Course: 2.2 高级保理 | The Way To Learn

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高级保理

The difference of perfect squares can be generalized as a factoring technique. By extension, any difference between terms that are raised to an even power like $\begin{aligned} a^{6} - b^{6} \end{aligned}$ can be factored using the difference of perfect squares technique. This is because even powers can always be written as perfect squares: $\begin{aligned} a^{6} - b^{6} = (a^{3})^{2} - (b^{3})^{2} \end{aligned}$ .
::完全正方形的差别可以作为一种保理技术而普遍化。推而广之,任何对等权力(如:6-b6)的术语之间的差别都可以使用完全正方形技术的差别来计算。这是因为即使权力也总是可以写成完美的正方形:a6-b6=(a3)2-(b3)2。

What about the sum or difference of terms with matching odd powers? How can those be factored?
::与奇异力量相匹配的条件的总和或差别如何?如何计算这些条件?

More Factoring Techniques
::更多保理技术

Factoring a trinomial of the form $\begin{aligned} a x^{2} + b x + c \end{aligned}$ is much more difficult when $\begin{aligned} a \neq 1 \end{aligned}$ . There are four techniques that can be used to factor such expressions.
::当 a\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ xx2+bx+c\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Guess and Check
::猜测和检查

The educated guess and check method can be time consuming but if the first and last coefficient only have a few factors, there are a finite number of possibilities. Take the expression :
::受过教育的猜测和检查方法可以耗费时间,但如果第一个和最后一个系数只有几个因素,则有一定数量的可能性。

$\begin{aligned} 6 x^{2} - 13 x - 28 \end{aligned}$
::6x2-13x-28

The 6 can be factored into the following four pairs:
::6可分为以下四对:

1, 6

2, 3

-1, -6

-2, -3

The -28 can be factored into the following twelve pairs:
::-28可乘以以下12对:

1, -28 or -28, 1
::1,2-28或-28, 1

-1, 28 or 28, -1
::- 1, 28 或 28, - 1

2, -14 or -14, 2
::2, -14或 -14, 2

-2, 14 or 14, -2
::-2、14或14-2

4, -7 or -7, 4
::4,7或7,4

-4, -7 or -7, -4
::4 -7 或 - 7 - 4

The correctly factored expression will need a pair from the top list and a pair from the bottom list. This is 48 possible combinations to try.
::正确系数表达式需要从顶部列表中选择一对, 从底部列表中选择一对。这是48种可能的组合来尝试。

If you try the first pair from each list and multiply out you will see that the first and the last coefficients are correct but the $\begin{aligned} b \end{aligned}$ coefficient does not.
::如果您尝试从每个列表中选择第一对并乘出, 你会看到第一个和最后一个系数是正确的, 但 b 系数并不正确。

$\begin{aligned} (1 x + 1) (6 x - 28) = 6 x^{2} - 28 x + 6 x - 28 \end{aligned}$
:1x+1)(6x-28)=6x2-28x+6x-28)

A systematic approach to every one of the 48 possible combinations is the best way to avoid missing the correct pair. In this case it is:
::系统处理48种可能的组合中的每一种是避免错失正确组合的最佳办法。

$\begin{aligned} (2 x - 7) (3 x + 4) = 6 x^{2} + 8 x - 21 x - 28 = 6 x^{2} - 13 x - 28 \end{aligned}$
:2x-7(3x+4)=6x2+8x-21x-28=6x2-13x-28)

This method can be extremely long and rely heavily on good guessing which is why other methods are preferable.
::这一方法可能非常长,而且严重依赖良好的猜测,这就是为什么其他方法更可取的原因。

Factoring by Grouping
::分组计数

The next factoring technique is factoring by grouping . Suppose you start with an expression already in factored form :
::下一个乘数计算技术是按组进行乘数计算。假设您从一个已经以乘数表示的表达式开始 :

$\begin{aligned} 12 x^{2} + 4 x z + 3 x y + y z \end{aligned}$
::12x2+4xz+3xy+yz 12x2+4xz+3xy+yz

Notice that the first two terms are divisible by both 4 and $\begin{aligned} x \end{aligned}$ and the last two terms are divisible by $\begin{aligned} y \end{aligned}$ . First, factor out these common factors and then notice that there emerges a second layer of common factors. The binomial $\begin{aligned} (3 x + z) \end{aligned}$ is now common to both terms and can be factored out just as before.
::注意前两个术语可以以4和x两种方式拆分,最后两个术语可以以y方式拆分。首先,考虑这些共同因素,然后注意出现第二层共同因素。二进制(3x+z)现在在这两个术语中都是共同的,可以和以前一样考虑。

\begin{aligned} 12 x^{2} + 4 x z + 3 x y + y z & = 4 x (3 x + z) + y (3 x + z) \\ = (3 x + z) (4 x + y) \end{aligned}

::12x2+4xz+3xy+yz=4x(3x+z)+y(3x+z)=(3x+z)=(3x+z)(4x+y)

To verify your work, multiply the binomials and compare it with the original expression.
::为了验证您的工作, 将二进制数乘以, 并将其与原始表达式比较。

$\begin{aligned} (4 x + y) (3 x + z) = 12 x^{2} + 4 x z + 3 x y + y z \end{aligned}$
:4x+y)(3x+z)=12x2+4xz+3xy+yz

Usually when you multiply the factored form of a polynomial , two terms can be combined because they are like terms . In this case, there are no like terms that can be combined.
::通常情况下,当您乘以一个多边式的因数形式时,两个词可以合并,因为它们类似术语。在这种情况下,没有类似术语可以合并。

Quadratic Formula
::二次曲线公式

An alternative method to factor uses the quadratic formula as a clue even though this is an expression and not an equation set equal to zero.
::系数的替代方法是使用二次公式作为线索,即使这是一个表达式,而不是一个等于零的方程式。

$\begin{aligned} 6 x^{2} - 13 x - 28 \end{aligned}$
::6x2-13x-28

$\begin{aligned} a = 6, b = - 13, c = - 28 \end{aligned}$
::a=6,b13,c28

\begin{aligned} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{13 \pm \sqrt{169 - 4 \cdot 6 \cdot - 28}}{2 \cdot 6} & = \frac{13 \pm 29}{12} \\ = \frac{13 + 29}{12} & = \frac{42}{12} = \frac{7}{2} \\ o r \\ = \frac{13 - 29}{12} & = - \frac{16}{12} = - \frac{4}{3} \end{aligned}

::xbb2-4ac2a=13169-462826=132912=13+2912=13+2912=4212=72或13-2912161243

This means that when set equal to zero, this expression is equivalent to
::这意味着当设定为零等于零时,该表达式等于

$\begin{aligned} (x - \frac{7}{2}) (x + \frac{4}{3}) = 0 \end{aligned}$
:x-72(x+43)=0)

Multiplying by 2 and multiplying by 3 only changes the left hand side of the equation because the right hand side will remain 0. This has the effect of shifting the coefficient from the denominator of the fraction to be in front of the $\begin{aligned} x \end{aligned}$ .
::乘以 2 乘以 2 乘以 3 只改变方程式的左手侧, 因为右手侧将保持 0 。其效果是将系数从分数分母的分母移到 x 前面。

$\begin{aligned} 6 x^{2} - 13 x - 28 = (x - \frac{7}{2}) (x + \frac{4}{3}) = (2 x - 7) (3 x + 4) \end{aligned}$
::6x2-13x-28=(x-72)(x+43)=(2x-7)(3x+4)

Factoring Algorithm
::乘数算法

Another useful and efficient technique is the procedural factoring algorithm. The proof of the algorithm is beyond the scope of this book, but is a reliable technique for getting a handle on tricky factoring questions of the form: $\begin{aligned} 6 x^{2} - 13 x - 28 \end{aligned}$
::另一种有用和有效的技术是程序保理算法。算法的证明超出了本书的范围,但是一种可靠的技术,可以处理形式:6x2-13x-28的棘手保理问题。

We will factor $\begin{aligned} 6 x^{2} - 13 x - 28 \end{aligned}$ using the factoring algorithm to introduce it to you.
::我们将使用乘数算法将乘数 6x2 -13x -28 介绍给你。

First, multiply the first coefficient (6) with the last coefficient (28) and set the first coefficient to 1:
::首先,将第一个系数(6)乘以最后一个系数(28),并将第一个系数定为1:

$\begin{aligned} x^{2} - 13 x - 168 \end{aligned}$
::x2 - 13x - 168

Second, factor as you normally would with $\begin{aligned} a = 1 \end{aligned}$ :
::第二,与您通常使用 a=1 的系数相同:

$\begin{aligned} (x - 21) (x + 8) \end{aligned}$
:x-21)(x+8)

Third, divide the second half of each binomial by the coefficient that was multiplied in step 1:
::第三,将每一二进制的后半除以第1步乘以的系数:

$\begin{aligned} (x - \frac{21}{6}) (x + \frac{8}{6}) \end{aligned}$
:x-216)(x+86)

Fourth, simplify each fraction completely:
::第四,完全简化每一部分:

$\begin{aligned} (x - \frac{7}{2}) (x + \frac{4}{3}) \end{aligned}$
:x-72(x+43))

Lastly, move the denominator of each fraction to become the coefficient of $\begin{aligned} x \end{aligned}$ :
::最后,将每一分数的分母移动成x的系数:

$\begin{aligned} (2 x - 7) (3 x + 4) \end{aligned}$
:2x-7(3x+4))

Sum or Difference of Matching Odd Powers
::匹配奇差功率之和或差数

The last method of advanced factoring does not involve expressions of the form $\begin{aligned} a x^{2} + b x + c \end{aligned}$ . Instead, it involves the patterns that arise from factoring the sum or difference of terms with matching odd powers. The patterns are:
::最后一种高级保理学方法不涉及表x2+bx+c的表达方式。相反,它涉及因乘以与奇异功率相匹配的语句之和或差别而产生的模式。

$\begin{aligned} a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) \end{aligned}$
::a3+b3=(a+b)(a2-ab+b2)

$\begin{aligned} a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \end{aligned}$
::a3-b3=(a-b)(a2+ab+b2)

This method is shown in the examples below and the pattern is fully explored in the Review.
::下面的例子显示了这一方法,《审查报告》充分探讨了这一模式。

Examples
::实例

Example 1
::例1

Earlier, you were asked how the sum and difference of terms with matching odd powers can be factored. The sum or difference of terms with matching odd powers can be factored in a precise pattern because when multiplied out, all intermediate terms cancel each other out.
::早些时候,有人问您如何计算与奇异功率相匹配的条件之和和差异。与奇异功率相匹配的条件之和或差异可以用精确的模式来计算,因为当乘以时,所有中间条件都相互抵消。

\begin{aligned} a^{5} + b^{5} = (a + b) (a^{4} - a^{3} b + a^{2} b^{2} - a b^{3} + b^{4}) \end{aligned}

::a5+b5=(a+b)(a4-a3b+a2b2-ab3+b4)

When $\begin{aligned} a \end{aligned}$ is distributed: $\begin{aligned} a^{5} - a^{4} b + a^{3} b^{2} - a^{2} b^{3} + a b^{4} \end{aligned}$
::a 分配时间: a5-a4b+a3b2-a2b3+ab4

When $\begin{aligned} b \end{aligned}$ is distributed: $\begin{aligned} + a^{4} b - a^{3} b^{2} + a^{2} b^{3} - a b^{4} + b^{5} \end{aligned}$
::b 分配时:+a4b-a3b2+a2b3-ab4+b5

Notice all the inside terms cancel: $\begin{aligned} a^{5} + b^{5} \end{aligned}$
::通知所有内部条件取消: a5+b5

Example 2
::例2

Show that $\begin{aligned} a^{3} - b^{3} \end{aligned}$ factors into the result given in the Sum or Difference of Matching Odd Powers section.
::显示匹配奇数功率总和或差数一节中给出的结果中的 a3-b3 系数。

Factoring,
::保理,

\begin{aligned} a^{3} - b^{3} & = (a - b) (a^{2} + a b + b^{2}) \\ = a^{3} + a^{2} b + a b^{2} - a^{2} b - a b^{2} - b^{3} \\ = a^{3} - b^{3} \end{aligned}

::a3-b3=(a-b(a2+ab+b2)=a3+a2b+a2b2-a2b2-a2b-b2-b3=a3-b3=a3-b3)

Example 3
::例3

Show that $\begin{aligned} a^{3} + b^{3} \end{aligned}$ factors into the result given in the Sum or Difference of Matching Odd Powers section.
::显示 A3+b3 匹配奇数功率总和或差数部分中给定结果的 a3+b3 系数。

Factoring,
::保理,

\begin{aligned} a^{3} + b^{3} & = (a + b) (a^{2} - a b + b^{2}) \\ = a^{3} - a^{2} b + a b^{2} + b a^{2} - a b^{2} + b^{3} \\ = a^{3} + b^{3} \end{aligned}

::a3+b3=(a+b(a2-ab+b2)=a3-a2b+ab2+b2+ba2+b2+b2+b2+b3=a3+b3+b3=a3+b3+b3

Example 4
::例4

Factor the following expression without using the quadratic formula or trial and error:
::在不使用二次公式或试验和错误的情况下,将下列表达式乘以:

$\begin{aligned} 8 x^{2} + 30 x + 27 \end{aligned}$
::8x2+30x+27

Using the factoring algorithm:
::使用保理算法:

\begin{aligned} 8 x^{2} + 30 x + 27 & \to x^{2} + 30 x + 216 \\ \to (x + 12) (x + 18) \\ \to (x + \frac{12}{8}) (x + \frac{18}{8}) \\ \to (x + \frac{3}{2}) (x + \frac{9}{4}) \\ \to (2 x + 3) (4 x + 9) \end{aligned}

::8x2+30x+27x2+30x+30x+216}(x+12)(x+12)(x+18)}(x+128)(x+128)(x+1888)}(x+32)(x+32)(x+94)__(2x+3)(4x+9)

Summary

Guess and Check method involves listing all possible factor pairs and systematically trying combinations to find the correct factored expression.
::猜测和检查方法包括列出所有可能的因子配对和系统尝试组合,以找到正确的因子表达式。

Factoring by Grouping involves identifying common factors in the terms and factoring them out.
::分组的乘数涉及在术语中找出共同因素,并将这些因素考虑在内。

Quadratic Formula can be used as a clue for factoring polynomials, even when the expression is not set equal to zero.
::二次曲线公式可以用作计算多数值的线索,即使表达式不设为为零。

The Factoring Algorithm involves multiplying the first and last coefficients, factoring as usual, dividing the second half of each binomial, simplifying fractions, and moving the denominator to become the coefficient of the variable.
::保理算法涉及乘以第一个和最后一个系数,与往常一样计入系数,将每个二进制系数的后半除以,简化分数,并移动分母成为变量的系数。

Sum or Difference of Matching Odd Powers involves the patterns that arise from factoring the sum or difference of terms with matching odd powers
::匹配奇数功率的总和或差额涉及将条件与奇数功率相匹配之和或条件差异乘数所产生的模式。

Review
::回顾

Factor each expression completely.
::将每个表达式都完全考虑进去。

1. $\begin{aligned} 2 x^{2} - 5 x - 12 \end{aligned}$
::1. 2x2-5x-12

2. $\begin{aligned} 12 x^{2} + 5 x - 3 \end{aligned}$
::2. 12x2+5x-3

3. $\begin{aligned} 10 x^{2} + 13 x - 3 \end{aligned}$
::3. 10x2+13x-3

4. $\begin{aligned} 18 x^{2} + 9 x - 2 \end{aligned}$
::4. 18x2+9x-2

5. $\begin{aligned} 6 x^{2} + 7 x + 2 \end{aligned}$
::5. 6x2+7x+2

6. $\begin{aligned} 8 x^{2} + 34 x + 35 \end{aligned}$
::6. 8x2+34x+35

7. $\begin{aligned} 5 x^{2} + 23 x + 12 \end{aligned}$
::7. 5x2+23x+12

8. $\begin{aligned} 12 x^{2} - 11 x + 2 \end{aligned}$
::8. 12x2-11x+2

Expand the following expressions. What do you notice?
::展开下面的表达式。您注意到什么 ?

9. $\begin{aligned} (a + b) (a^{8} - a^{7} b + a^{6} b^{2} - a^{5} b^{3} + a^{4} b^{4} - a^{3} b^{5} + a^{2} b^{6} - a b^{7} + b^{8}) \end{aligned}$
::9. (a+b)(a8-a7b+a6b2-a5b3+a4b4-a3b5+a2b6-AB7+b8)

10. $\begin{aligned} (a - b) (a^{6} + a^{5} b + a^{4} b^{2} + a^{3} b^{3} + a^{2} b^{4} + a b^{5} + b^{6}) \end{aligned}$
::10. (a-b)(a6+a5b+a4b2+a3b3+a2b4+AB5+b6)

11. Describe in words the pattern of the signs for factoring the difference of two terms with matching odd powers.
::11. 以文字说明将两个条件的差别与奇数功率相配的标记模式。

12. Describe in words the pattern of the signs for factoring the sum of two terms with matching odd powers.
::12. 用文字说明用两个条件和奇数功率乘以两个条件之和的标志模式。

Factor each expression completely.
::将每个表达式都完全考虑进去。

13. $\begin{aligned} 27 x^{3} - 64 \end{aligned}$
::13. 27x3-64

14. $\begin{aligned} x^{5} - y^{5} \end{aligned}$
::14. x5-y5 14 x5-y5

15. $\begin{aligned} 32 a^{5} - b^{5} \end{aligned}$
::15. 32a5-b5

16. $\begin{aligned} 32 x^{5} + y^{5} \end{aligned}$
::16. 32x5+y5

17. $\begin{aligned} 8 x^{3} + 27 \end{aligned}$
::17. 8x3+27

18. $\begin{aligned} 2 x^{2} + 2 x y + x + y \end{aligned}$
::18. 2x2+2xy+x+y

19. $\begin{aligned} 8 x^{3} + 12 x^{2} + 2 x + 3 \end{aligned}$
::19. 8x3+12x2+2x+3

20. $\begin{aligned} 3 x^{2} + 3 x y - 4 x - 4 y \end{aligned}$
::20. 3x2+3xy-4x-4yy

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

高级保理

More Factoring Techniques ::更多保理技术

Guess and Check ::猜测和检查

Factoring by Grouping ::分组计数

Quadratic Formula ::二次曲线公式

Factoring Algorithm ::乘数算法

Sum or Difference of Matching Odd Powers ::匹配奇差功率之和或差数

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)