Course: 9.13 分组保理 | The Way To Learn

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分组计数
Factoring by Grouping
::分组计数

Sometimes, we can factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factor by grouping.
::有时,我们可以通过将一组术语的共同单数乘以一个包含四个或四个以上术语的多元数系数。这个方法称为组合系数。

The following examples illustrates how this process works.
::以下例子说明了这一进程是如何运作的。

1. Factor $\begin{aligned} 2 x + 2 y + a x + a y \end{aligned}$ .
::1. Pric2x+2y+ax+ay。

There is no factor common to all the terms. However, the first two terms have a common factor of 2 and the last two terms have a common factor of $\begin{aligned} a \end{aligned}$ . Factor 2 from the first two terms and factor $\begin{aligned} a \end{aligned}$ from the last two terms:
::然而,前两个任期的共同系数为2,后两个任期的共同系数为a。

$\begin{aligned} 2 x + 2 y + a x + a y = 2 (x + y) + a (x + y) \end{aligned}$

::2x+2y+Ax+ay=2(x+y)+a(x+y)

Now we notice that the binomial $\begin{aligned} (x + y) \end{aligned}$ is common to both terms. We factor the common binomial and get:
::现在我们注意到二进制( X+y) 是两个条件的共同点。我们将共同的二进制( b进制) 乘以:

$\begin{aligned} (x + y) (2 + a) \end{aligned}$

:x+y)(2+a)

2. Factor $\begin{aligned} 3 x^{2} + 6 x + 4 x + 8 \end{aligned}$ .
::2. Pric3x2+6x+4x+8。

We factor 3x from the first two terms and factor 4 from the last two terms:
::我们从前两个任期乘以3x,从后两个任期乘以4:

$\begin{aligned} 3 x (x + 2) + 4 (x + 2) \end{aligned}$

::3x(x+2)+4(x+2)

Now factor $\begin{aligned} (x + 2) \end{aligned}$ from both terms: $\begin{aligned} (x + 2) (3 x + 4) \end{aligned}$ .
::现因数(x+2),两个词都来自x+2)(3x+4)。

Now the polynomial is factored completely.
::现在,多元性被完全计算在内。

Factor Quadratic Trinomials Where a ≠ 1
::QQQQQQQQQQQQQQIals 其中 1

Factoring by grouping is a very useful method for factoring quadratic trinomials of the form $\begin{aligned} a x^{2} + b x + c \end{aligned}$ , where $\begin{aligned} a \neq 1 \end{aligned}$ .
::分组计算是计算x2+bx+c形式的四边三角(a1)的一个非常有用的方法。

A quadratic like this doesn’t factor as $\begin{aligned} (x \pm m) (x \pm n) \end{aligned}$ , so it’s not as simple as looking for two numbers that multiply to $\begin{aligned} c \end{aligned}$ and add up to $\begin{aligned} b \end{aligned}$ . Instead, we also have to take into account the coefficient in the first term .
::像这样的二次曲线没有(x)cm(x)和(x)和这样的系数,所以它不像寻找两个乘以c和加到b的数值那么简单。相反,我们也必须考虑到第一个学期的系数。

To factor a quadratic polynomial where $\begin{aligned} a \neq 1 \end{aligned}$ , we follow these steps:
::乘以一个四边形的多面体, 我们遵循这些步骤: A++1, 我们遵循这些步骤:

We find the product $\begin{aligned} a c \end{aligned}$ .
::我们找到产品 ac。

We look for two numbers that multiply to $\begin{aligned} a c \end{aligned}$ and add up to $\begin{aligned} b \end{aligned}$ .
::我们寻找两个乘以 ac 并加到 b 的数字。

We rewrite the middle term using the two numbers we just found.
::我们用刚刚找到的两个数字重写中学期

We factor the expression by grouping.
::我们通过分组来计算表达方式。

Let’s apply this method to the following examples.
::让我们将这种方法应用到下面的例子中。

Factoring by Grouping
::分组计数

Factor the following quadratic trinomials by grouping.
::以组合方式乘以随后的二次三角曲线。

a) $\begin{aligned} 3 x^{2} + 8 x + 4 \end{aligned}$
:a) 3x2+8x+4

b) $\begin{aligned} 6 x^{2} - 11 x + 4 \end{aligned}$
:b) 6x2-11x+4

Solution:
::解决方案 :

Let’s follow the steps outlined above:
::让我们遵循上述步骤:

a) $\begin{aligned} 3 x^{2} + 8 x + 4 \end{aligned}$
:a) 3x2+8x+4

$\begin{aligned} 3 x^{2} + 8 x + 4 \end{aligned}$
::3x2+8x+4

Step 1: $\begin{aligned} a c = 3 \cdot 4 = 12 \end{aligned}$
::第1步:ac=34=12

Step 2: The number 12 can be written as a product of two numbers in any of these ways:
::第2步:第12号可以用下列任何一种方式作为两个数字的产物:

$\begin{aligned} 12 & = 1 \cdot 12 & and & 1 + 12 = 13 \\ 12 & = 2 \cdot 6 & and & 2 + 6 = 8 T h i s i s t h e c o r r e c t c h o i c e . \\ 12 & = 3 \cdot 4 & and & 3 + 4 = 7 \end{aligned}$

::12=112和1+12=1312=26和2+6=8 这是正确的选择 12=34和3+4=7

Step 3: Re-write the middle term: $\begin{aligned} 8 x = 2 x + 6 x \end{aligned}$ , so the problem becomes:
::第3步:重写中期: 8x=2x+6x, 问题就变成:

$\begin{aligned} 3 x^{2} + 8 x + 4 = 3 x^{2} + 2 x + 6 x + 4 \end{aligned}$

::3x2+8x+4=3x2+2x+6x+4

Step 4: Factor an $\begin{aligned} x \end{aligned}$ from the first two terms and a 2 from the last two terms:
::第4步:将前两个任期的x乘以前两个任期的xx,最后两个任期的2乘以后两个任期的x:

$\begin{aligned} x (3 x + 2) + 2 (3 x + 2) \end{aligned}$

::x(3x+2)+2(3x+2)

Now factor the common binomial $\begin{aligned} (3 x + 2) \end{aligned}$ :
::现在乘以共同的二进制( 3x+2) :

$\begin{aligned} (3 x + 2) (x + 2) T h i s i s t h e a n s w e r . \end{aligned}$

:3x+2(x+2)) 这是答案。

To check if this is correct we multiply $\begin{aligned} (3 x + 2) (x + 2) \end{aligned}$ :
::要检查是否正确, 我们乘以( 3x+2)( x+2) :

$\begin{aligned} 3 x + 2 \\ \underline{x + 2} \\ 6 x + 4 \\ \underline{3 x^{2} + 2 x} \\ 3 x^{2} + 8 x + 4 \end{aligned}$

::3x+2x+2_6x+43x2+2x3x2+8x4

The solution checks out.
::解决方案已经检查了。

b) $\begin{aligned} 6 x^{2} - 11 x + 4 \end{aligned}$
:b) 6x2-11x+4

$\begin{aligned} 6 x^{2} - 11 x + 4 \end{aligned}$
::6x2 - 11x+4

Step 1: $\begin{aligned} a c = 6 \cdot 4 = 24 \end{aligned}$
::第1步:ac=64=24

Step 2: The number 24 can be written as a product of two numbers in any of these ways:
::第2步:第24个数字可以用下列任何一种方式作为两个数字的产物:

$\begin{aligned} 24 & = 1 \cdot 24 & and & 1 + 24 = 25 \\ 24 & = - 1 \cdot (- 24) & and & - 1 + (- 24) = - 25 \\ 24 & = 2 \cdot 12 & and & 2 + 12 = 14 \\ 24 & = - 2 \cdot (- 12) & and & - 2 + (- 12) = - 14 \\ 24 & = 3 \cdot 8 & and & 3 + 8 = 11 \\ 24 & = - 3 \cdot (- 8) & and & - 3 + (- 8) = - 11 (C o r r e c t c h o i c e) \\ 24 & = 4 \cdot 6 & and & 4 + 6 = 10 \\ 24 & = - 4 \cdot (- 6) & and & - 4 + (- 6) = - 10 \end{aligned}$

::24=124和1+24=25241(24)和1+(24)=2524=212和2+12=14242(12)和-2+2+(12)=1424=38和3+8=11243(8)和-3+(8)+(8)=8)\11(修正选择)24=46和4+6=1024}(6)和-4+(6)+(6)__10

Step 3: Re-write the middle term: $\begin{aligned} - 11 x = - 3 x - 8 x \end{aligned}$ , so the problem becomes:
::步骤3:重写中期: - 11x3x-8x, 问题就变成:

$\begin{aligned} 6 x^{2} - 11 x + 4 = 6 x^{2} - 3 x - 8 x + 4 \end{aligned}$

::6x2-11x+4=6x2-3x-8x+4

Step 4: Factor by grouping: factor a $\begin{aligned} 3 x \end{aligned}$ from the first two terms and a -4 from the last two terms:
::第4步:按组别确定的因素:前两个任期的系数a 3x,后两个任期的系数a-4:

$\begin{aligned} 3 x (2 x - 1) - 4 (2 x - 1) \end{aligned}$

::3x( 2x- 1) - 4x( 2x- 1)

Now factor the common binomial $\begin{aligned} (2 x - 1) \end{aligned}$ :
::现在将共同的二进制( 2x-1) 乘以 :

$\begin{aligned} (2 x - 1) (3 x - 4) T h i s i s t h e a n s w e r . \end{aligned}$

:2x-1)(3x-4) 这是答案。

Example
::示例示例示例示例

Example 1
::例1

Factor $\begin{aligned} 5 x^{2} - 6 x + 1 \end{aligned}$ by grouping.
::组合法5x2-6x+1系数。

Let’s follow the steps outlined above:
::让我们遵循上述步骤:

$\begin{aligned} 5 x^{2} - 6 x + 1 \end{aligned}$
::5x2-6x+1

Step 1: $\begin{aligned} a c = 5 \cdot 1 = 5 \end{aligned}$
::第1步:ac=51=5

Step 2: The number 5 can be written as a product of two numbers in any of these ways:
::第2步:第5步可以用下列任何一种方式写成两个数字的产物:

$\begin{aligned} 5 & = 1 \cdot 5 & and & 1 + 5 = 6 \\ 5 & = - 1 \cdot (- 5) & and & - 1 + (- 5) = - 6 (C o r r e c t c h o i c e) \end{aligned}$

::5=15和1+5=651(-5)和-1+(-5-5)_6(更正选择)

Step 3: Re-write the middle term: $\begin{aligned} - 6 x = - x - 5 x \end{aligned}$ , so the problem becomes:
::步骤3:重写中期: - 6xxx- 5x, 问题就变成:

$\begin{aligned} 5 x^{2} - 6 x + 1 = 5 x^{2} - x - 5 x + 1 \end{aligned}$

::5x2-6x+1=5x2-x-5x1

Step 4: Factor by grouping: factor an $\begin{aligned} x \end{aligned}$ from the first two terms and $\begin{aligned} a - 1 \end{aligned}$ from the last two terms:
::第4步:按组分列的系数:前两个任期的乘数 x;后两个任期的乘数1-1:

$\begin{aligned} x (5 x - 1) - 1 (5 x - 1) \end{aligned}$

:x5x-1)-1-1(5x-1)

Now factor the common binomial $\begin{aligned} (5 x - 1) \end{aligned}$ :
::现在将共同的二进制( 5x-1) 乘以 :

$\begin{aligned} (5 x - 1) (x - 1) T h i s i s t h e a n s w e r . \end{aligned}$

:5x-1)(x-1) 这是答案。

Review
::回顾

Factor by grouping.
::分组因素。

$\begin{aligned} 6 x^{2} - 9 x + 10 x - 15 \end{aligned}$
::6x2-9x+10x-15

$\begin{aligned} 5 x^{2} - 35 x + x - 7 \end{aligned}$
::5x2-35x+x-7

$\begin{aligned} 9 x^{2} - 9 x - x + 1 \end{aligned}$
::9x2-9x-x+1

$\begin{aligned} 4 x^{2} + 32 x - 5 x - 40 \end{aligned}$
::4x2+32x-5x-40

$\begin{aligned} 2 a^{2} - 6 a b + 3 a b - 9 b^{2} \end{aligned}$
::2a2-6ab+3ab-9b2

$\begin{aligned} 5 x^{2} + 15 x - 2 x y - 6 y \end{aligned}$
::5x2+15x-2xy-6yy

Factor the following quadratic trinomials by grouping.
::以组合方式乘以随后的二次三角曲线。

$\begin{aligned} 4 x^{2} + 25 x - 21 \end{aligned}$
::4x2+25x-21

$\begin{aligned} 6 x^{2} + 7 x + 1 \end{aligned}$
::6x2+7x+1

$\begin{aligned} 4 x^{2} + 8 x - 5 \end{aligned}$
::4x2+8x-5

$\begin{aligned} 3 x^{2} + 16 x + 21 \end{aligned}$
::3x2+16x+21

$\begin{aligned} 6 x^{2} - 2 x - 4 \end{aligned}$
::6x2-2-2x-4

$\begin{aligned} 8 x^{2} - 14 x - 15 \end{aligned}$
::8x2 - 14x - 15

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。

Section outline

General

分组计数

Factoring by Grouping ::分组计数

Factor Quadratic Trinomials Where a ≠ 1 ::QQQQQQQQQQQQQQIals 其中 1

Factoring by Grouping ::分组计数

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Factoring by Grouping
::分组计数

Factor Quadratic Trinomials Where a ≠ 1
::QQQQQQQQQQQQQQIals 其中 1

Factoring by Grouping
::分组计数

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)