Course: 9.12 全额保理 | The Way To Learn

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 全额保理

Collapse Expand
全额保理
Factoring Completely
::全额保理

We say that a polynomial is factored completely when we can’t factor it any more. Here are some suggestions that you should follow to make sure that you factor completely:
::我们说,当我们再也不能考虑它的时候,多元性就完全被考虑在内。以下是一些你应该遵循的建议,以确保你完全考虑:

Factor all common monomials first.
::先考虑所有共同的单项协议的因数。

Identify special products such as difference of squares or the square of a binomial . Factor according to their formulas.
::根据公式确定特殊产品,如方形或二进制方形的差数。

If there are no special products, factor using the methods we learned in the previous sections.
::如果没有特殊产品,就采用我们在前几节中学到的方法。

Look at each factor and see if any of these can be factored further.
::审视每个因素,看看其中是否有任何因素可以进一步考虑。

Factoring Completely - Learn by Example
::完全乘数 - 按示例学习

1. $\begin{aligned} 6 x^{2} - 30 x + 36 \end{aligned}$
::1. 6x2-30x+36

Factor out the common monomial . In this case 6 can be divided from each term :
::在此情况下,6可以从每个术语中除以:

$\begin{aligned} 6 (x^{2} - 5 x + 6) \end{aligned}$

::6(x2-5x+6)

There are no special products. We factor $\begin{aligned} x^{2} - 5 x + 6 \end{aligned}$ as a product of two binomials: $\begin{aligned} (x) (x) \end{aligned}$
::没有特殊产品。我们将 x2 - 5x+6 乘以两个二进制的产物x) (x) 乘以 x2 - 5x+6 x)

The two numbers that multiply to 6 and add to -5 are -2 and -3, so:
::乘以 6 和加到 -5 的两个数字是 -2 和 - 3, 所以:

$\begin{aligned} 6 (x^{2} - 5 x + 6) = 6 (x - 2) (x - 3) \end{aligned}$

::6(x2-5x+6)=6(x-2)(x-3)

If we look at each factor we see that we can factor no more.
::如果我们看一下每一个因素,我们就会看到,我们无法再考虑更多的因素。

The answer is $\begin{aligned} 6 (x - 2) (x - 3) \end{aligned}$ .
::答案是6(x-2)(x-3)。

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

Factor out common monomials: $\begin{aligned} 2 x^{2} - 8 = 2 (x^{2} - 4) \end{aligned}$
::参数显示共同单单数: 2x2- 8=2( x2- 4)

We recognize $\begin{aligned} x^{2} - 4 \end{aligned}$ as a difference of squares. We factor it as $\begin{aligned} (x + 2) (x - 2) \end{aligned}$ .
::我们承认 x2 - 4 是方形的差数, 我们将其乘以( x+2)( x-2) 。

If we look at each factor we see that we can factor no more.
::如果我们看一下每一个因素,我们就会看到,我们无法再考虑更多的因素。

The answer is $\begin{aligned} 2 (x + 2) (x - 2) \end{aligned}$ .
::答案是2(x+2)(x-2)。

3. $\begin{aligned} x^{3} + 6 x^{2} + 9 x \end{aligned}$
::3. x3+6x2+9x

Factor out common monomials: $\begin{aligned} x^{3} + 6 x^{2} + 9 x = x (x^{2} + 6 x + 9) \end{aligned}$
::系数 : x3+6x2+9x=x(x2+6x+9)

We recognize $\begin{aligned} x^{2} + 6 x + 9 \end{aligned}$ as a perfect square and factor it as $\begin{aligned} (x + 3)^{2} \end{aligned}$ .
::我们确认x2+6x+9是一个完美的平方,将之乘以(x+3)2。

If we look at each factor we see that we can factor no more.
::如果我们看一下每一个因素,我们就会看到,我们无法再考虑更多的因素。

The answer is $\begin{aligned} x (x + 3)^{2} \end{aligned}$ .
::答案是 x( x+3) 2。

4. $\begin{aligned} - 2 x^{4} + 162 \end{aligned}$
::4.-2x4+162

Factor out the common monomial. In this case, factor out -2 rather than 2. (It’s always easier to factor out the negative number so that the highest degree term is positive.)
::将共同的单数计算出来。在此情况下, 将 - 2 而不是 2 计算出来( 总是更容易将负数计算出来, 以便最高比例的术语是正数。 )

$\begin{aligned} - 2 x^{4} + 162 = - 2 (x^{4} - 81) \end{aligned}$

::-2x4+162+2(x4-81)

We recognize expression in parenthesis as a difference of squares. We factor and get:
::我们确认括号中的表达形式是方形的差别。

$\begin{aligned} - 2 (x^{2} - 9) (x^{2} + 9) \end{aligned}$

::-2(x2--9)(x2+9)

If we look at each factor we see that the first parenthesis is a difference of squares. We factor and get:
::如果我们看每一因素,我们可以看到第一个括号是方形的差别。我们考虑到并获得:

$\begin{aligned} - 2 (x + 3) (x - 3) (x^{2} + 9) \end{aligned}$

::-2(x+3)(x-3)(x2+9)

If we look at each factor now we see that we can factor no more.
::如果我们现在审视每个因素,我们就会看到,我们无法再考虑更多的因素。

The answer is $\begin{aligned} - 2 (x + 3) (x - 3) (x^{2} + 9) \end{aligned}$ .
::答案是-2(x+3)(x-3)(x2+9)。

5. $\begin{aligned} x^{5} - 8 x^{3} + 16 x \end{aligned}$
::5. x5-8x3+16x

Factor out the common monomial: $\begin{aligned} x^{5} - 8 x^{3} + 14 x = x (x^{4} - 8 x^{2} + 16) \end{aligned}$
::乘以常见单单数 : x5- 8x3+14x=x( x4- 8x2+16)

We recognize $\begin{aligned} x^{4} - 8 x^{2} + 16 \end{aligned}$ as a perfect square and we factor it as $\begin{aligned} x (x^{2} - 4)^{2} \end{aligned}$ .
::我们承认 x4-8x2+16 是完美的方形, 我们将其乘以 x( x2- 4) 2 。

We look at each term and recognize that the term in " data-term="Parentheses" role="term" tabindex="0"> parentheses is a difference of squares.
::我们审视每个术语,认识到括号中的术语是方形的区别。

We factor it and get $\begin{aligned} ((x + 2) (x - 2))^{2} \end{aligned}$ , which we can rewrite as $\begin{aligned} (x + 2)^{2} (x - 2)^{2} \end{aligned}$ .
::我们将其乘数乘以(x+2)(x-2)2, 重写为(x+2)(2)(x-2)2。

If we look at each factor now we see that we can factor no more.
::如果我们现在审视每个因素,我们就会看到,我们无法再考虑更多的因素。

The final answer is $\begin{aligned} x (x + 2)^{2} (x - 2)^{2} \end{aligned}$ .
::最后一个答案是 x(x+2)2(x-2)2。

Factor out a Common Binomial
::考虑一个共同的二进制

The first step in the factoring process is often factoring out the common monomials from a polynomial. But sometimes have common terms that are binomials. For example, consider the following expression:
::保理工过程的第一步往往是从一个多义中将共同的单项性考虑出来。但有时有共同的二元性术语。例如,考虑以下表达方式:

$\begin{aligned} x (3 x + 2) - 5 (3 x + 2) \end{aligned}$

::x( 3x+2) - 5( 3x+2)

Since the term $\begin{aligned} (3 x + 2) \end{aligned}$ appears in both terms of the polynomial, we can factor it out. We write that term in front of a set of parentheses containing the terms that are left over:
::因为这个词(3x+2)两个词都出现在多义词中, 我们可以将它考虑在内。我们用一组括号在括号前写该词, 括号中包含剩下的词 :

$\begin{aligned} (3 x + 2) (x - 5) \end{aligned}$

:3x+2(x-5)

This expression is now completely factored.
::此表达式现已被完全考虑到。

Factoring out Common Binomials - Learn by Example
::将共同二义义因子因素推算出来 - 按示例学习

1. $\begin{aligned} 3 x (x - 1) + 4 (x - 1) \end{aligned}$
::1. 3x(x-1)+4(x-1)

$\begin{aligned} 3 x (x - 1) + 4 (x - 1) \end{aligned}$ has a common binomial of $\begin{aligned} (x - 1) \end{aligned}$ .
::3x(x-1)+4(x-1)有一个共同的二进制(x-1)。

When we factor out the common binomial we get $\begin{aligned} (x - 1) (3 x + 4) \end{aligned}$ .
::当我们把共同的二进制(x-1)(3x+4)考虑在内时,我们就会得到(x-1)(3x+4)。

2. $\begin{aligned} x (4 x + 5) + (4 x + 5) \end{aligned}$
::2. x(4x+5)+(4x+5)

$\begin{aligned} x (4 x + 5) + (4 x + 5) \end{aligned}$ has a common binomial of $\begin{aligned} (4 x + 5) \end{aligned}$ .
::x( 4x+5) +( 4x+5) +( 4x+5) 具有共同的二进制( 4x+5) 。

When we factor out the common binomial we get $\begin{aligned} (4 x + 5) (x + 1) \end{aligned}$ .
::当我们考虑到共同的二元论时,我们就会得到(4x+5)(x+1)。

Example
::示例示例示例示例

Factor completely: $\begin{aligned} 24 x^{3} - 28 x^{2} + 8 x \end{aligned}$ .
::完全因数: 24x3- 28x2+8x

First, notice that each term has $\begin{aligned} 4 x \end{aligned}$ as a factor. Start by factoring out $\begin{aligned} 4 x \end{aligned}$ :
::首先,注意每个术语有4x作为因数。首先,从考虑到4x开始:

$\begin{aligned} 24 x^{3} - 28 x^{2} + 8 x = 4 x (6 x^{2} - 7 x + 2) \end{aligned}$
::24x3-28x2+8x=4x(6x2-7x+2)

Next, factor the trinomial in the parenthesis. Since $\begin{aligned} a \neq 1 \end{aligned}$ find $\begin{aligned} a \cdot c \end{aligned}$ : $\begin{aligned} 6 \cdot 2 = 12 \end{aligned}$ . Find the factors of 12 that add up to -7. Since 12 is positive and -7 is negative, the two factors should be negative:
::其次,将括号中的三角乘数乘以三。自 a1 发现 ac: 62=12。查找12的乘数加到 - 7. 由于12是正数和 -7是负数, 这两个因素应该是负数:

$\begin{aligned} 12 & = - 1 \cdot - 12 & and & - 1 + - 12 = - 13 \\ 12 & = - 2 \cdot - 6 & and & - 2 + - 6 = - 8 \\ 12 & = - 3 \cdot - 4 & and & - 3 + - 4 = - 7 \end{aligned}$

::12112 和 -112131226 和 -2681234 和 -3447

Rewrite the trinomial using $\begin{aligned} - 7 x = - 3 x - 4 x \end{aligned}$ , and then :
::使用 - 7x3x- 4x 重写三角词, 然后 :

$\begin{aligned} 6 x^{2} - 7 x + 2 & = 6 x^{2} - 3 x - 4 x + 2 \\ = 3 x (2 x - 1) - 2 (2 x - 1) \\ = (3 x - 2) (2 x - 1) \end{aligned}$

::6x2-7x+2=6x2-3x-4x+2=3x(2x-1)-2x-1=3x-2(2x-1)-1=3x-2(2x-1)-1

The final factored answer is:
::最后一个因素答案是:

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

Review
::回顾

Factor completely.
::完全的因数。

$\begin{aligned} 2 x^{2} + 16 x + 30 \end{aligned}$
::2x2+16x+30

$\begin{aligned} 5 x^{2} - 70 x + 245 \end{aligned}$
::5x2-70x+245

$\begin{aligned} - x^{3} + 17 x^{2} - 70 x \end{aligned}$
::-x3+17x2-70x

$\begin{aligned} 2 x^{4} - 512 \end{aligned}$
::2x4-512

$\begin{aligned} 25 x^{4} - 20 x^{3} + 4 x^{2} \end{aligned}$
::25x4 - 20x3+4x2

$\begin{aligned} 12 x^{3} + 12 x^{2} + 3 x \end{aligned}$
::12x3+12x2+3x

$\begin{aligned} 12 c^{2} - 75 \end{aligned}$
::12c2-75

$\begin{aligned} 6 x^{2} - 600 \end{aligned}$
::6x2-600 6x2-600

$\begin{aligned} - 5 t^{2} - 20 t - 20 \end{aligned}$
::- 5t2-20t-20

$\begin{aligned} 6 x^{2} + 18 x - 24 \end{aligned}$
::6x2+18x-24

$\begin{aligned} - n^{2} + 10 n - 21 \end{aligned}$
::-n2+10n-21

$\begin{aligned} 2 a^{2} - 14 a - 16 \end{aligned}$
::2a2-14a-16

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

全额保理

Factoring Completely ::全额保理

Factoring Completely - Learn by Example ::完全乘数 - 按示例学习

Factor out a Common Binomial ::考虑一个共同的二进制

Factoring out Common Binomials - Learn by Example ::将共同二义义因子因素推算出来 - 按示例学习

Example ::示例示例示例示例

Review ::回顾

Review (Answers) ::回顾(答复)

Factoring Completely
::全额保理

Factoring Completely - Learn by Example
::完全乘数 - 按示例学习

Factor out a Common Binomial
::考虑一个共同的二进制

Factoring out Common Binomials - Learn by Example
::将共同二义义因子因素推算出来 - 按示例学习

Example
::示例示例示例示例

Review
::回顾

Review (Answers)
::回顾(答复)