Section 6.6 Solving Quadratic Equations by Factoring
9.
()()
2
8150
530
xx
xx
++=
++=
5 0 or 3 0
5 3
xx
xx
+= +=
=− =−
Check −5:
{
}
{
}
10.
()()
2
560
320
xx
xx
++=
++=
30 or 20
3 2
xx
xx
+= +=
=− =−
Check −3:
{
}
{
}
11.
()()
2
2150
350
xx
xx
−−=
+−=
{
}
{
}
12.
()()
2
42 0
760
xx
xx
+− =
+−=
7 0 or 6 0
7 6
xx
xx
+= −=
=− =
The solution set is
{
}
7,6 .−
()()
920
xx
+−=
9 0 or 2 0
9 2
xx
xx
+= −=
=− =
The solution set is
{
}
9, 2 .−
2
The solution set is
{
}
1, 10 .
17.
2
40
xx
+=
Chapter 6 Factoring Polynomials
18.
()
2
60
60
xx
xx
−=
−=
0 or 6 0
6
xx
x
=−=
=
{
}
20.
()
2
30
30
xx
xx
+=
+=
0 or 3 0
3
xx
x
=+=
=−
The solution set is
{
}
0, 3 .−
22.
()
2
2
8
80
80
0 or 8 0
8
xx
xx
xx
xx
x
=
−=
−=
=−=
=
The solution set is
{
}
0,8 .
23.
2
25
xx
=
24.
()
2
2
35
350
350
0 or 3 5 0
xx
xx
xx
xx
=
−=
−=
=−=
()
350
350
0 or 3 5 0
3 5
5
3
xx
xx
xx
x
x
+=
+=
=+=
=−
=−
2 3
3
2
x
x
=−
=−
The solution set is
3,0 .
2
−
27.
()
2
2
440
20
20
xx
x
x
++=
−=
+=
Section 6.6 Solving Quadratic Equations by Factoring
29.
()
2
2
2
12 36
12 36 0
60
60
xx
xx
x
x
=−
−+=
−=
−=
31.
()
2
2
2
4129
41290
23 0
230
23
3
2
xx
xx
x
x
x
x
=−
−+=
−=
−=
=
=
33.
()()
2
2
274
2740
21 40
2 1 0 or 4 0
xx
xx
xx
xx
=+
−−=
+−=
+= − =
3 4 1
4
3
xx
x
==−
=
The solution set is
4
1, .
3
−
35.
2
2
518
5180
xx
xx
=−
+− =
Chapter 6 Factoring Polynomials
37.
()()
2
49 0
770
x
xx
−=
+−=
7 0 or 7 0
7 7
xx
xx
+= −=
=− =
The solution set is
{
}
7,7 .−
{
}
39.
()()
2
4250
25250
x
xx
−=
+−=
2 5 0 or
25
5
2
x
x
x
+=
=−
=−
250
25
5
2
x
x
x
−=
=
=
The solution set is
55
,.
22
−
{
}
41.
()()
2
2
81 25
81 25 0
95950
x
x
xx
=
−=
+−=
9 5 0 or
x
+=
950
x
−=
42.
()()
2
2
25 49
25 49 0
57570
x
x
xx
=
−=
+−=
570 or
57
x
x
+=
=−
570
57
x
x
−=
=
()()
2
2
421
4210
370
xx
xx
xx
−=
−−=
+−=
30 or
3
x
x
+=
=−
70
7
x
x
−=
=
The solution set is
{
}
3, 7 .−
44.
()
2
318
318
xx
xx
−=
−=
()()
44150
25230
xx
xx
+−=
+−=
250 or
25
5
x
x
+=
=−
230
23
3
x
x
−=
=
Section 6.6 Solving Quadratic Equations by Factoring
46.
()
2
38 5
38 5
xx
xx
+=−
+=−
47.
()( )
{
}
{
}
2
1414
3414
xx
xx
−+=
+−=
48.
()()
()()
2
2
3830
524 30
560
320
xx
xx
xx
xx
−+=−
+−=−
++=
++=
30
3
x
x
+=
=−
or
20
2
x
x
+=
=−
The solution set is
{
}
3, 2 .−−
50.
()( )
2
33 5 7
314157
xx
xx
++=
++=
51.
()()
2
816 1
81616
yy y
yyy
+= −
+= −
52.
()( )
()()
2
2
9425
9820
20 0
540
yy y
yyy
yy
yy
+= +
+=+
+− =
+−=
5 0 or 4 0
5 4
yy
yy
+= −=
=− =
The solution set is
{
}
5, 4 .−
Chapter 6 Factoring Polynomials
{
}
55.
()
2
2
2
64 48 9
64 48 9 0
83 0
830
83
3
8
ww
ww
w
w
w
w
=−
−+=
−=
−=
=
=
The solution set is
3.
8
{
}
57.
()
()
()()()
2
4560
4320
4 0 or 3 0 or 2 0
4 3 2
xxx
xxx
xx x
xx x
−++=
−++=
−= += +=
==−=−
The solution set is
{
}
3, 2, 4 .−−
{
}
()
()()
2
40
220
xx
xx x
−=
+−=
0 or 2 0 or 2 0
2 2
xx x
xx
=+=−=
=− =
The solution set is
{
}
2, 0, 2 .−
61.
()
()()
32
2
320
320
210
yyy
yy y
yy y
++=
++=
++=
The solution set is
{
}
3, 0,1 .−
63.
() ( )
22
2 4 50 46xxxx x−+= +−
()
222
222
22
2 8 16 50 46
21632 4
31632 4
xx xx xx
xx xxx
xx xx
−+ + = + −
−++=+
−+=+
Section 6.6 Solving Quadratic Equations by Factoring
64.
222
(4)(5)(23)(1) (225)13
9202 32 2513
xx xx xx
xx xx x x
−−++−= −−
−++ +−= − −
The solution set is
{
}
15, 2 .−−
65.
()()
() ()
()()
2
25 260
23 22 0
540
xx
xx
xx
−−−+=
−− −− =
−−=
5 0 or 4 0
5 4
xx
xx
−= −=
==
The solution set is
{
}
4,5 .
{
}
67.
2
16 20 300htt=− + +
Substitute 0 for h and solve for t.
2
016 20300
tt
=− + +
68.
2
16 20 300htt=− + +
Substitute 304 for h and solve for t.
4 1 1
10.25
4
tt
t
==
==
The ball’s height will be 304 feet at two times: 0.25
second and 1 second after it is thrown. These
solutions correspond to the points
()()
0.25,304 and 1,304
on the graph.
69. Substitute 276 for h and solve for t.
2
Reject
3
4
t=−
since time cannot be negative. The
ball’s height will be 276 feet 2 seconds after it is
Chapter 6 Factoring Polynomials
71.
2
16 72htt=− +
Substitute 32 for h and solve for t.
2
2
32 16 72
16 72 32 0
tt
tt
=− +
−+=
72.
2
21282Sx x=−+
Substitute 72 for S and solve for x.
2
2
21282
72 2 12 82
Sx x
xx
=−+
=−+
73.
2
21282Sx x=−+
Substitute 72 for S and solve for x.
2
2
21282
66 2 12 82
Sx x
xx
=−+
=−+
74. International travelers spent $72 billion 1 year and 5
years after 2000. This corresponds to the point
()
1, 72
and the point
()
5, 72
on the graph.
()( )
5 62830
xx
−−=
6 0 or 2 83 0
6 2 83
83
41.5
2
xx
xx
x
−= − =
==
==
()( )
5102750
xx
−−=
10 0
10
x
x
−=
=
or
2750
275
75 or 37.5
x
x
−=
=
Section 6.6 Solving Quadratic Equations by Factoring
80.
2
2
tt
N−
=
Substitute 36 for N and solve for t.
2
36 2
tt
−
=
81.
2
2
tt
N−
=
Substitute 45 for N and solve for t.
2
2
45 2
245 2 2
tt
tt
−
=
−
⋅=
82. Let x = the width of the garden.
Then 5x+ = the length.
()()
2
5 300
5 300
lw A
xx
xx
⋅=
+=
+=
83. Let x = the width of the parking lot.
Then 3x+ = the length.
()()
2
3 180
3 180
lw A
xx
xx
⋅=
+=
+=
()
()
()
2
2
2
12160
2
1
221260
2
2 1 120
2120
xx
xx
xx
xx
−=
−=⋅
−=
−=
be negative. Then 8 and 2 1 15xx=−=, so the base
is 8 inches and the height is 15 inches.
Chapter 6 Factoring Polynomials
85. Use the formula for the area of a triangle where x is
the base and 1x+ is the height.
()
1
2
1115
2
bh A
xx
=
+=
86. Let x = the width of the path surrounding the garden
(as shown in the figure).
Area of vacant lot =
()()
215212xx++
()()
2
2 15 2 12 378
42430180378
xx
xxx
++=
+++=
87. a. Area of border
()()
Area of
Area of a large rectangle flower bed
2
2
2 12 2 10 10 12
4 20 24 120 120
444
xx
xxx
xx
=+ +−⋅
=+++−
=+
b. Find the width of the border for which the area
of the border would be 168 square feet.
()
2
2
2
4 44 168
4 44 168 0
411420
xx
xx
xx
+=
+−=
+−=
92. does not make sense; Explanations will vary.
Sample explanation: The quadratic will be
2
25120.xx−−=
93. makes sense
94. does not make sense; Explanations will vary.
have fewer than 2 solutions
98. true
99. If −3 and 5 are solutions of the quadratic equation,
then
()
33xx−− = + and 5x− must be factors of
the polynomial on the left side when the quadratic
Section 6.6 Solving Quadratic Equations by Factoring
100.
()
()
32
32
16 16 0
16 16 0
xx x
xx x
−− +=
−+−+=
101. 2
920
31
xx
−+
=
Because
0
31= (and there is no other power of 3
that is equal to 1),
2
9200.xx−+= Solve this
equation.
102.
()
3
2
55 1xx−+ =
The only number that can be cubed (raised to the
third power) to give 1 is 1. Therefore, the given
equation is equivalent to the quadratic equation
103.
2
2yx x=−−
To match this equation with its graph, find the
intercepts.
To find the y-intercept, let x = 0 and solve for .y
The x intercepts are −1 and 2.
The only graph with y–intercept −2 and x–
intercepts −1 and 2 is graph, c, so this is the graph
The y-intercept is −2.
To find the x−intercepts, let 0y= and solve for
.x
2
02
xx
=+−
105.
2
4yx=−
To match this equation with its graph, find the
intercepts.
To find the y-intercepts, let x = 0 and solve for .y
2
04 4y=−=−
The only graph with y–intercept −4 and x-intercepts
−2 and 2 is graph d, so this is the graph of
2
4.yx=−