3: Factoring Trinomials of the Form x2 + bx + c
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6.3 • Factoring Trinomials of the Form x 2 ϩ bx ϩ c
243
The third line contains the numbers that we want. Thus
x2 Ϫ 11x ϩ 24 ϭ (x Ϫ 3) (x Ϫ 8)
Classroom Example
Factor x2 ϩ 2x Ϫ 24.
EXAMPLE 3
Factor x2 ϩ 3x Ϫ 10.
Solution
To factor x2 ϩ 3x Ϫ10, we want to find two numbers whose product is Ϫ10 and whose sum is 3.
Product
1(Ϫ10)
Ϫ1(10)
2(Ϫ5)
Ϫ 2(5)
ϭ Ϫ10
ϭ Ϫ 10
ϭ Ϫ 10
ϭ Ϫ 10
Sum
1 ϩ (Ϫ10) ϭ Ϫ9
Ϫ1 ϩ 10 ϭ 9
2 ϩ (Ϫ5) ϭ Ϫ 3
Ϫ2 ϩ 5 ϭ 3
The bottom line is the key line. Thus
x2 ϩ 3x Ϫ 10 ϭ (x ϩ 5)(x Ϫ 2)
Classroom Example
Factor n2 Ϫ 4n Ϫ 32.
EXAMPLE 4
Factor x2 Ϫ 2x Ϫ 8.
Solution
We are looking for two numbers whose product is Ϫ8 and whose sum is Ϫ2.
Product
1(Ϫ8)
Ϫ1(8)
2(Ϫ4)
Ϫ 2(4)
ϭ Ϫ8
ϭ Ϫ8
ϭ Ϫ8
ϭ Ϫ8
Sum
1 ϩ (Ϫ8) ϭ Ϫ7
Ϫ1 ϩ 8 ϭ 7
2 ϩ (Ϫ4) ϭ Ϫ 2
Ϫ2 ϩ 4 ϭ 2
The third line has the information we want.
x2 Ϫ 2x Ϫ 8 ϭ (x Ϫ 4)(x ϩ 2)
The tables in the last four examples illustrate one way of organizing your thoughts for such
problems. We show complete tables; that is, for Example 4, we include the bottom line even
though the desired numbers are obtained in the third line. If you use such tables, keep in mind
that as soon as you get the desired numbers, the table need not be completed any further.
Furthermore, you may be able to find the numbers without using a table. The key ideas are the
product and sum relationships.
Classroom Example
Factor x2 ϩ 13x Ϫ 14.
EXAMPLE 5
Factor x2 Ϫ 13x ϩ 12.
Solution
Product
(Ϫ1)(Ϫ12) ϭ 12
Sum
(Ϫ1) ϩ (Ϫ12) ϭ Ϫ13
We need not complete the table.
x2 Ϫ 13x ϩ 12 ϭ (x Ϫ 1)(x Ϫ 12)
244
Chapter 6 • Factoring, Solving Equations, and Problem Solving
In the next example, we refer to the concept of absolute value. Recall that the absolute
value of any nonzero real number is positive. For example,
0 40 ϭ 4 and 0Ϫ40 ϭ 4
Classroom Example
Factor x2 Ϫ 6x Ϫ 7.
EXAMPLE 6
Factor x2 Ϫ x Ϫ 56.
Solution
Notice that the coefficient of the middle term is Ϫ1. Therefore, we are looking for two
numbers whose product is Ϫ56; because their sum is Ϫ1, the absolute value of the negative
number must be one larger than the absolute value of the positive number. The numbers are
Ϫ8 and 7, and we have
x2 Ϫ x Ϫ 56 ϭ (x Ϫ 8)(x ϩ 7)
Classroom Example
Factor m2 ϩ m ϩ 3.
EXAMPLE 7
Factor x2 ϩ 10x ϩ 12.
Solution
Product
Sum
1(12) ϭ 12 1 ϩ 12 ϭ 13
2(6) ϭ 12
2ϩ6ϭ8
3(4) ϭ 12
3ϩ4ϭ7
Since the table is complete and no two factors of 12 produce a sum of 10, we conclude that
x2 ϩ 10x ϩ 12
is not factorable using integers.
In a problem such as Example 7, we need to be sure that we have tried all possibilities before
we conclude that the trinomial is not factorable.
Back to Solving Equations
The property ab ϭ 0 if and only if a ϭ 0 or b ϭ 0 continues to play an important role as
we solve equations that involve the factoring ideas of this section. Consider the following
examples.
Classroom Example
Solve x2 ϩ 15x ϩ 26 ϭ 0.
EXAMPLE 8
Solve x2 ϩ 8x ϩ 15 ϭ 0.
Solution
x2 ϩ 8x ϩ 15 ϭ 0
Factor the left side
(x ϩ 3)(x ϩ 5) ϭ 0
x ϩ 3 ϭ 0 or x ϩ 5 ϭ 0
Use ab ϭ 0 if and only if a ϭ 0 or b ϭ 0
x ϭ Ϫ3 or
x ϭ Ϫ5
The solution set is {Ϫ5, Ϫ3}.
6.3 • Factoring Trinomials of the Form x 2 ϩ bx ϩ c
Classroom Example
Solve x2 Ϫ 8x Ϫ 9 ϭ 0.
EXAMPLE 9
245
Solve x2 ϩ 5x Ϫ 6 ϭ 0.
Solution
x2 ϩ 5x Ϫ 6 ϭ 0
(x ϩ 6)(x Ϫ 1) ϭ 0
x ϩ 6 ϭ 0 or x Ϫ 1 ϭ 0
x ϭ Ϫ6 or
xϭ1
The solution set is {Ϫ6, 1}.
Classroom Example
Solve m2 Ϫ 8m ϭ 33.
EXAMPLE 10
Solve y2 Ϫ 4y ϭ 45.
Solution
y2 Ϫ 4y ϭ 45
y2 Ϫ 4y Ϫ 45 ϭ 0
(y Ϫ 9)(y ϩ 5) ϭ 0
y Ϫ 9 ϭ 0 or y ϩ 5 ϭ 0
y ϭ 9 or
y ϭ Ϫ5
The solution set is {Ϫ5, 9}.
Don’t forget that we can always check to be absolutely sure of our solutions. Let’s check
the solutions for Example 10. If y ϭ 9, then y2 Ϫ 4y ϭ 45 becomes
92 Ϫ 4(9) 45
81 Ϫ 36 45
45 ϭ 45
If y ϭ Ϫ5, then y2 Ϫ 4y ϭ 45 becomes
(Ϫ5) 2 Ϫ 4(Ϫ5) 45
25 ϩ 20 45
45 ϭ 45
Back to Problem Solving
The more we know about factoring and solving equations, the more easily we can solve word
problems.
Classroom Example
Find two consecutive odd integers
whose product is 35.
EXAMPLE 11
Find two consecutive integers whose product is 72.
Solution
Let n represent one integer. Then n ϩ 1 represents the next integer.
n(n ϩ 1) ϭ 72 The product of the two integers is 72
n2 ϩ n ϭ 72
n2 ϩ n Ϫ 72 ϭ 0
(n ϩ 9)(n Ϫ 8) ϭ 0
n ϩ 9 ϭ 0 or n Ϫ 8 ϭ 0
n ϭ Ϫ9 or n ϭ 8
If n ϭ Ϫ9, then n ϩ 1 ϭ Ϫ9 ϩ 1 ϭ Ϫ8. If n ϭ 8, then n ϩ 1 ϭ 8 ϩ 1 ϭ 9. Thus the
consecutive integers are Ϫ9 and Ϫ8 or 8 and 9.
246
Chapter 6 • Factoring, Solving Equations, and Problem Solving
Classroom Example
A triangular lot has a height that is 8
yards longer than the base. The area
of the lot is 24 square yards. Find
the base and height of the lot.
EXAMPLE 12
A rectangular plot is 6 meters longer than it is wide. The area of the plot is 16 square meters.
Find the length and width of the plot.
w
w+6
Figure 6.3
Solution
We let w represent the width of the plot, and then w ϩ 6 represents the length (see Figure 6.3).
Using the area formula A ϭ lw, we obtain
w(w ϩ 6) ϭ 16
w2 ϩ 6w ϭ 16
w2 ϩ 6w Ϫ 16 ϭ 0
(w ϩ 8)(w Ϫ 2) ϭ 0
w ϩ 8 ϭ 0 or w Ϫ 2 ϭ 0
w ϭ Ϫ8 or
wϭ2
The solution of Ϫ8 is not possible for the width of a rectangle, so the plot is 2 meters wide
and its length (w ϩ 6) is 8 meters.
The Pythagorean theorem, an important theorem pertaining to right triangles, can also
serve as a guideline for solving certain types of problems. The Pythagorean theorem states
that in any right triangle, the square of the longest side (called the hypotenuse) is equal to
the sum of the squares of the other two sides (called legs); see Figure 6.4. We can use this
theorem to help solve problems.
c
b
a2 + b2 = c2
a
Figure 6.4
Classroom Example
Suppose that the lengths of the three
sides of a right triangle are consecutive even integers. Find the lengths
of the three sides.
EXAMPLE 13
Suppose that the lengths of the three sides of a right triangle are consecutive whole numbers.
Find the lengths of the three sides.
Solution
Let s represent the length of the shortest leg. Then s ϩ 1 represents the length of the other leg,
and s ϩ 2 represents the length of the hypotenuse. Using the Pythagorean theorem as a guideline, we obtain the following equation:
6.3 • Factoring Trinomials of the Form x 2 ϩ bx ϩ c
247
Sum of squares of two legs ϭ Square of hypotenuse
6447448
64748
s2 ϩ (s ϩ 1) 2
ϭ
(s ϩ 2) 2
Solving this equation yields
s2 ϩ s2 ϩ 2s ϩ 1 ϭ s2 ϩ 4s ϩ 4
2s2 ϩ 2s ϩ 1 ϭ s2 ϩ 4s ϩ 4
s2 ϩ 2s ϩ 1 ϭ 4s ϩ 4
s2 Ϫ 2s ϩ 1 ϭ 4
s2 Ϫ 2s Ϫ 3 ϭ 0
(s Ϫ 3)(s ϩ 1) ϭ 0
s Ϫ 3 ϭ 0 or s ϩ 1 ϭ 0
s ϭ 3 or
s ϭ Ϫ1
Remember (a ϩ b)2 ϭ a2 ϩ 2ab ϩ b2
Add Ϫs2 to both sides
The solution of Ϫ1 is not possible for the length of a side, so the shortest side is of length 3.
The other two sides (s ϩ 1 and s ϩ 2) have lengths of 4 and 5.
Concept Quiz 6.3
For Problems 1–10, answer true or false.
1. Any trinomial of the form x2 ϩ bx ϩ c can be factored (using integers) into the product
of two binomials.
2. To factor x2 Ϫ 4x Ϫ 60 we look for two numbers whose product is Ϫ60 and whose
sum is Ϫ4.
3. A trinomial of the form x2 ϩ bx ϩ c will never have a common factor other than 1.
4. If n represents an odd integer, then n ϩ 1 represents the next consecutive odd integer.
5. The Pythagorean theorem only applies to right triangles.
6. In a right triangle the longest side is called the hypotenuse.
7. The polynomial x2 ϩ 25x ϩ 72 is not factorable.
8. The polynomial x2 ϩ 27x ϩ 72 is not factorable.
9. The solution set of the equation x2 ϩ 2x Ϫ 63 ϭ 0 is {Ϫ9, 7}.
10. The solution set of the equation x2 Ϫ 5x Ϫ 66 ϭ 0 is {Ϫ11, Ϫ6}.
Problem Set 6.3
For Problems 1–30, factor each trinomial completely. Indicate any that are not factorable using integers. (Objective 1)
19. y2 Ϫ y Ϫ 72
20. y2 Ϫ y Ϫ 30
21. x2 ϩ 21x ϩ 80
22. x2 ϩ 21x ϩ 90
23. x2 ϩ 6x Ϫ 72
24. x2 Ϫ 8x Ϫ 36
25. x2 Ϫ 10x Ϫ 48
26. x2 Ϫ 12x Ϫ 64
27. x2 ϩ 3xy Ϫ 10y2
28. x2 Ϫ 4xy Ϫ 12y2
29. a2 Ϫ 4ab Ϫ 32b2
30. a2 ϩ 3ab Ϫ 54b2
1. x ϩ 10x ϩ 24
2. x ϩ 9x ϩ 14
3. x ϩ 13x ϩ 40
4. x ϩ 11x ϩ 24
5. x Ϫ 11x ϩ 18
6. x Ϫ 5x ϩ 4
7. n Ϫ 11n ϩ 28
8. n Ϫ 7n ϩ 10
9. n ϩ 6n Ϫ 27
10. n ϩ 3n Ϫ 18
11. n2 Ϫ 6n Ϫ 40
12. n2 Ϫ 4n Ϫ 45
For Problems 31– 50, solve each equation. (Objective 2)
13. t ϩ 12t ϩ 24
14. t ϩ 20t ϩ 96
31. x2 ϩ 10x ϩ 21 ϭ 0
32. x2 ϩ 9x ϩ 20 ϭ 0
15. x2 Ϫ 18x ϩ 72
16. x2 Ϫ 14x ϩ 32
33. x2 Ϫ 9x ϩ 18 ϭ 0
34. x2 Ϫ 9x ϩ 8 ϭ 0
17. x2 ϩ 5x Ϫ 66
18. x2 ϩ 11x Ϫ 42
35. x2 Ϫ 3x Ϫ 10 ϭ 0
36. x2 Ϫ x Ϫ 12 ϭ 0
2
2
2
2
2
2
2
2
2
2
2
2
248
Chapter 6 • Factoring, Solving Equations, and Problem Solving
37. n2 ϩ 5n Ϫ 36 ϭ 0
38. n2 ϩ 3n Ϫ 18 ϭ 0
39. n2 Ϫ 6n Ϫ 40 ϭ 0
40. n2 Ϫ 8n Ϫ 48 ϭ 0
41. t2 ϩ t Ϫ 56 ϭ 0
42. t2 ϩ t Ϫ 72 ϭ 0
43. x2 Ϫ 16x ϩ 28 ϭ 0
44. x2 Ϫ 18x ϩ 45 ϭ 0
45. x2 ϩ 11x ϭ 12
46. x2 ϩ 8x ϭ 20
47. x(x Ϫ 10) ϭ Ϫ16
48. x(x Ϫ 12) ϭ Ϫ35
49. Ϫx2 Ϫ 2x ϩ 24 ϭ 0
50. Ϫx2 ϩ 6x ϩ 16 ϭ 0
For Problems 51– 68, set up an equation and solve each
problem. (Objectives 3 and 4)
51. Find two consecutive integers whose product is 56.
52. Find two consecutive odd whole numbers whose product is 63.
53. Find two consecutive even whole numbers whose product is 168.
54. One number is 2 larger than another number. The sum
of their squares is 100. Find the numbers.
55. Find four consecutive integers such that the product of
the two larger integers is 22 less than twice the product
of the two smaller integers.
56. Find three consecutive integers such that the product of
the two smaller integers is 2 more than ten times the
largest integer.
57. One number is 3 smaller than another number. The square
of the larger number is 9 larger than ten times the smaller
number. Find the numbers.
59. Suppose that the width of a certain rectangle is
3 inches less than its length. The area is numerically
6 less than twice the perimeter. Find the length and
width of the rectangle.
60. The sum of the areas of a square and a rectangle is
64 square centimeters. The length of the rectangle is 4
centimeters more than a side of the square, and the
width of the rectangle is 2 centimeters more than a side
of the square. Find the dimensions of the square and the
rectangle.
61. The perimeter of a rectangle is 30 centimeters, and the
area is 54 square centimeters. Find the length and width
of the rectangle. [Hint: Let w represent the width; then
15 Ϫ w represents the length.]
62. The perimeter of a rectangle is 44 inches, and its area
is 120 square inches. Find the length and width of the
rectangle.
63. An apple orchard contains 84 trees. The number of trees
per row is five more than the number of rows. Find the
number of rows.
64. A room contains 54 chairs. The number of rows is
3 less than the number of chairs per row. Find the number of rows.
65. Suppose that one leg of a right triangle is 7 feet shorter
than the other leg. The hypotenuse is 2 feet longer than
the longer leg. Find the lengths of all three sides of the
right triangle.
66. Suppose that one leg of a right triangle is 7 meters longer
than the other leg. The hypotenuse is 1 meter longer than
the longer leg. Find the lengths of all three sides of the
right triangle.
67. Suppose that the length of one leg of a right triangle is
2 inches less than the length of the other leg. If the
length of the hypotenuse is 10 inches, find the length of
each leg.
68. The length of one leg of a right triangle is 3 centimeters more than the length of the other leg. The length of
the hypotenuse is 15 centimeters. Find the lengths of the
two legs.
58. The area of the floor of a rectangular room is 84 square
feet. The length of the room is 5 feet more than its
width. Find the length and width of the room.
Thoughts Into Words
69. What does the expression “not factorable using integers”
mean to you?
70. Discuss the role that factoring plays in solving
equations.
71. Explain how you would solve the equation
(x Ϫ 3)(x ϩ 4) ϭ 0
and also how you would solve
(x Ϫ 3)(x ϩ 4) ϭ 8.
6.4 • Factoring Trinomials of the Form ax 2 ϩ bx ϩ c
249
Further Investigations
For Problems 72 – 75, factor each trinomial and assume
that all variables appearing as exponents represent positive integers.
72. x2a ϩ 10xa ϩ 24
73. x2a ϩ 13xa ϩ 40
74. x2a Ϫ 2xa Ϫ 8
75. x2a ϩ 6xa Ϫ 27
76. Suppose that we want to factor n2 ϩ 26n ϩ 168 so that
we can solve the equation n2 ϩ 26n ϩ 168 ϭ 0. We
need to find two positive integers whose product is 168
and whose sum is 26. Since the constant term, 168, is
rather large, let’s look at it in prime factored form:
168 ϭ 2
n2 ϩ 26n ϩ 168 ϭ 0
(n ϩ 12)(n ϩ 14) ϭ 0
n ϩ 12 ϭ 0 or n ϩ 14 ϭ 0
n ϭ Ϫ12 or
n ϭ Ϫ14
The solution set is {Ϫ14, Ϫ12}.
Solve each of the following equations.
(a) n2 ϩ 30n ϩ 216 ϭ 0
(b) n2 ϩ 35n ϩ 294 ϭ 0
и2и2и3и7
Now we can mentally form two numbers by using all of
these factors in different combinations. Using two 2s and
the 3 in one number and the other 2 and the 7 in another
number produces 2 и 2 и 3 ϭ 12 and 2 и 7 ϭ 14.
Answers to the Concept Quiz
1. False
2. True
3. True
4. False
9. True
10. False
6.4
Therefore, we can solve the given equation as follows:
5. True
(c) n2 Ϫ 40n ϩ 384 ϭ 0
(d) n2 Ϫ 40n ϩ 375 ϭ 0
(e) n2 ϩ 6n Ϫ 432 ϭ 0
(f) n2 Ϫ 16n Ϫ 512 ϭ 0
6. True
7. True
8. False
Factoring Trinomials of the Form ax 2 ؉ bx ؉ c
OBJECTIVES
1
Factor trinomials where the leading coefﬁcient is not 1
2
Solve equations that involve factoring
Now let’s consider factoring trinomials where the coefficient of the squared term is not 1. We
present here an informal trial and error technique that works quite well for certain types of
trinomials. This technique simply relies on our knowledge of multiplication of binomials.
Classroom Example
Factor 3x2 ϩ 10x ϩ 8.
EXAMPLE 1
Factor 2x2 ϩ 7x ϩ 3.
Solution
By looking at the first term, 2x2, and the positive signs of the other two terms, we know that
the binomials are of the form
(2x ϩ _____)(x ϩ _____)
Since the factors of the constant term, 3, are 1 and 3, we have only two possibilities to try:
(2x ϩ 3)(x ϩ 1)
or
(2x ϩ 1)(x ϩ 3)
By checking the middle term of both of these products, we find that the second one yields the
correct middle term of 7x. Therefore,
2x2 ϩ 7x ϩ 3 ϭ (2x ϩ 1)(x ϩ 3)
250
Chapter 6 • Factoring, Solving Equations, and Problem Solving
Classroom Example
Factor 15y2 Ϫ 13y ϩ 2.
EXAMPLE 2
Factor 6x2 Ϫ 17x ϩ 5.
Solution
First, we note that 6x2 can be written as 2x и 3x or 6x и x. Second, since the middle term of the
trinomial is negative, and the last term is positive, we know that the binomials are of the form
(2x Ϫ _____)(3x Ϫ _____)
or
(6x Ϫ _____)(x Ϫ _____)
Since the factors of the constant term, 5, are 1 and 5, we have the following possibilities:
(2x Ϫ 5)(3x Ϫ 1) (2x Ϫ 1)(3x Ϫ 5)
(6x Ϫ 5)(x Ϫ 1)
(6x Ϫ 1)(x Ϫ 5)
By checking the middle term for each of these products, we find that the product (2x Ϫ 5)(3x Ϫ 1)
produces the desired term of Ϫ17x. Therefore,
6x2 Ϫ 17x ϩ 5 ϭ (2x Ϫ 5)(3x Ϫ 1)
Classroom Example
Factor 18n2 ϩ 12n Ϫ 16.
EXAMPLE 3
Factor 8x2 Ϫ 8x Ϫ 30.
Solution
First, we note that the polynomial 8x 2 Ϫ 8x Ϫ 30 has a common factor of 2. Factoring out the
common factor gives us 2(4x 2 Ϫ 4x Ϫ 15). Now we need to factor 4x 2 Ϫ 4x Ϫ 15.
We note that 4x2 can be written as 4x и x or 2x и 2x. The last term, Ϫ15, can be written
as (1)(Ϫ15), (Ϫ1)(15), (3)(Ϫ5), or (Ϫ3)(5). Thus we can generate the possibilities for the
binomial factors as follows:
✔
Using 1 and ؊15
Using ؊1 and 15
(4x Ϫ 15)(x ϩ 1)
(4x ϩ 1)(x Ϫ 15)
(2x ϩ 1)(2x Ϫ 15)
(4x Ϫ 1)(x ϩ 15)
(4x ϩ 15)(x Ϫ 1)
(2x Ϫ 1)(2x ϩ 15)
Using 3 and ؊5
Using ؊3 and 5
(4x ϩ 3)(x Ϫ 5)
(4x Ϫ 5)(x ϩ 3)
(2x Ϫ 5)(2x ϩ 3)
(4x Ϫ 3)(x ϩ 5)
(4x ϩ 5)(x Ϫ 3)
(2x ϩ 5)(2x Ϫ 3)
By checking the middle term of each of these products, we find that the product indicated with
a check mark produces the desired middle term of Ϫ4x. Therefore,
8x2 Ϫ 8x Ϫ 30 ϭ 2(4x2 Ϫ 4x Ϫ 15) ϭ 2(2x Ϫ 5)(2x ϩ 3)
Let’s pause for a moment and look back over Examples 1, 2, and 3. Obviously, Example 3
created the most difficulty because we had to consider so many possibilities. We have suggested one possible format for considering the possibilities, but as you practice such problems, you
may develop a format that works better for you. Regardless of the format that you use, the key
idea is to organize your work so that you consider all possibilities. Let’s look at another example.
Classroom Example
Factor 9x2 ϩ 10x ϩ 4.
EXAMPLE 4
Factor 4x2 ϩ 6x ϩ 9.
Solution
First, we note that 4x2 can be written as 4x и x or 2x и 2x. Second, since the middle term is
positive and the last term is positive, we know that the binomials are of the form
(4x ϩ _____)(x ϩ _____)
or
(2x ϩ _____)(2x ϩ _____)