回答 - Ch 7:乘数代数表达式
章节大纲
-
Section 7.2 Factoring Out the Greatest Common Factor
::第7.2节 考虑最大共同因素Review
::回顾-
3
a
2
(
12
+
3
a
−
2
a
5
)
::3a2(12+3a-2a5) -
No common factors
::无共同因素 -
::3x(x2-7) -
::5x4(x2+3) -
::2x2x2+5x-1 -
::-2x4(5x2-6x+2-2)或2x2x4(-5x2+6x-2) -
::12xy(1+2y+3y2) -
::a(5a2-7) -
::15y10(3y2+2) -
::4xy( 4yz+x2) -
:x+3)(2x+4)
-
: -
: -
: -
:
::探索更多-
S
A
=
2
π
r
(
r
+
h
)
::SA=2r(r+h) -
S
(
p
)
=
5
p
(
2
+
p
)
::S(p)=5p(2+p) -
Either, when
x
=
0
or
x
=
400
::或, 当 x=0 或 x=400 时
Section 7.3 Factoring Special Quadratics
::第7.3节Review
::回顾-
(
x
+
1
)
(
x
−
1
)
: -
(
x
+
2
)
2
: -
(
4
x
−
3
)
2
: -
-
3
(
x
−
6
)
2
::-3(x-6)2 -
(
12
x
+
7
)
(
12
x
−
7
)
: -
(
14
x
+
5
)
2
: -
Not factorable
::不可考虑因素 -
2
(
9
x
+
2
)
2
::2(9x+2)2 -
(
15
+
x
)
(
15
−
x
)
: -
(
11
−
6
x
)
2
: -
(
x
−
(
10
−
10
√
2
)
)
: -
4
(
8
x
+
13
)
(
8
x
−
13
)
::4(8x+13)(8x-13)
Explore More
::探索更多-
Spencer should have expanded the brackets instead of just squaring everything inside the brackets. He should have expanded
(
2
x
−
5
)
(
2
x
−
5
)
to get the right answer.
::Spencer应该扩大括号,而不是仅仅将括号中的所有内容都分到一起。 他应该扩大(2x-5)(2x-5),以获得正确的答案。 -
f
(
b
)
=
b
2
−
a
2
;
f
(
-
b
)
=
(
-
b
)
2
−
a
2
=
b
2
−
a
2
: -
Using the zero factor property, the hole would have a radius of 3. This does not make sense, for if the hole had a radius of 3, the washer would not exist.
::使用零因数属性,洞的半径为3,这没有意义,因为如果洞的半径为3, 洗衣机就不会存在。
Section 7.4 Factoring the Sum and Difference of Cubes
::第7.4节 计算立方体的总和和差额Review
::回顾-
(
x
−
3
)
(
x
2
+
3
x
+
9
)
: -
(
x
+
4
)
(
x
2
−
4
x
+
16
)
: -
4
(
2
x
−
1
)
(
4
x
2
+
2
x
+
1
)
::4( 2x- 1)( 4x2+2x+1) -
(
4
x
+
7
)
(
16
x
2
−
28
x
+
49
)
: -
(
8
−
9
x
)
(
64
+
72
x
+
81
x
2
)
: -
x
(
5
x
+
2
)
(
25
x
2
−
10
x
+
4
)
: -
81
(
2
x
+
1
)
(
4
x
2
−
2
x
+
1
)
::81(2x+1)(4x2-2-2x+1) -
5
x
3
(
x
−
3
)
(
x
2
+
3
x
+
9
)
::5x3(x-3)(x2+3x+9) -
2
x
4
(
7
x
−
8
)
(
49
x
2
+
56
x
+
64
)
::2x4(7x-8)(49x2+56x+64) -
(
5
x
+
1
)
(
25
x
2
−
5
x
+
1
)
: -
(
4
−
9
x
)
(
16
+
36
x
+
81
x
2
)
: -
x
(
2
x
−
7
)
(
4
x
2
+
14
x
+
49
)
: -
5
x
2
(
x
+
5
)
(
x
2
−
5
x
+
25
)
::5x2(x+5)(x2-5x+25) -
2
(
7
x
+
10
)
(
49
x
2
−
70
x
+
100
)
::2(7x+10)(49x2-70x+100)
Explore More
::探索更多-
V
=
x
(
42
−
2
x
)
(
36
−
2
x
)
;
V
(
1
)
=
1
,
360
i
n
3
;
V
(
3
)
=
3
,
240
i
n
3
;
V
(
5
)
=
4
,
160
i
n
3
.
::V=x(42-2x)(36-2x);V(1)=1,360in3;V(3)=3,240in3;V(5)=4,160in3。 -
2
x
,
x
−
4
, and
x
2
+
4
x
+
16
::2xx, x-4, 和 x2+4x+16 -
b
2
−
4
a
c
=
m
2
−
4
m
2
=
-
3
m
2
::b2-4ac=m2-4m2=-3m2
Section 7.5 Factoring Quadratics When the Lead Coefficient Equals 1
::第7.5节 含铅系数等于1时的保理四方Review
::回顾-
(
x
−
2
)
(
x
+
1
)
: -
(
x
−
4
)
(
x
+
6
)
: -
x
(
x
−
6
)
: -
(
x
+
3
)
(
x
+
3
)
: -
Not factorable with whole number coefficients
::无法用整数系数计算 -
(
x
−
6
)
(
x
−
5
)
: -
(
x
−
2
)
(
x
+
15
)
: -
(
x
+
4
)
(
x
+
7
)
: -
(
x
−
2
)
(
x
−
6
)
: -
(
x
−
11
)
(
x
+
4
)
: -
(
x
−
10
)
(
x
+
2
)
: -
(
x
+
3
)
(
x
+
1
)
: -
Not factorable
::不可考虑因素 -
(
x
−
9
)
(
x
+
4
)
: -
x
(
x
+
1
)
::x(x+1)
Explore More-
Answers will vary. Should include something about the sum
b
having many more options than the factors of
c
.
::答案会有所不同。 答案应该包括总和 b 中比 c 因素有更多的选择。 -
x
2
−
12
x
+
36
=
(
x
−
6
)
(
x
−
6
)
and
x
2
−
64
=
(
x
+
8
)
(
x
−
8
)
.
Preference will vary.
::x2-12x+36=(x-6)(x-6)和 x2-64=(x+8)(x-8)-(8))。
Section 7.6 Factoring Quadratics When the Lead Coefficient is Not Equal to 1
::第7.6节 当铅系数不等于1时的保理四方Review
::回顾-
(
5
x
+
3
)
(
x
+
3
)
: -
3
x
(
2
x
−
7
)
::3x(2x-7) -
(
2
x
+
1
)
(
5
x
−
3
)
: -
(
3
x
−
4
)
(
x
+
2
)
: -
(
2
x
+
1
)
(
2
x
+
3
)
: -
6
(
2
x
2
−
2
x
−
3
)
::6( 2x2 - 2x3) 6( 2x2 - 2x3) -
(
8
x
+
1
)
(
2
x
−
1
)
: -
5
(
x
−
3
)
(
x
−
4
)
::5(x-3)(x-4) 5(x-3)(x-4) -
(
2
x
+
1
)
(
x
+
3
)
: -
3
(
x
2
+
x
+
9
)
::3(x2+x9) -
2
(
4
x
+
1
)
(
x
−
2
)
::2(4x+1)(x-2) -
(
10
x
−
3
)
(
x
+
3
)
: -
(
2
x
+
3
)
(
2
x
+
3
)
: -
5
x
(
3
x
+
7
)
::5x(3x+7) -
(
2
x
−
3
)
(
3
x
−
5
)
:
Explore More
::探索更多-
(
3
x
+
4
)
by
(
3
x
+
4
)
: -
4.5 seconds
::4.5秒 -
Factoring
out
a
leaves
x
2
+
b
a
x
+
c
a
.
To find the factors, you can use the quadratic formula.
::正在计算叶子 x2+bax+ca。 要找到系数, 您可以使用二次公式 。
Section 7.7 Factoring by Grouping
::第7.7节 分组计保Review
::回顾-
(
x
−
4
)
(
x
2
+
3
)
: -
(
x
+
6
)
(
x
+
3
)
(
x
−
3
)
: -
(
3
x
−
4
)
(
x
2
+
5
)
: -
(
2
x
−
3
)
(
x
−
2
)
(
x
2
+
2
x
+
4
)
: -
(
x
+
1
)
(
2
x
+
5
)
(
2
x
−
5
)
: -
(
2
x
+
9
)
(
2
x
2
−
5
)
: -
(
2
x
+
3
)
(
3
x
−
5
)
(
4
x
2
−
6
x
+
9
)
: -
(
5
x
+
2
)
(
3
x
2
−
2
)
: -
(
4
x
+
5
)
(
x
+
5
)
(
x
−
5
)
: -
(
3
x
−
2
)
(
x
2
+
4
)
: -
(
x
−
6
)
(
3
x
+
2
)
(
3
x
−
2
)
: -
(
x
+
3
)
(
x
−
3
)
(
x
2
+
3
x
+
9
)
: -
(
x
−
2
)
(
x
+
2
)
(
x
−
2
)
:
Explore More-
x
,
x
−
9
,
3
x
3
−
2
::x, x- 9, 3x3-2 -
x
3
+
6
x
2
−
12
x
−
72
=
(
x
−
6
)
(
x
2
−
12
)
. Therefore,
x
=
6
or
x
=
√
12
.
::x3+6x2-12x-72=(x-6)(x2-12) 因此, x=6 或 x12。
Section 7.8 Factoring Algebraic Expressions That Are Quadratic in Form
::第7.8节 刻度表层的乘数代数表达式Review
::回顾-
(
x
+
2
)
(
x
−
2
)
(
x
2
−
2
)
: -
(
x
4
+
5
)
(
x
2
−
3
)
(
x
2
+
3
)
: -
(
x
2
−
15
)
(
x
2
−
3
)
: -
(
4
x
2
+
1
)
(
x
2
−
3
)
: -
(
2
x
5
+
1
)
(
3
x
5
+
8
)
: -
2
x
(
3
x
2
−
2
)
(
x
2
+
5
)
::2x(3x2-2)(x2+5) -
(
x
+
2
)
(
x
−
2
)
(
2
x
2
+
3
)
: -
3
x
2
(
4
x
2
+
3
)
(
x
2
+
5
)
::3x2(4x2+3)(x2+5) -
(
3
x
3
−
1
)
(
x
3
+
6
)
: -
(
√
x
−
1
)
(
√
x
−
1
)
:
Explore More-
5
x
,
2
x
+
1
,
x
−
3
::5x, 2x+1, x-3
Section 7.9 Connections: Proving Formulas Geometrically
::第7.9节 连接: 以几何方式检验公式-
A
=
a
2
+
2
a
b
+
b
2
::A=a2+2ab+b2 A=a2+2ab+b2 -
A
=
R
1
+
R
2
+
R
3
+
R
4
::A=R1+R2+R3+R4 -
R
1
+
R
2
+
R
3
+
R
4
=
a
2
+
2
a
b
+
b
2
::R1+R2+R3+R4=a2+2ab+b2 -
A
=
a
2
−
b
2
::A=a2 - b2 -
A
=
(
a
+
b
)
(
a
−
b
)
::A=(a+b)(a-b) -
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
::a2-b2=(a+b)(a-b) -
a
3
+
b
3
::a3+b3 键 -
a
2
(
a
+
b
)
−
(
a
2
b
(
a
−
b
)
+
b
(
a
−
b
)
2
)
: -
(
a
+
b
)
3
−
b
2
(
a
−
b
)
−
b
(
a
−
b
)
2
: -
a
3
+
b
3
=
(
a
+
b
)
(
a
2
−
2
a
b
+
b
2
)
::a3+b3=(a+b)(a2-2ab+b2) -
V
=
a
3
−
b
3
::V=a3-3-b3 级 -
Prism 1:
a
2
(
a
−
b
)
;
Prism 2:
b
a
(
a
−
b
)
;
Prism 3:
b
(
a
−
b
)
2
::棱晶1:a2(a-b);棱晶2:ba(a-b);棱晶3:b(a-b)2 -
V
=
(
a
−
b
)
(
a
2
+
2
a
b
+
b
2
)
::V=(a-b)(a2+2ab+b2)
-
3
a
2
(
12
+
3
a
−
2
a
5
)