + b2 =
a2 + 2ab + b2;
2
(2x3 + 6y)2 = (2x3) + 2(2x3)(6y) + (6y)2 = 4x6 + 24x3y + 36y2
Use a special product rule to find each product.
14. (9x + 5y)2
15. (6x - 4y)2
81x2 + 90xy + 25y2
2
36x2 - 48xy + 16y2
2
17. (4x - 3y3)(4x + 3y3)
16. (2x - 5y)(2x + 5y)
4
16x2 - 9y6
2
4x - 25y
2
2
19. (4x - 7y3)
18. (8x + 3y)
4
2
2
2
2
16x - 56xy3 + 49y6
64x + 48x y + 9y
359
Chapter 6
Lesson 6
20. The sum and difference rule is useful for mental-math calculations. Explain how
you can use the rule and mental math to calculate 32 · 28. (Hint: 32 · 28 = (30 + 2)(30 - 2).)
32 · 28 = (30 + 2)(30 - 2); by the sum and difference rule, this is equal to
Write each product using the sum and difference rule. Then find the product
using mental math.
21. 37 · 43
22. 26 · 34
(40 - 3)(40 + 3); 1591
(30 - 4)(30 + 4); 884
23. 22 · 18
24. 27 · 13
(20 + 2)(20 - 2); 396
(20 + 7)(20 - 7); 351
25. 45 · 55
26. 99 · 101
(50 - 5)(50 + 5); 2475
(100 - 1)(100 + 1); 9,999
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© Houghton Mifflin Harcourt Publishing Company
302 - 22, which is easy to calculate by mental math: 302 - 22 = 900 - 4 = 896.
Chapter 6
Chapter 6
360
Lesson 6
360
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
Assign these pages to help your students practice
and apply important lesson concepts. For
additional exercises, see the Student Edition.
Answers
Additional Practice
1. x2 + 4x + 4
2. m2 + 8m + 16
3. 9 + 6a + a2
4. 4x2 + 20x + 25
5. 9a2 + 12a + 4
6. 36 + 60b + 25b2
7. b2 - 6b + 9
8. 64 - 16y + y2
9. a2 − 20a + 100
10. 9x2 - 42x + 49
11. 16m2 - 72m + 81
12. 36 - 36n + 9n2
13. x2 - 9
14. 64 − y2
15. x2 - 36
16. 25x2 − 4
17. 100x2 - 49y2
18. x4 - 9y2
19. a. 36 - x2; b. 4 - x2; c. 32
20. a. 16 - x2; b. 20
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Problem Solving
1. x2 - 16; $128
2. 0.75x2 - x - 65; 2575 square feet
3. x2 - 144 in2
4. D
5. G
6. A
7. G
Chapter 6
361
Lesson 6
Name
Class
Date
Notes
6-6
© Houghton Mifflin Harcourt Publishing Company
Additional Practice
361
Chapter 6
Lesson 6
© Houghton Mifflin Harcourt Publishing Company
© Houghton Mifflin Harcourt Publishing Company
Problem Solving
Chapter 6
Chapter 6
362
Lesson 6
362
Lesson 6