节: 分组计数 | 7.7 按集团计数

章节大纲

The Charlot equation in chemistry can be used to determine the pH of buffer solutions, like the bicarbonate buffer solution in blood—the liquid part, which is made up of carbonic acid, bicarbonate ion, and carbon dioxide ^1-3 . If the equation to model this is $\begin{align*}x^3+6x^2-12x-72=0\end{align*}$ , where x is the hydrogen ion concentration, we can determine this concentration by factoring. We learn how to factor like this one in this section.
::化学中的碳化方程式可以用来确定缓冲溶液的pH值,例如血液中的碳酸双碳酸缓冲溶液-液体部分,由碳酸、碳酸双碳酸离子和二氧化碳1-3组成。如果模型的方程式是 x3+6x2-12x-72=0,其中x是氢离子浓度,我们可以通过乘数来确定这种浓度。我们学会了如何将这一元素纳入本节。

Factoring By Grouping
::逐组计数

In the previous section, we introduced factoring by grouping of quadratic expressions. We will expand this idea to other expressions here.
::在前一节中,我们引入了按等方表达式分组的乘数。我们将将这个概念扩展至这里的其他表达式。

Example 1
::例1

Factor $\begin{align*}x^4+7x^3-8x-56\end{align*}$ by grouping.
::组合法乘以 x4+7x3-8x-56。

Solution: First, try to group the 1st two and last two terms together. Factor out any common factors in the groups.
::解决方案: 首先, 尝试将前两个和最后两个条件组合在一起。将组合中的任何共同因素考虑在内。

$\begin{aligned} \underset{x^{3} (x + 7)}{\underset{⏟}{x^{4} + 7 x^{3}}} \underset{- 8 (x + 7)}{\underset{⏟}{- 8 x - 56}} \end{aligned}$
$\begin{align*}\underbrace{x^4+7x^3}_{x^3{\color{green}(x+7)}}\underbrace{-8x-56}_{ \text{-}8{\color{green}(x+7)}}\end{align*}$
::x4+7x33x3(x+7)-8x-568(x+7)

Notice that what is inside the " data-term="Parentheses" role="term" tabindex="0"> parentheses is the same . If you can factor by grouping, this should happen. Now we factor out the common binomial .
::注意括号内的内容是相同的。如果您可以通过分组来计算, 应该会发生这种情况。现在我们可以计算出共同的二进制。

$\begin{aligned} x^{3} (x + 7) - 8 (x + 7) \\ (x + 7) (x^{3} - 8) \end{aligned}$
$\begin{align*}& x^3(x+7)-8(x+7)\\ & (x+7)(x^3-8)\end{align*}$
::x3(x+7)-8(x+7)(x+7)(x7)(x3-8)

Next, we check to see if the factors can be factored any further. The 2nd factor is a difference of cubes. Using the formula ,
::下一步,我们检查一下这些系数是否可以进一步计算。第二个系数是立方体的差别。使用公式,

$\begin{aligned} (x + 7) (x^{3} - 8) \\ (x + 7) (x - 2) (x^{2} + 2 x + 4) . \end{aligned}$
$\begin{align*}& (x+7)(x^3-8)\\ & (x+7)(x-2)(x^2+2x+4).\end{align*}$
:x+7)(x3-8)(x+7)(x-2)(x2+2x+4)。)

by CK-12 demonstrates how to factor by grouping.
::在 CK-12 中显示如何通过分组来计算系数。

Example 2
::例2

Factor $\begin{align*}x^3+5x^2-x-5\end{align*}$ by grouping.
::组合法乘以 x3+5x2-x-5。

Solution: Grouping the 1st two and the last two terms, we have
::解决方案:将第一和第二个和最后两个任期分组,我们有:

$\begin{aligned} x^{3} + 5 x^{2} - x - 5 \\ x^{2} (x + 5) - 1 (x + 5) \\ (x + 5) (x^{2} - 1) \end{aligned}$
$\begin{align*}& {\color{green}{x^3+5x^2}}{\color{violet}{-x-5}}\\ & {\color{green}{x^2(x+5)}}{\color{violet}{-1(x+5)}}\\ & (x+5)(x^2-1)\end{align*}$
::x3+5x2 - x5x2 - x5x2(x+5) - 1(x+5)(x+5)(x5)(x2-1)

Look to see if we can factor any further. The 2nd factor is a difference of squares .
::看看我们能否进一步考虑。第二个因素是方形的差别。

$\begin{aligned} (x + 5) (x^{2} - 1) \\ (x + 5) (x - 1) (x + 1) \end{aligned}$
$\begin{align*}& (x+5)(x^2-1)\\ &(x+5)(x-1)(x+1)\end{align*}$
:x+5)(x2-1)(x+5)(x-1)(x+5)(x-1)(x+1)(x+1))

This video by CK-12 demonstrates how to factor polynomials by grouping.
::CK-12的这段影片展示了如何通过分组来考虑多语种因素。

Example 3
::例3

Factor $\begin{align*}2x^3-3x^2+8x-12\end{align*}$ .
::系数 2x3-3x2+8x-12.

Solution: We group the 1st two and 2nd two terms.
::解决方案:我们组第2和第2个任期。

$\begin{aligned} 2 x^{3} - 3 x^{2} + 8 x - 12 \\ x^{2} (2 x - 3) + 4 (2 x - 3) \\ (2 x - 3) (x^{2} + 4) \end{aligned}$
$\begin{align*}{\color{green}{2x^3-3x^2}}{\color{violet}{+8x-12}}\\ {\color{green}{x^2(2x-3)}}{\color{violet}{+4(2x-3)}} \\ (2x-3)(x^2+4) \end{align*}$
::2x3-3x2+8x-12x2(2x-3)+4(2x-3)(2x-3)(2x-3)(x2+4)

Now, determine if you can factor further. No, $\begin{align*}x^2+4\end{align*}$ is a sum of squares and a prime factor.
::现在, 确定您能否进一步系数。否, x2+4 是平方和质因子的和。

Example 4
::例4

Factor $\begin{align*}2x^4-5x^3+2x-5\end{align*}$ .
::系数 2x4 - 5x3+2x-5。

Solution: Again, we group the 1st two terms and the last two terms and factor out GCFs where we have them.
::解决方案:我们再次将头两个任期和最后两个任期组合在一起,并将我们拥有的绿色气候基金排除在外。

$\begin{aligned} 2 x^{4} - 5 x^{3} + 2 x - 5 \\ x^{3} (2 x - 5) + 1 (2 x - 5) \\ (2 x - 5) (x^{3} + 1) sum of cubes \\ (2 x - 5) (x + 1) (x^{2} + x + 1) \end{aligned}$
$\begin{align*}& {\color{green}{2x^4-5x^3}}{\color{violet}{+2x-5}}\\ & {\color{green}{x^3(2x-5)}}{\color{violet}{+1(2x-5)}}\\ & (2x-5)(x^3+1) \qquad \qquad \ \color{gray}{\text{sum of cubes}}\\ & (2x-5)(x+1)(x^2+x+1)\end{align*}$
::2x4-5x3+2x2x-2x-5)+1(2x-5)+1(2x-5)(2x-5)(x5)(x3+1)立方体(2x-5)(x+1)和立方体(2x-5)(x+1)(x2+x+1)

This video by CK-12 provides additional examples of by grouping terms.
::CK-12的这段影片通过分组术语提供了更多例子。

Example 5
::例5

Factor the expression from the Charlot equation from the Introduction: $\begin{align*}x^3+6x^2-12x-72\end{align*}$ .
::导言中的字符罗方程式表达式的乘数: x3+6x2-12x-72。

Solution: We group the 1st two terms and the last two terms.
::解决方案:我们将第一个任期和最后两个任期分组。

$\begin{aligned} x^{3} + 6 x^{2} - 12 x - 72 \\ = x^{2} (x + 6) - 12 (x + 6) \\ = (x + 6) (x^{2} - 12) \end{aligned}$
$\begin{align*}&{\color{green}{x^3+6x^2}}{\color{violet}{-12x-72}}\\ &={\color{green}{x^2(x+6)}}{\color{violet}{-12(x+6)}}\\ &=(x+6)(x^2-12)\end{align*}$
::x3+6x2- 12x- 72=x2( x+6)- 12( x+6)=( x+6)( x2- 12)

Summary
::摘要

To factor by grouping, look for terms that have a GCF, and factor the GCF out of those terms. If you can factor by grouping, what remains after you factor will also have a GCF, and you can factor that out to factor completely.
::通过分组来考虑因素,寻找具有全球合作框架的术语,并将全球合作框架从这些术语中考虑出来。如果能够通过分组来考虑因素,剩下的因素也会有一个全球合作框架,并且可以完全考虑因素。

Review
::回顾

Factor the polynomials below using factoring by grouping. Factor each polynomial completely.
::乘以下方的多元值乘以乘数组合。乘以每个多元值的完全值。

1. $\begin{align*}x^3-4x^2+3x-12\end{align*}$
::1. x3 - 4x2+3x- 12

2. $\begin{align*}x^3+6x^2-9x-54\end{align*}$
::2. x3+6x2-9x-54

3. $\begin{align*}3x^3-4x^2+15x-20\end{align*}$
::3. 3x3-4x2+15x-20

4. $\begin{align*} 2x^4-3x^3-16x+24\end{align*}$
::4. 2x4-3x3-16x+24

5. $\begin{align*}4x^3+4x^2-25x-25\end{align*}$
::5. 4x3+4x2-2-25x-25

6. $\begin{align*}4x^3+18x^2-10x-45\end{align*}$
::6. 4x3+18x2-10x-45

7. $\begin{align*}24x^4-40x^3+81x-135\end{align*}$
::7. 24x4-40x3+81x-135

8. $\begin{align*}15x^3+6x^2-10x-4\end{align*}$
::8. 15x3+6x2-10x-4

9. $\begin{align*}4x^3+5x^2-100x-125\end{align*}$
::9. 4x3+5x2-100x-125

10. $\begin{align*}3x^3-2x^2+12x-8\end{align*}$
::10. 3x3-2x2+12x-8

11. $\begin{align*}9x^3-54x^2-4x+24\end{align*}$
::11. 9x3-54x2-4x+24

12. $\begin{align*}x^4+3x^3-27x-81\end{align*}$
::12. x4+3x3-27x-81

13. $\begin{align*}x^3-2x^2-4x+8\end{align*}$
::13. x3-2x2-4x+8

Explore More
::探索更多

1. The volume of a rectangular prism is $\begin{align*}3x^5 - 27x^4 - 2x^2 + 18x\end{align*}$ . What are the lengths of the prism's sides?
::1. 矩形棱晶体积为3x5-27x4-2x2+18x。棱晶面的长度是多少?

2. Solve the Charlot equation from the Introduction using the zero product property.
::2. 使用零产品属性从导言中解决Charlot等式。

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。

PLIX
::PLIX

Try this interactive that reinforces the concepts explored in this section:
::尝试这一互动,强化本节所探讨的概念:

References
::参考参考资料

1. "Buffer Solution," last edited May 24, 2017,
::1. 2017年5月24日上次编辑的"缓冲解决方案"

2. "Blood," last edited May 22, 2017,
::2017年5月22日上次编辑的"血迹"

3. "Charlot Equation," last edited August 5, 2014,
::3. 2014年8月5日经编辑的《地平法》

章节大纲

Factoring By Grouping ::逐组计数

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

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

PLIX ::PLIX

References ::参考参考资料

Factoring By Grouping
::逐组计数

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

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

PLIX
::PLIX

References
::参考参考资料