课程： 1.7 保理聚合物

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保理聚合物
Introduction
::导言

What if you came across a polynomial like $\begin{aligned} 3 x^{2} - 27 \end{aligned}$ with multiple factors? How could you factor it completely ?
::万一你遇到一个3x2-27 等多因数的多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数多因数如何?

The process is related to the process of factoring whole numbers. If you were asked to find the prime factorization of 42, you might start with the factors 6 x 7 or 2 x 21. However, each of these sets of factors includes a number that can also be factored (6 and 21, respectively). The prime factorization of 42 is 2 x 3 x 7.
::该过程与整数的乘数计算过程有关。如果要求您找到42的乘数,您可以从因数 6 x 7 或 2 x 21 开始。然而,每组因数包括一个也可以乘数(分别为6 和 21),42 的乘数为2 x 3 x 7。

Factoring Polynomials
::保理聚合物

Remember that a polynomial is an expression with at least one algebraic term . Monomials have a single term and binomials have two terms . These algebraic terms contain variables which can have exponents and can be adjoined together by addition or subtraction .
::记住多数值是一个表达式, 至少有一个代数术语。单数值有一个单数术语, 二数值有两个术语。这些代数术语包含变量, 这些变量可以具有引号, 并且可以通过加法或减法连接在一起。

We say that a polynomial is factored completely when all prime factors have been found . The following steps will help in the process of factoring completely :
::我们说,当发现所有主要因素时,就完全考虑到一个多元性因素。以下步骤将有助于完全考虑因素的过程:

First, factor all common monomials.
::首先,所有共同的单项协议因素。

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

If there are no special products, determine the factors of the polynomial using trinomial factoring techniques.
::如果没有特殊产品,请使用三边保理技术确定多面性因素。

Double check each factor to see if any of these can be factored further.
::双倍检查每个因数,以确定其中任何因数是否可以进一步计算。

The following video works through three example problems involving completely:
::以下视频通过三个完全涉及的范例问题进行演练:

Greatest Common Factor
::最大共同因素

When factoring, start by pulling out the greatest common factor (or GCF).
::在进行保理时,首先取出最大的共同因素(或全球合作框架)。

Special Forms
::特别形式

Below are special forms that will help you multiply and factor binomials and trinomials that specifically fit these patterns. Note that these factors work only if they fit these rules exactly .
::下面是特殊的形式,可以帮助您进行乘法和因子二进制和三进制,具体符合这些模式。请注意,这些因素只有在符合这些规则时才起作用。

Special Forms for Factoring
::保理特殊表格

Perfect Square Trinomials:
$\begin{aligned} a^{2} + 2 a b + b^{2} = (a + b)^{2} \end{aligned}$
$\begin{aligned} a^{2} - 2 a b + b^{2} = (a - b)^{2} \end{aligned}$
::完美广场三角形:a2+2ab+b2=(a+b)2a2-2ab+b2=(a-b)2a2-2ab+b2=(a-b)2

Difference of two squares:
$\begin{aligned} a^{2} - b^{2} = (a + b) (a - b) \end{aligned}$
::两平方差: a2-b2=(a+b)(a-b)

Sum of two cubes:
$\begin{aligned} a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) \end{aligned}$
::两个立方体的总和: a3+b3=(a+b)(a2-ab+b2)

Difference of two cubes:
$\begin{aligned} a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \end{aligned}$
::两个立方体的差值: a3-b3=(a-b)(a2+ab+b2)

For help with factoring polynomials such as in the previous examples, see this video for two multistep examples, including factoring out the GCF and factoring perfect square trinomials:
::如前几个例子所示,如果能帮助考虑多种多边货币,请看这段录像中两个多步骤的例子,包括考虑考虑绿色气候基金和考虑完全的平方三角货币:

Factor Out a Common Binomial
::消除共同二进制因素

The 1st 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:
::保理学过程的第一项步骤往往是从多面性中将共同的单项性因素考虑在内。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 " data-term="Parentheses" role="term" tabindex="0"> 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 by Grouping
::分组计数

In some cases, there is not a a GCF for all the terms in a polynomial. If you have four terms with no GCF, then try factoring by grouping .
::在某些情况下,在多种语言中,没有一个通用的绿色气候基金。如果你有四个条件而没有绿色气候基金,那么通过分组来尝试乘法。

Step 1: Group the 1st two terms together, and then the last two terms together.
::第1步:将第一两个任期组合在一起,然后将最后两个任期组合在一起。

Step 2: Factor out a GCF from each separate binomial.
::步骤2:将绿色气候基金从每个单独的二元体中扣除。

Step 3: Factor out the common binomial.
::步骤3:考虑到共同的二元论。

Factoring Using the a-c Method
::使用a-c方法进行保理

When factoring a quadratic that doesn't fit a special form, you can use the a-c Method. T he standard form of a quadratic is $\begin{aligned} a x^{2} + b x + c \end{aligned}$ . The a-c Method starts by multiplying $\begin{aligned} a a n d c \end{aligned}$ . Then find the factors of that product that add up to $\begin{aligned} b \end{aligned}$ . You can then split the trinomial into 4 terms and use grouping to solve.
::当计算一个不符合特殊形态的二次方块时, 您可以使用 a- c 方法。二次方块的标准形式是 ax2+bx+c。 a- c 方法以乘以 a 和 c 开始。然后找到该产品乘以 b 的乘数。然后, 您可以将三角形分为 4 个术语, 并使用分组来解答。

Examples
::实例

Example 1
::例1

Factor the following using greatest common factor:
::以下列最大共同系数乘以下列因素:

a) $\begin{aligned} 3 x + 6 \end{aligned}$
:a) 3x+6

Solution:
::解决方案 :

$\begin{aligned} 3 x + 6 = 3 (x + 2) \end{aligned}$ , since $\begin{aligned} 6 = 3 \times 2 \end{aligned}$ and thus $\begin{aligned} 3 \end{aligned}$ is the greatest common factor.
::3x+6=3(x+2),因为6=3x2,因此,3是最大的共同系数。

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

Solution:
::解决方案 :

$\begin{aligned} 4 x^{2} + 2 x = 2 x (2 x + 1) \end{aligned}$ , since $\begin{aligned} 4 = 2 \times 2 a n d x^{2} = x \times x \end{aligned}$ , and thus $\begin{aligned} 2 x \end{aligned}$ is the GCF.
::4x2+2x=2x(2x+1),因为4=2x2和x2=xxxx,因此2x是绿色气候基金。

Example 2
::例2

Factor the following polynomials completely:
::完全乘以下列多元数 :

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

Solution:
::解决方案 :

a) Factor out common monomials: $\begin{aligned} 2 x^{2} - 8 = 2 (x^{2} - 4) \end{aligned}$
:a) 将共同单体:2x2-8=2x2-4)

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) 。

Each factor is prime. We have factored completely.
::每个因素都是首要的,我们已完全考虑到每个因素。

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

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

Solution:
::解决方案 :

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。

Each factor is prime. We have factored completely.
::每个因素都是首要的,我们已完全考虑到每个因素。

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

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

Solution:
::解决方案 :

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 the expression in parentheses as a difference of squares. We factor and get
::我们确认括号中的表达法是方形的差别。

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

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

If we look at each factor we see that the 1st parentheses is a difference of squares. We factor and get
::如果我们看一下每一个因素,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

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

::-2x+3(x-3)(x-3)(x2+9)。

Each factor is prime, which means each factor cannot be factored any further. We have factored completely.
::每个因素都是首要的,这意味着每个因素都不能进一步考虑,我们已完全考虑。

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)。

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

Solution:
::解决方案 :

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 $\begin{aligned} x (x^{2} - 4)^{2} \end{aligned}$ .
::我们认识到 x4-8x2+16 是完美的正方形, 我们将其乘以 x( x2- 4) 2 。

We look at each term and recognize that the term in 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。

Each factor is prime. We have factored completely.
::每个因素都是首要的,我们已完全考虑到每个因素。

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。

Example 3
::例3

Factor out the common binomials:
::计及共同的二进制 :

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

Solution:
::解决方案 :

$\begin{aligned} 3 x (x - 1) + 4 (x - 1) \end{aligned}$ 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 $\begin{aligned} (x - 1) (3 x + 4) \end{aligned}$ .
::当我们考虑到共同的二元论时,我们得到(x-1)(3x+4)。

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

Solution:
::解决方案 :

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

Factor completely: $\begin{aligned} 2 x^{3} + 3 x^{2} + 10 x + 15 \end{aligned}$
::完全因数: 2x3+3x2+10x+15

Solution:
::解决方案 :

Group the 1st two terms and the last two terms:
::组第1个任期和最后两个任期:

$\begin{aligned} (2 x^{3} + 3 x^{2}) + (10 x + 15) \end{aligned}$
:2x3+3x2)+(10x+15)

Factor out the GCF from each binomial:
::将绿色气候基金从每个二元性因素中计算出来:

$\begin{aligned} x^{2} (2 x + 3) + 5 (2 x + 3) \end{aligned}$
::x2( 2x+3)+5(2x+3)

Factor out the common binomial:
::考虑共同的二进制:

$\begin{aligned} (2 x + 3) (x^{2} + 5) \end{aligned}$
:2x+3)(x2+5)

Example 5
::例5

Factor completely using the a-c method:
::完全使用 a-c 方法的系数 :

a) $\begin{aligned} 24 x^{3} - 28 x^{2} + 8 x \end{aligned}$ .
:a) 24x3-28x2+8x。

Solution:
::解决方案 :

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 parentheses. 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:
::接下来,请将括号中的三角值乘以括号。自 {_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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

::121121213122622226681234447

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) 。

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

Solution:
::解决方案 :

Factor out the common monomial. In this case, 6 can be divided from each term:
::在此情况下,6个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作为两个二进制的产物, (x) (x) 乘以 x2 - 5x+6 。

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)。

Each factor is prime. We have factored completely.
::每个因素都是首要的,我们已完全考虑到每个因素。

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

Review
::回顾

Factor completely:
::完全因数 :

$\begin{aligned} 7 x - 14 \end{aligned}$
::7x-14

$\begin{aligned} 3 x + 9 x y \end{aligned}$
::3x+9xy 3x+9xy

$\begin{aligned} x y z + y^{2} z^{3} \end{aligned}$
::xyz+y2z3 xyz+y2z3

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

Please see the Appendix.
::请参看附录。

Resource
::资源资源资源资源资源资源资源资源资源资源资源资源

The WTAMU Virtual Math Lab has a detailed page on factoring polynomials here: .
::WTAMU虚拟数学实验室有详细页面,

This page contains many videos showing example problems being solved.
::此页面包含许多视频, 显示正在解决的示例问题。

章节大纲

常规

保理聚合物

Introduction ::导言

Factoring Polynomials ::保理聚合物

Greatest Common Factor ::最大共同因素

Special Forms ::特别形式

Special Forms for Factoring ::保理特殊表格

Factoring by Grouping ::分组计数

Factoring Using the a-c Method ::使用a-c方法进行保理

Examples ::实例

Solution: ::解决方案 :

Solution: ::解决方案 :

Review ::回顾

Review (Answers ) ::回顾(答复)

Resource ::资源资源资源资源资源资源资源资源资源资源资源资源

Introduction
::导言

Factoring Polynomials
::保理聚合物

Greatest Common Factor
::最大共同因素

Special Forms
::特别形式

Special Forms for Factoring
::保理特殊表格

Factoring by Grouping
::分组计数

Factoring Using the a-c Method
::使用a-c方法进行保理

Examples
::实例

Solution:
::解决方案 :

Solution:
::解决方案 :

Review
::回顾

Review (Answers )
::回顾(答复)

Resource
::资源资源资源资源资源资源资源资源资源资源资源资源