Section 6.3 Factoring Trinomials Whose Leading Coefficient is Not 1
92. a.
()()
2
35231 2xx x x+−= − +
93.
2
32
32
2
352
23 11 12 4
36
5 12
xx
xxxx
xx
xx
−+
−−+−
−
−+
94.
2
32
32
253
22 13 6
24
xx
xxxx
xx
+−
−+−+
−
95. a.
2
32xx++
b.
()()
21xx++
96.
()()
2
2 3121 1.xx x x++= + +
104. does not make sense; Explanations will vary.
Sample explanation: The polynomial can be
factored further because
2
56(2)(3).xx x x−+=− −
105. true
A sample change is:
321xx++
is prime.
109.
2
32xbx++
The possible factorizations that will give
2
3x
as the
first term and 2 as the last term are:
Chapter 6 Factoring Polynomials
110.
2
23xbx++
The possible factorizations that will give
2
2x
as the
first term and 3 as the last term are:
()()
2
23 12 53
xx xx
++=++
111.
10 5
3415xx−−
Since
()
2
510
xx=
, the first terms of the factors
112.
2
274
nn
xx−−
113.
4105
710
xy
xy
−=
+=−
Multiply the second equation by –4 and then add the
equations.
4105
428 40
29 145
xy
xy
y
−=
−− =
−=
114.
4
0.00086 8.6 10−
=×
115.
1
88
66
x
x−=−
Multiply both sides by the LCD of 6.
1
x
2
81 100
x
=−
22 2
32 2
(24)2(24)
24248
xx x x x
xxxxx
=−++−+
=− ++ −+
1. The GCF is
4
.x
()
54 4
1xx xx+= +
2.
()()
2
718 2 9xx x x+−=− +
3. The GCF is
2
.xy
The polynomial in the
parentheses is prime because there are no factors of
()()
()( )
22
7223721 3
7313
37 1
xx xxx
xx x
xx
−+=−−+
=−−−
=− −
Section 6.4 Factoring Special Forms
6. Factor
32
5315xxx+++
by grouping.
()
7. The GCF is .x
()
()()
32 2
2115 2115
21 5
xxxxxx
xx x
−+= −+
=−−
A
9. Factor
22
17 30xxyy−+
by trial and error. The
only factors of 30 whose sum is
17−
are
15−
and
2.−
()()
22
17 30 15 2xxyyxyxy−+ =− −
A
11. The GCF is 2.
()
22
16 70 24 2 8 35 12xx xx−+= −+
Factor the polynomial in parentheses by grouping.
12. Factor
22
3107xxyy++
by grouping.
3 and 7, so 21.acac== =
The only factors of 21 whose sum is 10 are 3 and 7.
13. First, factor out the greatest common factor.
32 2
6 8 30 2 (3 4 15)xx x xxx−+ + =− −−
Then, factor the trinomial.
32 2
6 8 30 2 (3 4 15)
2(3 5)( 3)
xx x xxx
xx x
−+ + =− −−
=− + −
Thus, factor using
()().AB ABAB−=+ −
222
81 9
(9)(9)
xx
xx
−= −
=+ −
b. Notice that the trinomial fits the form
22
.
A
B−
55
25 4 5 (2 )
(5 2 )(5 2 )
xx
xx
−=−
=+ −
b. Notice that the trinomial fits the form
22
.
A
B−
2 (3 1)(3 1)
xx x
=+−
b. First factor out the GCF.
22
72 18 18(4 )xx−=−
Chapter 6 Factoring Polynomials
A
4. First, factor the difference of two squares.
422
81 16 (9 4)(9 4)xxx−= + −
5. a. Notice that the trinomial fits the form
22
2.
A
AB B++
Thus, factor using
222
2().AABBAB++=+
22
14 49 ( 7)xx x++=+
A
c. Notice that the trinomial fits the form
A
6. Notice that the trinomial fits the form
22
2.
A
AB B++
7. Notice that the polynomial fits the form
33
.
A
B+
(3)( 39)
xxx
=+ −+
8. Notice that the polynomial fits the form
33
.
A
B−
Thus it factors as
22
()( ).ABA ABB−++
22
2
125 8 (5 ) 2
(5 2) (5 ) (5 )(2) 2
xx
xxx
+= +
=+ − +
1.
()()
A
BAB+−
2.
2
()AB+
3.
2
()AB−
6. 6x; 6x
7. 6
10. 3; +9
13. false
6.4 Exercise Set
1.
()()
222
25 5 5 5xx xx−=−=+ −
Section 6.4 Factoring Special Forms
5.
() ( )( )
2
22
492 32323xx xx−= − = + −
9.
() ( )( )
2
22
149 1 7 17 17xxxx−=− =+ −
10.
()()()
2
22
164 1 8 18 18xxxx−=− =+−
14.
() ( )( )
2
42222
25 5 5 5xx xx−= −= + −
16.
()
()()
2
422
22
49 25 7 5
7575
yy
yy
−= −
=+ −
19.
() ( )
()()
22
22
25 16 5 4
5454
xyx y
xyxy
−= −
=+ −
22.
() ()
22
14 4 7 2
xy x y
−= −
()()
()( )
()
()()
422
222
2
16 4 4
42
422.
xxx
xx
xxx
−= + −
=+ −
=++−
()
()()
2
111.
xxx
=++−
25.
()
()
()()
2
422
2
16 81 4 9
492323
xx
xxx
−= −
=+ + −
26.
()
2
422
81 1 9 1
xx
−= −
27.
()
()()
22
2182 923 3xxxx−= −= + −
22
Chapter 6 Factoring Polynomials
33.
()
32
3273 9xxxx+= +
34.
()
32
3153 5xxxx+= +
35.
()
()()
22
18 2 2 9 2 3 3yy yy− =−=+ −
38.
()
()()
32
3753 25
355
yyyy
yy y
−= −
=+−
39.
()
()()
32
18 2 2 9 1
23 13 1
xxxx
xx x
−= −
=+−
42.
22
444(1)
4( 1)( 1)
xx
xx
−+=− −
=− + −
43.
32
520 5(4)
5( 2)( 2)
yyyy
yy y
−+ =− −
=− + −
()
5
x
=−
49.
()
()
222
2
21 21 1
1
xx x x
x
−+=− +
=−
50.
()
222
2
44 22 2
xx x x
−+=− +
52.
()
()
2
24 144 2 12 12
12
xx x x
x
++=+ +
=+
53.
() ()
()
2
22
2
4412 221
21
xx x x
x
++= + +
=+
2
22
56.
() ()
()
2
22
2
64 16 1 8 2 8 1
81
yy y y
y
−+= − +
=−
57.
2
10 100xx−+
is prime.
To be a perfect square trinomial, the middle term
Section 6.4 Factoring Special Forms
62.
()()
()
2
222
2
18 81 2 9 9
9
xxyyx xyy
xy
−+ =− +
=−
65.
22
16 40 25xxyy−+
() ( )()
()
22
2
42455
45
xxyy
xy
=−⋅+
=−
68.
()
() ( )
22
22
2
18 24 8 2 9 12 4
23 232 2
xx xx
xx
++= ++
=+⋅+
71.
()
()
22
2
2422 21
21
yy yy
y
−+= −+
=−
72.
()
22
2402002 20100
yy yy
−+= −+
75.
22
2
624246(44)
6( 2)
xx xx
x
−+ −=− −+
=− −
76.
22
530455(69)
xx xx
−+ −=− −+
()
()
()
()
22
2
111
11
xxx
xxx
=+ −⋅+
=+ −+
333 2 2
Chapter 6 Factoring Polynomials
82.
()
()
()
()
333 2 2
2
64 4 4 4 4
4 4 16
xx xxx
xxx
−=−=− +⋅+
=− ++
84.
()
()()
()
()
3
33
22
2
27 1 3 1
313 311
319 31
yy
yyy
yyy
−= −
=− +⋅+
=− ++
87.
33
64xy −
()
()()
()
()
33
22
22
4
444
4416
xy
xy xy xy
xy x y xy
=−
=− +⋅+
=− ++
88.
()
()()
()
()
3
33 3
22
22
27 3
333
339
xy xy
xy xy xy
xy x y xy
−= −
=− +⋅+
=− ++
90.
()
()
43
33
64 64
4
yy y y
yy
−= −
=−
() ()
()
()
2
2
2
23 2 3 32 2
23 2 9 6 4
yyy
yyy
=− +⋅+
=− ++
92.
()
33
128 250 2 64 125
yy
−=−
94.
() ()
( )() ()
()
()
33
33
22
22
827 2 3
23 2 23 3
234 6 9
xyx y
xy x xy y
xyx xyy
+= +
=+ −⋅+
=+ −+
95.
() ( )
()() ()
()
()
33
33
22
22
125 64 5 4
54 5 54 4
5 4 25 20 16
xyx y
xy x xy y
xy x xy y
−= −
=− +⋅+
=− + +
Section 6.4 Factoring Special Forms
98.
()
2
2
2
93
16 4
25 5
33
44
55
xx
xx
−= −
=+ −
100.
43
3
3
2
2
2
1
88
1
2
111
222
11
224
y
yyy
yy
yy y y
y
yy y
−= −
=−
=− +⋅+
=− ++
102.
()
()
()()
32
22
0.64 0.64
0.8
0.8 0.8
xx x x
xx
xx x
−= −
=−
=+ −
105.
2
32
32
2
21
353
3
2 5
xx
xxxx
xx
xx
++
−−−−
−
−
32
2
2
2
6 3
6 12
9 18
9 18
xx
xx
xx
x
x
−
−
−
−
−
The quotient
2
69xx++
factors further.
()
2
2
69 3xx x++=+
.
Thus,
()()
2
32
4318 2 3xxx x x+−−=− +
.
Area of the smaller square =
22 4⋅=
Area of the shaded area is area of the larger square
minus the area of the smaller square:
()()
2
9 43232xxx−= + −
109. Area of large square =
2
x
Chapter 6 Factoring Polynomials
111. – 114. Answers will vary.
115. does not make sense; Explanations will vary.
Sample explanation: The original expression has a
118. makes sense
119. false; Changes to make the statement true will vary.
A sample change is:
2
25x+
is prime.
120. true
123. The error in the proof is in step 7 where you are
asked to divide by
0.ab−=
Division by 0 is not permitted because division by
zero is undefined.
125.
() ( )
()()
22
22
25 5
55
nnn n
nnnn
xyx y
xyxy
−=−
=+ −
127.
()()
2
32 31xx+− ++
()
2
2
31
x
=+−
Therefore, for
2
91xkx++
to be a perfect square
trinomial, k must be −6 or 6.
129.
2
64 16xxk−+
Let r be the number such that
2
.rk=
Then,
()
2
22
64 16 8 2 8 .xxkx xrr−+= −⋅⋅+
The expression on the right should be changed to
()()
2323xx+−
.
131. The graphs coincide.
2
2
to
()
21x−
.
133. The graphs do not coincide.
()
()
333 2
11 1 1xx xxx−= − = − + +
The polynomial on the right side should be changed
Section 6.5 A General Factoring Strategy
135.
()()
22
10 5 2 14 5 1xx xx−+− −−
()( )
()
()()
22
22
2
10 5 2 14 5 1
10 14 5 5 2 1
43
xx xx
xx xx
x
=−++−++
=−+−+++
=− +
137.
32
3753( 25)3(5)(5)x x xx xx x−= −= + −
6.5 Check Points
1. First use common factoring.
4222
545 5(9)xxxx−= −
4( 4)( 4)
xx x
=+−
Finally, use difference of two squares again.
54
22
2
4644(16)
4( 4)( 4)
4 ( 4)( 2)( 2)
xxxx
xx x
xx x x
−= −
=+−
= ++−
4. Use factor by grouping.
2
3( 5)
xx
=−
3483(16)xy y yx y−= −
Then use difference of two squares.
4544
2222
3483(16)
3( 4 )( 4 )
xy y yx y
yx y x y
−= −
=+ −
Chapter 6 Factoring Polynomials
6.5 Concept and Vocabulary Check
1. b
2. e
6.5 Exercise Set
1.
2
735 7(5)xxxx−+ =− −
2.
2
624 6(4)xxxx−+ =− −
6.
32
64 1 (4 1)(16 4 1)xxxx−= − + +
7.
2
55 5( )( )
(5 )( )
xyxxy xyxxy
xx y
+++= ++ +
=+ +
13.
33 22
27 8 (3 2)(9 6 4)xy xy xy xy+= + − +
14.
33 22
216 125 (6 5)(36 30 25)x y xy x y xy+= + − +
15.
2
615(35)(23)xx x x+− = + −
19.
()
32
777 1xxxx+= +
20.
()
32
6246 4xxxx+= +
21.
()
()()
22
55305 6
52 3
xx xx
xx
−−= −−
=+ −
23.
()
()()
()
()()
44
22
2
21622 81
29 9
2933
xx
xx
xxx
−= −
=+ −
=++−
44
Section 6.5 A General Factoring Strategy
27.
()
()
32 2
2
324 483 816
34
xxxxxx
xx
−+= −+
=−
29.
()
()
()
52 23
22
22 2 1
21 1
xx xx
xx x x
+= +
=+−+
31.
()
2
68234xxxx+= +
32.
()
2
21 35 7 3 5xxxx−= −
35.
()
()
43222
2
2
7147 7 21
71
yyyyyy
yy
++= ++
=+
()()
24 1 2 1
yy
=+ −
40.
()
()()
22
32 4 6 2 16 2 3
28 3 2 1
yy yy
yy
+−= +−
=− +
42.
()
3273 9rrrr−= −
43.
()()
2
4852521ww w w+−= + −
47.
2
64x+ is prime because it is the sum of two
squares with no common factor other than 1.
48.
2
36y+ is prime because it is the sum of two
()
()()
()
()
2
2
24 2
24
yy y
yy
=+−+
=+ −
Chapter 6 Factoring Polynomials
53.
() ( )
()
2
22
2
16 24 9 4 2 4 3 3
43
yy y y
y
++= + ⋅+
=+
56.
()
()()
32 2
721147 32
721
yyyyyy
yy y
−+= −+
=−−
58.
()
()()
()
()()
54
22
2
16 16
44
422
yyyy
yy y
yy y y
−= −
=+ −
= ++−
61.
43222
9186 3(362)xxxxxx++= ++
62.
43222
10 20 15 5 (2 4 3)xxxxxx++= ++
67.
2
964y+ is prime because it is the sum of two
squares with no common factor other than 1.
()()()
23 5 5
yyy
=+ + −
70.
()
()
32 32
2
12 16 3 4 12 16 3 4
yyy yy y
+−−= + +−−
71.
()
()()
230682 1534
2172
rrrrrr
rr r
+−= +−
=+ −
72.
()
()( )
32 2
3 27 210 3 9 70
3514
rr rrrr
rr r
−−= −−
=+−
()()
()
()()
52 2
52
11
111
yy y
yy y y
=+−
= ++−
Section 6.5 A General Factoring Strategy
78.
()
()
()
()
43
33
2
27 27
3
339
xxxx
xx
xx x x
+= +
=+
=+ −+
81.
()
2
68 234xxyxxy+= +
82.
()
2
21 35 7 3 5xxyxxy−= −
83.
()()
7321 7 321
xy x y xy x y
−+−=−+−
86.
()()
22
412 6 2xxyyxyxy−− =− +
87.
32 2 42
72 12 24ab a ab+−
()
22 22
12 6 1 2aab ab=+−
91.
()
()()
42 22
2
48 3 3 16 1
34141
xy xy xy x
xy x x
−= −
=+−
92.
()
32 2 2 2
16 4 4 4 1
ab ab ab a
−= −
95.
55
77xy xy−
()
()()
()
()()
44
2222
22
7
7
7
xy x y
xy x y x y
xy x y x y x y
=−
=+−
=++−
98.
()
()()
22
18 57 30
3 6 19 10
3 2 5 3 2
xy xy xy
xy x xy y
xy x y x y
++
=++
=++
99.
2
2 44 242bx bx b++
()
2
222121
bx x
=++
Chapter 6 Factoring Polynomials
103.
3223
36 62 12xy xy xy−+
105.
2222
ay by ax bx−−+
()( )
()()
()
()
()()()
22 22
22 22
22
ay by ax bx
ya b xa b
abyx
ababyx
=−+−+
=−−−
=− −
=+ − −
107.
32
91514ax ax ax+−
()
()()
2
91514
3732
ax x x
ax x x
=+−
=+−
109.
43 22 22 2
26 2 2(3 )xxyxy xxxyy++ = ++
110.
43 22 22 2
39 3 3(3 )xxyxy xxxyy−+ = −+
112.
()
45 44
16 16
xyy yxy
−= −
()
()
()( )( )
2
110 7 6
15 62 1
xxx
xx x
=+ −−
=+ − +
114.
() ()()
()
()
()( )( )
2
2
12 1 4 1 5 1
112 4 5
16 52 1
xx xx x
xxx
xx x
−− −− −
=− −−
=− − +
()()
72 72
xaxa
=−+ −−
118.
() ()()
222
2
69 6 3
xax a
−− =−−
119.
()
()
()()
22
2
2
22
81625
816 5
45
xx a
xx a
xa
++−
=++−
=+ −
Section 6.5 A General Factoring Strategy
121.
()()
()( )
3
76 23
242
11
11
yyyy yy
yy y y
+= + = +
=+−+
123.
()
()()
22
256 16 16 16
16 4 4
tt
tt
−= −
=+−
124. Factoring out the height from the volume gives
125. Area of outer circle =
2
b
π
Area of inner circle =
2
a
π
Area of shaded ring =
22
ba
ππ
−
()
()()
22 22
ba ba
baba
ππ π
π
−= −
=+ −
126. – 128. Answers will vary.
129. makes sense
134. false; Changes to make the statement true will vary.
A sample change is:
22
4 100 4( 25)xx+= +
135. false; Changes to make the statement true will vary.
A sample change is: Some polynomials are
completely factored after one step.
()
()()
()
()
()( )( )
22
22
2
5141
514
5122
yyy y
yy y
yy y y
=−−−
=−−
=−+−
() ()
2
520 5100xx+− ++
()()()
()
()
22
2
2
52 51010
510
5
xx
x
x
=+ − + +
=+−
=−
141.
()
() ( )
2222
22
327 3 9
33
nnnn
nn
xyxy
xy
−=−
=−
Chapter 6 Factoring Polynomials
143. The graphs do not coincide.
144. The graphs coincide.
This verifies that the factorization
()()
2
6104231 2xx xx+−= − +
is correct.
146. The graphs do not coincide.
32
210 210xxx+−−
()
32
25 5
xxx
=+−−
147.
()
()()
2
22
9163 4
3434
xx
xx
−= −
=+ −
148. 52 10xy−=
To find the x-intercept, let 0.y=
()
52010
510
x
x
−=
=
149. Let x= the measure of the first angle.
20
x
=
Measure of first angle = x = 20°
Measure of second angle = 3x = 60°
Measure of third angle = x + 80 = 100°
()
11 2
3
1
02
3
0
=−
=
=
0
=
152.
2
2
(2)(3)6 3266
12
(4)(3)
xx xxx
xx
xx
−+−=+−−−
=+−
=+ −
6.6 Check Points
Section 6.6 Solving Quadratic Equations by Factoring
2. All the terms are on one side and zero is on the
other side. Thus, factor.
{
}
2
650
xx
−+=
3. Move all terms to one side and obtain zero on the
other side. Then factor.
2
2
42
420
2(2 1) 0
xx
xx
xx
=
−=
−=
4. Move all terms to one side and obtain zero on the
other side. Then factor.
2
2
2
10 25
10 25 0
(5) 0
xx
xx
x
=−
−+=
−=
Because both factors are the same, it is only
5. Move all terms to one side and obtain zero on the
other side. Then factor.
2
16 25
x
=
55
44
xx
=− =
The solution set is
55
,.
44
−
6. Write the equation in standard form by finding the
(9)(2)0
xx
−+=
90 or 20
92
xx
xx
−= +=
==−
The solution set is
{
}
2,9 .−
7.
2
16 48 160
htt
=− + +
Chapter 6 Factoring Polynomials
8. Let x= the width of the sign.
Let
3x+=
the length of the sign.
The area of 54 square units can be represented as
follows.
Alw
=⋅
6.6 Concept and Vocabulary Check
1. quadratic equation
2.
0 or 0AB==
6.6 Exercise Set
1.
()
70xx+=
0 or 7 0
7
xx
x
=+=
=−
The solution set is
{
}
7,0 .−
{
}
{
}
4.
()()
380
3 0 or 8 0
3 8
xx
xx
xx
−+=
−= +=
==−
The solution set is
{
}
8, 3 .−
7 0 or 3 2 0
7 3 2
2
3
xx
xx
x
+= −=
=− =
=
The solution set is
2
7, .
−
2
x
=−
The solution set is
9,4 .
2
−
8.
()( )
853110
50 or 3 110
5 3 11
xx
xx
xx
−+=
−= + =
==−