Section: 以四边形形式显示的代数表达式乘数 | 7.8 刻度表中的乘数代数表达式

Section outline

A large part of algebra is learning techniques to solve equations. In this chapter, we have focused on factoring 2nd- and 3rd- degree polynomial expressions, because if we can factor them, we can solve equations where the expression is equal to zero. We have a lot of techniques for dealing with lower-degree , but for polynomials with a degree of five or higher, the process is not straightforward. This problem is known as the Abel-Ruffini Theorem : There is no solution to polynomial equations of degree five or higher with arbitrary coefficients by adding, subtracting, multiplying, and dividing and by taking roots ¹ .
::代数的很大一部分是学习解方程的技巧。在本章中, 我们把注意力集中在二度和三度多元表达式的乘数上, 因为如果我们能将其乘数乘以二度和三度多元表达式, 我们就能解算表达式等于零的方程式。我们有很多处理低度表达式的技术, 但对于五度或以上多度的多元论者, 这一过程并不简单。这个问题被称为 Abel- Ruffini 理论: 以任意系数计算, 5级或以上 5 级或 5 级或 5 级的多等式是无法解决的。添加、减、乘、乘、分法和根根1 解决不了。

Not surprisingly, it is important to mathematicians to know which equations can be solved and which cannot. From this theorem grew an area of mathematics called Galois Theory, named after mathematician Évariste Galois (pictured below), which addresses this question and more ² . It is still an area of active research.
::毫不奇怪,数学家必须知道哪些方程式可以解答,哪些无法解答。从这个理论学说中,数学学界发展了一个数学领域,叫做Galois Theory,以数学家 Évariste Galois (下面的图象)命名,它解决了这个问题和更多的问题。它仍然是一个积极研究的领域。

Galois determined that some equations can be solved, like $\begin{aligned} 0 = x^{6} - 16 x^{3} + 64 \end{aligned}$ . We see how to factor an expression like $\begin{aligned} x^{6} - 16 x^{3} + 64 \end{aligned}$ in this section.
::Galois确定有些方程式可以解答, 比如 0=x6 - 16x3+64。我们可以看到如何在本节中将 X6 - 16x3+64 等表达式乘以。

Factoring Expressions That Are Quadratic in Form
::表中的二次曲线乘数表达式

Expressions that are quadratic in form can be written in the form $\begin{aligned} a u^{2} + b u + c \end{aligned}$ , where u is a placeholder for another expression. We call this placeholder a dummy variable . Expressions that are quadratic in form involve the sum or difference of an expression and its square. This is not always easy to recognize, so the dummy variable may be helpful in identifying expressions that are quadratic in form.
::以 au2+bu+c 形式写成的二次表达式可以以窗体 au2+bu+c 的形式写成,其中u 是另一种表达式的占位符。我们将此占位符称为假变量。形式为二次表达式的表示式涉及表达式及其正方形的和或差。这并不总是容易识别,因此假变量可能有助于识别形式为二次表达式的表达式。

As you go through this section, you may see less of a need for the dummy variable. Think of it like training wheels on a bicycle. You take the training wheels off when you feel comfortable riding the bike. Likewise, you can stop using a dummy variable when you feel comfortable factoring without it. For those of you who want to be cautious, we will use a dummy variable throughout this section.
::在您通过此节时, 您可能会看到对假变量的需求更少。把它想象成在自行车上训练轮子。当您觉得自己可以骑自行车时, 就可以把培训轮子拿掉。同样, 您也可以在没有它时, 当您感到舒适时, 停止使用假变量。对于你们中想要谨慎的人, 我们将在整个章节中使用一个假变量。

Example 1
::例1

Factor $\begin{aligned} 2 x^{4} - x^{2} - 15 \end{aligned}$ .
::2x4 -x2 - 15。

Solution: This expression involves an $\begin{aligned} x^{2} \end{aligned}$ and an $\begin{aligned} (x^{2})^{2} = x^{4} \end{aligned}$ . Let's use a dummy variable for $\begin{aligned} x^{2} \end{aligned}$ — say, $\begin{aligned} u = x^{2} . \end{aligned}$ Then, $\begin{aligned} x^{4} = (x^{2})^{2} = u^{2} . \end{aligned}$ Replacing these in the expression gives us an expression that is quadratic in form: $\begin{aligned} 2 u^{2} - u - 15 \end{aligned}$ . Factoring by grouping , we have
::解析度 : 此表达式包含 x2 和 a (x2) 2=x4 。让我们对 x2 使用一个假变量来表示 x2 - 例如, u=x2 。然后, x4= (x2) 2=u2 。将这些变量替换为表达式中的表达式, 给我们以二次表达式的形式表示 : 2u2 - u- 15 。通过分组进行计算, 我们找到了

$\begin{aligned} 2 u^{2} - u - 15 \\ = & 2 u^{2} - 6 u + 5 u - 15 \\ = & 2 u (u - 3) + 5 (u - 3) \\ = & (u - 3) (2 u + 5) . \end{aligned}$

::2u2-u-15=2u2-6u+5u-15=2u(u-3)+5(u-3)=(u-3)(2u+5)。

Lastly, we substitute the original expression for u : $\begin{aligned} (x^{2} - 3) (2 x^{2} + 5) \end{aligned}$ . Neither of these are factorable further, so we have factored completely.
::最后,我们用原来的词语来代替ux2-3)(2x2+5),这两个词都不是进一步考虑的因素,因此我们完全考虑。

by pcoetech demonstrates how to factor expressions that are in quadratic form .
::通过 pcoetech 演示如何使用二次形的系数表达式。

Example 2
::例2

Factor $\begin{aligned} x^{6} - 16 x^{3} + 64 \end{aligned}$ .
::系数 x6-16x3+64。

Solution: First we need to see if the expression might be quadratic in form. $\begin{aligned} x^{6} = (x^{3})^{2}, \end{aligned}$ so this expression is quadratic in form. Using a dummy variable, $\begin{aligned} u = x^{3}, \end{aligned}$ we have
::解析度 : 首先, 我们需要首先查看表达式是否为形态的四方形。 x6= (x3) 2, 所以这个表达式为形态的四方形。使用一个假变量 u=x3, 我们发现

$\begin{aligned} u^{2} - 16 u + 64 & = (u - 8)^{2} \\ = (x^{3} - 8)^{2} replacing u in the factors \\ = (x^{3} - 8) (x^{3} - 8) two factors are difference of cubes \\ = (x - 2) (x^{2} + 2 x + 4) (x - 2) (x^{2} + 2 x + 4) \\ = (x - 2)^{2} (x^{2} + 2 x + 4)^{2} . \end{aligned}$

::u2-16u+64=(u-8)2=(x3-8)(x3-8)(x3-8)(x3-8)2替代因数u=(x3-8)(x3-8)两个因数是立方体=(x-2)(x2+2+2x+4)(x-2)(x-2)(x2+2x+4)=(x-2)(x2+2x+4)=(x-2)(x2+2+4)2的差数。

This video by CK-12 demonstrates how to factor expressions that are in quadratic form.
::CK-12的这段视频展示了以二次形表示的系数。

Example 3
::例3

F actor $\begin{aligned} 6 x^{5} - 51 x^{3} - 27 x \end{aligned}$ .
::系数 6x5-51-51x3-27.x

Solution: First, pull out the GCF among the three terms .
::解决办法:首先,在三个条件中拿出绿色气候基金。

$\begin{aligned} 6 x^{5} - 51 x^{3} - 27 x = 3 x (2 x^{4} - 17 x^{2} - 9) \end{aligned}$

::6x5-51x3-27x=3x(2x4-17x2-9)

The expression inside the " data-term="Parentheses" role="term" tabindex="0"> parentheses is quadratic in form. Using a dummy variable, $\begin{aligned} u = x^{2} \end{aligned}$ , we have
::括号内表达式的表达式以四方形形式出现。使用 u=x2 的模拟变量,我们发现

$\begin{aligned} 2 u^{2} - 17 u - 9 \\ = 2 u^{2} - 18 u + u - 9 \\ = 2 u (u - 9) + 1 (u - 9) \\ = (u - 9) (2 u + 1) \\ = (x^{2} - 9) (2 x^{2} + 1) . \end{aligned}$

::2u2-17u-9=2u2-18u-18u+18u-9=2u(u-9)+1(u-9)=(u-9)(2u+1)=(x2-9)(2x2+1)。

Factor $\begin{aligned} x^{2} - 9 \end{aligned}$ further, since it is a difference of two squares, and include the GCF factor again.
::系数x2-9又进一步,因为它有两平方之差,并再次包括绿色气候基金系数。

$\begin{aligned} 3 x (x^{2} - 9) (2 x^{2} + 1) \\ = 3 x (x - 3) (x + 3) (2 x^{2} + 1) \end{aligned}$

::3x(x2- 9) (2x2+1) = 3x(x-3)(x+3) (2x2+1)

Summary
::摘要

To recognize an expression that is quadratic in form, note that the variable part of one term needs to be the square of the variable part in another term.
::要承认形式上为四边形的表达式,请注意,一个术语的可变部分必须是另一个术语中可变部分的正方形。

To factor an expression that is quadratic in form, use a dummy variable to rewrite the expression as a quadratic expression, factor the expression, and then replace the dummy variable at the end of the process.
::将表达式乘以形式为二次形的表达式,使用假变量将表达式重写为二次形表达式,乘以表达式,然后在进程结束时替换假变量。

Review
::回顾

Factor the following expressions completely:
::将以下表达式完全乘以 :

1. $\begin{aligned} x^{4} - 6 x^{2} + 8 \end{aligned}$
::1. x4-6x2+8

2. $\begin{aligned} x^{8} - 4 x^{4} - 45 \end{aligned}$
::2. x8 - 4x4 - 45

3. $\begin{aligned} x^{4} - 18 x^{2} + 45 \end{aligned}$
::3. x4-18x2+45

4. $\begin{aligned} 4 x^{4} - 11 x^{2} - 3 \end{aligned}$
::4. 4x4-11x2-3

5. $\begin{aligned} 6 x^{10} + 19 x^{5} + 8 \end{aligned}$
::5. 6x10+19x5+8

6. $\begin{aligned} 6 x^{5} + 26 x^{3} - 20 x \end{aligned}$
::6. 6x5+26x3-20x

7. $\begin{aligned} 2 x^{4} - 5 x^{2} - 12 \end{aligned}$
::7. 2x4-5x2-12

8. $\begin{aligned} 12 x^{6} + 69 x^{4} + 45 x^{2} \end{aligned}$
::8. 12x6+69x4+45x2

9. $\begin{aligned} 3 x^{6} + 17 x^{3} - 6 \end{aligned}$
::9. 3x6+17x3-6

10. $\begin{aligned} x - 2 \sqrt{x} + 1 \end{aligned}$
::10. x-2x+1

Explore More
::探索更多

1. The volume of a rectangular prism is $\begin{aligned} 10 x^{3} - 25 x^{2} - 15 x \end{aligned}$ . What are the lengths of the prism's sides?
::1. 矩形棱晶体积为10x3-25x2-15x。棱晶面的长度是多少?

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

References
::参考参考资料

1. "Abel-Ruffini Theorem," last edited May 18, 2017,
::1. “Abel-Ruffini Theorem”, 2017年5月18日编辑,

2. "Évariste Galois," last edited May 18, 2017,
::2017年5月18日,

Section outline

Factoring Expressions That Are Quadratic in Form ::表中的二次曲线乘数表达式

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers for Review and Explore More Problems ::回顾和探讨更多问题的答复

References ::参考参考资料

Factoring Expressions That Are Quadratic in Form
::表中的二次曲线乘数表达式

Example 1
::例1

Example 2
::例2

Example 3
::例3

Summary
::摘要

Review
::回顾

Explore More
::探索更多

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

References
::参考参考资料