6.7 以分组方式进行保理
Section outline
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The volume of a rectangular prism is 3 x 5 − 27 x 4 − 2 x 2 + 18 x . What are the lengths of the prism's sides?
::矩形棱柱的体积是 3x5 - 27x4 - 2x2+18x。 棱柱面的长度是多少?Factoring by Grouping
::分组计数You have already been introduced to factoring by grouping . We will expand this idea to other here.
::您已经通过分组被引入保理。 我们将把这个想法推广到这里的其他人 。Factor the polynomial by grouping: x 4 + 7 x 3 − 8 x − 56
::通过组合( x4+7x3- 8x- 56) 乘以多圆性: x4+7x3- 8x-56First, group the first two and last two terms together. Pull out any common factors.
::首先,将前两个和最后两个条件组合在一起,找出任何共同因素。x 4 + 7 x 3 ⏟ x 3 ( x + 7 ) − 8 x − 56 ⏟ − 8 ( x + 7 )
::x4+7x3_x3( x+7)-8x-56_8( x+7)Notice what is inside the parenthesis is the same . This should always happen when factoring by grouping. Pull out this common factor .
::注意括号内的内容是相同的。 当按组计时, 总是会发生这种情况。 拔出这个共同因素 。x 3 ( x + 7 ) − 8 ( x + 7 ) ( x + 7 ) ( x 3 − 8 )
::x3(x+7)-8(x+7)(x+7)(x7)(x3-8)Look at the factors. Can they be factored any further? Yes. The second factor is a difference of cubes. Use the formula .
::查看各种因素。 它们是否可以被进一步考虑? 是的。 第二个因素是立方体的差别。 使用公式 。( x + 7 ) ( x 3 − 8 ) ( x + 7 ) ( x − 2 ) ( x 2 + 2 x + 4 )
:x+7)(x3-8)(x+7)(x-2)(x2+2x+4))
Factor the polynomial by grouping: x 3 + 5 x 2 − x − 5
::通过组别( x3+5x2- x- 5) 来乘以多数值Follow the steps from above.
::遵循从上到下的步骤。x 3 + 5 x 2 − x − 5 x 2 ( x + 5 ) − 1 ( x + 5 ) ( x + 5 ) ( x 2 − 1 )
::x3+5x2 - x5x2 - x5x2(x+5) - 1(x+5)(x+5)(x5)(x2-1)Look to see if we can factor either factor further. Yes, the second factor is a difference of squares .
::看看我们能否进一步考虑其中任何一个因素。是的,第二个因素是方形的差别。( x + 5 ) ( x 2 − 1 ) ( x + 5 ) ( x − 1 ) ( x + 1 )
:x+5)(x2-1)(x+5)(x-1)(x+5)(x-1)(x+1)(x+1))
Now, let's find all real-number solutions of 2 x 3 − 3 x 2 + 8 x − 12 = 0 .
::现在,让我们找到所有2x3-3x2+8x-12=0的真数字解决方案。Follow the steps from above.
::遵循从上到下的步骤。2 x 3 − 3 x 2 + 8 x − 12 = 0 x 2 ( 2 x − 3 ) + 4 ( 2 x − 3 ) = 0 ( 2 x − 3 ) ( x 2 + 4 ) = 0
::2x3-3x2+8x-12=0x2(2x-3)+4(2x-3)=0(2x-3)(x2+4)=0Now, determine if you can factor further. No, x 2 + 4 is a sum of squares and not factorable. Setting the first factor equal to zero, we get x = 3 2 .
::现在, 确定您能否进一步系数 。 否, x2+4 是平方之和, 不可计算。 设定第一个系数为零, 我们得到 x=32 。Examples
::实例Example 1
::例1Earlier, you were asked to find the lengths of the prism's sides.
::早些时候,有人要求你找出棱镜两面的长度。We need to factor 3 x 5 − 27 x 4 − 2 x 2 + 18 x to find the lengths of the prism's sides.
::我们需要乘以 3x5 - 27x4 - 2x2+18x 才能找到棱晶两面的长度 。First, pull out the common factor. x ( 3 x 4 − 27 x 3 − 2 x + 18 )
::首先,拉出共同系数 x( 3x4- 27x3-2x+18)Next, factor ( 3 x 4 − 27 x 3 − 2 x + 18 ) by grouping the first two and last two terms together.
::下一个系数(3x4-27x3-2x3+18),将前两个和最后两个术语组合在一起。( 3 x 4 − 27 x 3 − 2 x + 18 ) = ( 3 x 4 − 27 x 3 ) + ( − 2 x + 18 ) = 3 x 3 ( x − 9 ) − 2 ( x − 9 )
:3x4-27x3-2x+18)=(3x4-2x3+3)+(-2x+18)=3x3(x-9)-(2x-9)
Now pull out the common factor.
::现在拿出共同因素。3 x 3 ( x − 9 ) − 2 ( x − 9 ) = ( 3 x 3 − 2 ) ( x − 9 )
::3x3(x-9)-(2x-9)=(3x3-2)(x-9)The expression can't be factored further, so 3 x 5 − 27 x 4 − 2 x 2 + 18 x = x ( 3 x 3 − 2 ) ( x − 9 ) and the lengths of the sides of the rectangular prism are x , 3 x 3 − 2 , and x − 9 .
::不能进一步计算表达式, 所以 3x5- 27x4- 2x2+18x=xx( 3x3-2)( x- 9) , 矩形棱柱两侧的长度是 x, 3x3-2, 和 x- 9 。Each of these problems is done in the same way: Group the first two and last two terms together, pull out any common factors, what is inside the parenthesis is the same, factor it out, then determine if either factor can be factored further.
::每一个问题都是以同样的方式解决的:将前两个和最后两个条件组合起来,找出任何共同因素,括号内的内容相同,因素出来,然后确定其中任何一个因素是否可以进一步考虑。Example 2
::例2Factor by grouping x 3 + 7 x 2 − 2 x − 14
::通过组别 x3+7x2- 2x- 14 进行乘数
x 3 + 7 x 2 − 2 x − 14 x 2 ( x + 7 ) − 2 ( x + 7 ) ( x + 7 ) ( x 2 − 2 )
::x3+7x2 - 2x- 14x2x2(x+7)-2(x+7)(x+7)(x2-2)x 2 − 2 is not a difference of squares because 2 is not a square number . Therefore , this cannot be factored further.
::x2-2 不是平方差, 因为 2 不是平方数字, 因此无法进一步计算 。Example 3
::例3Factor by grouping 2 x 4 − 5 x 3 + 2 x − 5
::按组别 2x4 - 5x3+2x- 5 乘以系数2 x 4 − 5 x 3 + 2 x − 5 x 3 ( 2 x − 5 ) + 1 ( 2 x − 5 ) ( 2 x − 5 ) ( x 3 + 1 ) Sum of cubes, factor further . ( 2 x − 5 ) ( x + 1 ) ( x 2 + x + 1 )
::2x4-5x3+2x2x-2x-5)+1(2x-5)+1(2x-5)(2x-5)(x5)(x3+1)立方体总和,再加系数。 (2x-5)(x+1)(x2+1)+x2+1)Example 4
::例4Find all the real-number solutions of 4 x 3 − 8 x 2 − x + 2 = 0 .
::查找所有 4x3 - 8x2 - x+2=0 的真数解决方案 。Factor by grouping.
::分组因素 。4 x 3 − 8 x 2 − x + 2 = 0 4 x 2 ( x − 2 ) − 1 ( x − 2 ) = 0 ( x − 2 ) ( 4 x 2 − 1 ) = 0 ( x − 2 ) ( 2 x − 1 ) ( 2 x + 1 ) = 0 x = 2 , 1 2 , − 1 2
::4x3-8x2-x2-x2+2=04x2x2x-2-1(x-2)=0(x-2)(4x2-1)=0(x-2)(2x-2)(2x-1)(2x+1)=0x=2、12,-12Review
::回顾Factor the following polynomials using factoring by grouping. Factor each polynomial completely.
::使用乘数组合的乘数乘以后面的多数值乘数。乘以每个多数值组合的乘数。-
x
3
−
4
x
2
+
3
x
−
12
::x3 - 4x2+3x- 12 -
x
3
+
6
x
2
−
9
x
−
54
::x3+6x2-9x-54 -
3
x
3
−
4
x
2
+
15
x
−
20
::3x3-4x2+15x-20 -
2
x
4
−
3
x
3
−
16
x
+
24
::2x4-3x3-16x+24 -
4
x
3
+
4
x
2
−
25
x
−
25
::4x3+4x2-2-25x-25 -
4
x
3
+
18
x
2
−
10
x
−
45
::4x3+18x2-10x-45 -
24
x
4
−
40
x
3
+
81
x
−
135
::24x4-40x3+81x-135 -
15
x
3
+
6
x
2
−
10
x
−
4
::15x3+6x2-10x-4 -
4
x
3
+
5
x
2
−
100
x
−
125
::4x3+5x2-100x-125 -
3
x
3
−
2
x
2
+
12
x
−
8
::3x3-2x2+12x-8
Find all the real-number solutions of the polynomials below.
::找到下面所有多面体的 真实数字解决方案 。-
9
x
3
−
54
x
2
−
4
x
+
24
=
0
::9x3 - 54x2 - 4x+24=0 -
x
4
+
3
x
3
−
27
x
−
81
=
0
::x4+3x3-27x-81=0 -
x
3
−
2
x
2
−
4
x
+
8
=
0
::x3 - 2x2 - 4x+8=0 -
Challenge
Find ALL the solutions of
x
6
−
9
x
4
−
x
2
+
9
=
0
.
::挑战查找所有 x6- 9x4- x2+9=0 的解决方案 。 -
Challenge
Find ALL the solutions of
x
3
+
3
x
2
+
16
x
+
48
=
0
.
::挑战查找 X3+3x2+16x+48=0的所有解决方案 。
Review (Answers)
::回顾(答复)Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。 -
x
3
−
4
x
2
+
3
x
−
12