7-3
Factoring x 2 + bx + c
Going Deeper
Essential question: How can you factor x2 + bx + c?
TE ACH
Standards for
Mathematical Content
1
A-SSE.1.2 Use the structure of an expression to
identify ways to rewrite it.*
ENGAGE
Questioning Strategies
• When you factor a trinomial, what type of
expression do you expect? the product of two
Prerequisites
Multiplying binomials
binomials
Math Background
• Using FOIL, which two terms produce the
x2 -term in the trinomial? the first Using FOIL,
which two terms produce the constant term in
the trinomial? the last Using FOIL, which terms
produce the x-term in the trinomial? the sum of
Students have multiplied binomials to create
trinomials. In this lesson, students will begin to
do the opposite. They will factor trinomials of
the form x2 + bx + c to find the binomial factors
of a trinomial. In the next lesson, they will factor
trinomials whose leading coefficient is
not 1. Factoring a trinomial is an important
algebraic skill. At its most basic, it allows students
to see the binomial factors that make up a
trinomial. But also it will allow students to solve
some quadratic equations, find the zeros of some
quadratic functions, and aid in graphing quadratic
functions; all operations they will perform when
they study quadratic equations and functions.
the product of the inner terms and the product of
the outer terms
• When factoring a trinomial, why do you factor
the constant into pairs of factors rather than find
prime factors? The constant term comes from
multiplying the constant terms of two binomials.
So, you need two factors that may not be prime.
IN T RO DUC E
Completing several warm-up exercises will help
students see how to factor trinomials
Ask students to use FOIL to multiply:
A. (x + 4)(x + 3) x2 + 7x + 12
B. (x - 5)(x + 2) x2 - 3x - 10
C. (x + 3)2 x2 + 6x + 9
x2
x x x x x
Next, have students factor the following numbers into
all possible different pairs of integer factors.
x
1 1 1 1 1
A. 12 1 and 12, 2 and 6, 3 and 4, -1 and -12, -2 and
-6, -3 and -4
Next, have students place x-tiles and 1-tiles to the
left of and above the rectangle. These tiles represent
the binomials that are multiplied to obtain the
trinomial. In this case, the binomials are x + 5
across the top and x + 1 arranged vertically on the
left side. Students can then determine that
x2 + 6x + 5 = (x + 5)(x + 1).
B. -10 -1 and 10, -2 and 5, 2 and -5, 1 and -10
C. 9 1 and 9, 3 and 3, -1 and -9, -3 and -3
Then ask students how one of each of these pairs of
factors is related to each multiplication.
Chapter 7
381
Lesson 3
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Differentiated Instruction
If students are having difficulty in factoring
trinomials, algebra tiles can help them visualize
the process. Give students algebra tiles to represent
the trinomial x2 + 6x + 5 (one x2-tile, six x-tiles,
and five 1-tiles). You can either arrange the tiles for
students or have the students arrange the tiles to
make a rectangle as shown below. Students should
write the trinomial x2 + 6x + 5 that is depicted by
the tiles.
Name
Class
Notes
7-3
Date
Factoring x 2 + bx + c
Going Deeper
Essential question: How can you factor x2 + bx + c?
A-SSE.1.2
1
ENGAGE
Factoring Trinomials
You know how to multiply binomials: for example, (x + 3)(x - 5) = x2 - 2x - 15.
In this lesson, you will learn how to reverse this process and factor trinomials.
There are several important things you should remember from multiplying binomials.
• Using FOIL, the constant term in the trinomial is a result of multiplying the last
terms in the two binomials.
• Using FOIL, the x-term results from adding the products of the outside terms
and inside terms.
You can factor x2 + 10x + 21 by working FOIL backward. Both signs in the trinomial are
plus signs, so you know both binomials are of the form x plus something. Therefore, you
can set up the factoring as shown below.
x2 + 10x + 21 = (x + ? )(x + ? )
To find the constant terms in the binomials, use the information above and follow the
steps below.
1) The constant term in the trinomial, 21, is the
product of the last terms in the two binomials.
Factor 21 into pairs. The factor pairs are shown
in the table at the right.
Factors of 21
Sum of Factors
1 and 21
22
3 and 7
10 !
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2) The correct factor pair is the one whose sum is
the coefficient of x in the trinomial.
3) Complete the binomial expression with the
appropriate numbers.
(
x2 + 10x + 21 = x + 3
)(x + 7 )
REFLECT
1a. You want to factor x2 - 6x + 8. What factoring pattern would you set up to begin
the process? Explain.
The first sign is - and the second is +. To get a negative x-term and a positive
constant, both binomials must be (x - ? ). The pattern is (x - ? )(x - ? ).
381
Chapter 7
Lesson 3
1b. You want to factor x2 - 2x - 15. What factoring pattern would you set up to begin
the process? Explain. Would this pattern also work for x2 + 2x - 15? Explain.
Both signs are -. To get a negative x-term and a negative constant, one binomial
must be (x + ? ) and the other must be (x - ? ). The pattern is (x + ? )(x - ? ).
(x + ? )(x - ? ).
1c. Use factoring patterns to factor x2 + 8x + 16 and x2 - 6x + 9. What do you notice
about the factored forms? What special type of trinomials are x2 + 8x + 16 and
x2 - 6x + 9?
(x + 4)(x + 4), (x - 3)(x - 3); they are squares of binomials.
2
A-SSE.1.2
EXAMPLE
A
Factoring Trinomials
2
Factor x + 3x - 10.
The constant is negative, so you know one binomial will have a subtraction sign.
x2 + 3x - 10 = (x + ? )(x - ? )
Complete the table at the right. Note that you
are finding the factors of -10, not 10. Since the
coefficient of x is positive, the factor with the
greater absolute value will be positive (and the
other factor will be negative).
(
x2 + 3x - 10 = x + 5
B
Factors of -10
Sum of Factors
-1 and 10
9
-2 and 5
3
)( x - 2 )
Factor x2 - 8x - 48.
The constant is negative, so you know one binomial will have a subtraction sign.
x2 - 8x - 48 = (x + ? )(x - ? )
Complete the table at the right. Since the
coefficient of x is negative, the factor with the
greater absolute value will be negative (and the
other factor will be positive).
(
x2 - 8x - 48 = x + 4
Chapter 7
Chapter 7
)(x - 12 )
382
Factors of -48
1 and -48
2 and
-24
Sum of Factors
-47
-22
3 and -16
-13
4 and -12
-8
6 and -8
-2
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© Houghton Mifflin Harcourt Publishing Company
Yes; the positive x-term and the negative constant also require the pattern to be
Lesson 3
382
Lesson 3
2
CLOS E
EXAMPLE
Questioning Strategies
• In part A, explain what would happen to the
coefficient of x if the greater of the factor pair
were negative. The coefficient of x would be
Essential Question
How can you factor x2 + bx + c?
Trinomials can be factored into two binomials
by finding factor pairs for the constant term
and determining which factor pair sums to the
coefficient of the x-term.
negative.
• If the problem in part A were to factor
x2 - 3x - 10, what would be the factored form?
(x - 5)(x + 2)
Summarize
Have students create a flowchart like the one below
for factoring trinomials of the form x2 + bx + c .
• How are the factored forms of x2 - 3x - 10 and
x2 + 3x - 10 alike? How are they different?
The factored forms have the same coefficients
for x and the same numbers as constants.
The difference is in which binomial involves
subtraction.
Identify b and c.
• After completing part B, determine the factors of
x2 + 8x - 48. (x - 4)(x + 12)
Find factor pairs for c.
EXTRA EXAMPLE
Factor.
Sum one factor pair for c.
2
A. y + 5y + 6 (y + 3)(y + 2)
B. z2 - 9z - 22 (z - 11)(z + 2)
Does the
sum = b?
Highlighting
the Standards
No
Yes
2 EXAMPLE
Chapter 7
Write the
factored form
of x2 + bx + c
PR ACTICE
Where skills are
taught
383
Where skills are
practiced
1 ENGAGE
EXS. 1–4
2 EXAMPLE
EXS. 5–24
Lesson 3
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and its Reflect question offer
an opportunity to address Mathematical
Practice Standard 1 (Make sense of problems
and persevere in solving them). Students will
need to analyze each trinomial and make
conjectures about the form of the binomial
factors. Then, they will need to determine
which factor pair to use to factor the
trinomial.
Notes
REFLECT
2a. Complete the table below. Assume that b, c, p, and q are positive numbers.
Trinomial
Form of Binomial Factors
x2 + bx + c
x2 - bx + c
x2 - bx - c or
x2 + bx - c
( x + p )( x
( x - p )( x
( x + p )( x
)
)
q)
+ q
- q
-
For the last row in the table, explain how to determine which factor contains
a + sign and which factor contains a - sign.
If the sign before bx is -, then the factor involving the larger of p or q will have
a - sign. If the sign before bx is +, then the factor involving the larger of p or q
will have a + sign.
PRACTICE
Complete the factorization of the polynomial.
(
)
) (d - 3 )
1. t2 + 6t + 5 = (t + 5) t + 1
(
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3. d2 + 5d - 24 = d + 8
(
)(
2. z2 - 121 = (z + 11) z -
(
4. x4 - 4 = x2 + 2
11
x2 - 2
)
)
Factor the polynomial.
5. y2 + 3y - 4
6. x2 - 2x + 1
( y + 4)(y - 1)
(x - 1)2
2
2
7. p - 2p - 24
8. g - 100
(p - 6)(p + 4)
(g - 10)(g + 10)
9. z2 - 7z + 12
10. q2 + 25q + 100
(q + 20)(q + 5)
(z - 3)(z - 4)
11. m2 + 8m + 16
12. n2 - 10n - 24
(m + 4)(m + 4)
(n - 12)(n + 2)
13. x2 + 25x
14. y2 - 13y - 30
(y - 15)(y + 2)
x(x + 25)
383
Chapter 7
Lesson 3
Factor the polynomial.
15. z2 - 9
16. p2 + 3p - 54
(p - 6)(p + 9)
(z + 3)(z - 3)
2
2
17. x + 11x - 42
18. g - 14g - 51
(g + 3)(g - 17)
2
2
19. n - 81
20. y - 25y
y(y - 25)
(n - 9)(n + 9)
21. x2 + 11x + 30
22. x2 - x - 20
(x + 6)(x + 5)
(x + 4)(x - 5)
23. x2 + 6x - 7
24. x2 + 2x + 1
(x + 1)(x + 1)
(x + 7)(x - 1)
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(x - 3)(x + 14)
Chapter 7
Chapter 7
384
Lesson 3
384
Lesson 3
Problem Solving
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
1. x − 8 m
2. rug: (x + 5)(x − 4); wall: (x + 2)(x + 15); rug:
16 ft by 25 ft, wall: 22 ft by 35 ft
Assign these pages to help your students practice
and apply important lesson concepts. For
additional exercises, see the Student Edition.
3. a. x + 5
b. 12 in. by 19 in.
c. x2 + x − 12
Answers
4. D
Additional Practice
2. (x + 1)(x + 8)
3. (x + 4)(x + 9)
4. (x + 7)(x + 2)
5. (x + 3)(x + 4)
6. (x + 6)(x + 3)
7. (x − 6)(x − 3)
8. (x − 4)(x − 1)
9. (x − 5)(x − 4)
10. (x − 2)(x − 10)
11. (x − 9)(x − 2)
12. (x − 8)(x − 4)
13. (x + 9)(x − 2)
14. (x + 12)(x − 2)
15. (x + 3)(x − 1)
16. (x + 5)(x − 3)
17. (x + 6)(x − 1)
18. (x + 8)(x − 3)
19. (x + 1)(x − 6)
20. (x + 5)(x − 7)
21. (x + 3)(x − 10)
22. (x + 7)(x − 8)
23. (x + 2)(x − 4)
24. (x + 4)(x − 5)
6. B
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1. (x + 2)(x + 5)
5. J
25. (n + 8)(n − 3)
n
n2 + 5n − 24
0
02 + 5(0) − 24 = −24
1
12 + 5(1) − 24 = −18
2
22 + 5(2) − 24 = −10
3
32 + 5(3) − 24 = 0
4
42 + 5(4) − 24 = 12
n
(n + 8)(n − 3)
0
(0 + 8)(0 − 3) = −24
1
(1 + 8)(1 − 3) = −18
2
(2 + 8)(2 − 3) = −10
3
(3 + 8)(3 − 3) = 0
4
(4 + 8)(4 − 3) = 12
Chapter 7
385
Lesson 3
Name
Class
Date
Notes
7-3
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Additional Practice
385
Chapter 7
Lesson 3
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© Houghton Mifflin Harcourt Publishing Company
Problem Solving
Chapter 7
Chapter 7
386
Lesson 3
386
Lesson 3