课程： 7.6 当主要系数不等于1时的计数四方

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当领导系数不等于 1 时计数四方
The height achieved by a person when shot out of a cannon at the circus is $\begin{aligned} - 16 t^{2} + 64 t + 36 \end{aligned}$ , where t is time in seconds and height is measured in feet. Approximately how long was the human cannonball in the air? We can use factoring to help us determine this, and we discuss factoring quadratic expressions where the leading coefficient is not equal to 1 in this section.
::一个人在马戏团射出大炮时所达到的高度是 - 16t2+64t+36, 其时间以秒计, 高度以脚计。估计空中的人体炮弹持续了多久? 我们可以使用乘数来帮助我们确定这一点, 我们讨论四边表达法, 其中主要系数不等于本节的 1 。

Factoring Quadratic Expressions When the Leading Coefficient is Not Equal to 1 By Grouping
::组合计算,当前高系数不等于1时的二次曲线表达式乘数

Factoring by grouping is an algebraic technique in which you collect terms that have a greatest common factor in groups, and then factor out the GCF from just those terms. If you can , there will be another GCF to factor out, and hopefully, what remains are prime factors.
::分组计算是一种代数技术,其中你收集了在组别中具有最大共同因素的术语,然后将绿色气候基金从这些术语中推算出来。如果可以的话,就会有另一个全球气候基金来考虑,希望还有什么是主要因素。

In the previous section, we used a shortcut to factor when the leading coefficient was 1. Since the leading coefficient is not equal to 1, we cannot use the shortcut here.
::在前一节中,我们用捷径来计算主要系数为1.,因为主要系数不等于1,我们在这里不能使用捷径。

Example 1
::例1

Factor $\begin{aligned} 4 x^{2} + 8 x - 5 \end{aligned}$ .
::因数 4x2+8x-5。

Solution: Let's consider this in steps.
::解决问题:让我们一步一步地考虑一下。

1. Find $\begin{aligned} a c \end{aligned}$ and the factors of this number that add up to $\begin{aligned} b \end{aligned}$ . Notice this is a similar 1st step to the previous section, where the leading coefficient or a was 1: $\begin{aligned} 1 \cdot c = c \end{aligned}$ . In this example, we are looking for numbers that multiply to -20 and add to 8.
::1. 查找 ac 和该数字的系数加到 b 。注意这是与上一节类似的第一步,前一节的主要系数或系数为1: 1,c=c。例举一例,我们正在寻找乘以 - 20和加到8的数字。

$\begin{aligned} 11 \cdot 20 & 20 - 1 = 19 \\ 2 \cdot 10 & 10 - 2 = 8 \\ 4 \cdot 5 & 5 - 4 = 1 \end{aligned}$
Since the middle term is positive, the factors that work are 10 and -2.
::112020-1=1921010-2=8455-4=1 由于中期为正数,工作因素为10和-2。

2. Next, we want to rewrite the trinomial with the x- term expanded, using the two factors from Step 1. In this example, we replace the 8 x with 10 x + -2 x .
::2. 其次,我们想用扩大的x期重写三部曲,使用第1步的两个因素,将8x改为10x+-2x,在此示例中,我们用10x+-2x取代8x。

$\begin{aligned} 4 x^{2} + 10 x + - 2 x - 5 \end{aligned}$

::4x2+10x+-2x-5

3. Subsequently, we want to group the 1st two and 2nd two terms together. Then we want to find the GCF of each group. Notice that after following these steps, we have a GCF of $\begin{aligned} 2 x + 5 \end{aligned}$ t hat we can factor out from both terms.
::3. 随后,我们要将第一二和第二二任期组合在一起,然后我们想找到每个集团的绿色气候基金,注意在采取这些步骤之后,我们有一个2x+5的绿色气候基金,我们可以从两个条件中加以考虑。

$\begin{aligned} 4 x^{2} + 10 x + - 2 x - 5 \\ ↙ ↘ \\ (4 x^{2} + 10 x) + (- 2 x - 5) \\ 2 x (2 x + 5) - 1 (2 x + 5) \\ (2 x + 5) (2 x - 1) \end{aligned}$

::4x2+10x+2-2x-5 (4x2+10x)+(2x-5)5x2x(2x+5)-1-2x+5(2x+5)(2x+5)-5(2x-1)

What happens if we list $\begin{aligned} - 2 x \end{aligned}$ before $\begin{aligned} 10 x \end{aligned}$ in Step 2?
::如果我们在第二步10x之前列出 -2x呢?

$\begin{aligned} 4 x^{2} - 2 x + 10 x - 5 \\ (4 x^{2} - 2 x) + (10 x - 5) \\ 2 x (2 x - 1) + 5 (2 x - 1) \\ (2 x - 1) (2 x + 5) \end{aligned}$

::4x2-2x+10x-5(4x2-2-2x)+(10x-5)2x(2x-1)+5(2x-1)-2x-1(2x-1)(2x-1)(2x-1)(2x-1)(2x+5)

This tells us it does not matter which x- term we list 1st in Step 2 above.
::这告诉我们,在上文步骤2中,我们列出哪个X-期是第1级,这无关紧要。

by Mathispower4u demonstrates how to factor trinomials when the leading coefficient is not equal to 1.
::通过 Mathispower4u 演示当主要系数不等于 1 时如何乘以三角系数。



Example 2
::例2

Factor $\begin{aligned} 15 x^{2} + 14 x + 3 \end{aligned}$ .
::15x2+14x+3 系数

Solution: There is no GCF, so we find the factors of $\begin{aligned} a c \end{aligned}$ that add up to $\begin{aligned} b \end{aligned}$ . In this example, we are looking for numbers that multiply to 45 and add to 14.
::解决办法:没有全球合作框架,因此,我们发现乙丙丙酯因素加起来等于b。在这个例子中,我们寻求的是乘以45的数字,加到14。

$\begin{aligned} 1 \cdot 45 & 1 + 45 = 46 \\ 3 \cdot 15 & 3 + 15 = 18 \\ 5 \cdot 9 & 5 + 9 = 14 \end{aligned}$

The factors of 45 that add up to -4 are -9 and 5, since the larger number in absolute value has to be negative. We split up the middle term to create two groups.
::45个因素加起来等于 - 4是 - 9 和 5,因为绝对值的较大数字必须是负数。我们将中期分成两个组, 组成两个组。

$\begin{aligned} 15 x^{2} + 14 x + 3 \\ ↙ ↘ \\ 15 x^{2} + 5 x + 9 x + 3 \end{aligned}$
Now, working with each of the groups, we have
::15x2+14x+3 +15x2+5x+9x+3 现在,我们与每个集团合作,

$\begin{aligned} (15 x^{2} + 5 x) + (9 x + 3) \\ 5 x (3 x + 1) + 3 (3 x + 1) \\ (3 x + 1) (5 x + 3) . \end{aligned}$

:15x2+5x)+(9x+3)5x(3x+1)+3(3x+1)+3(3x+1)(3x+1)(5x+3)。

Example 3
::例3

Factor $\begin{aligned} 24 x^{2} - 30 x - 9 \end{aligned}$ .
::24x2-30x-9因数

Solution: $\begin{aligned} 24 x^{2} - 30 x - 9 \end{aligned}$ also has a GCF of 3. Factoring this out, we have $\begin{aligned} 3 (8 x^{2} - 10 x - 3) \end{aligned}$ . $\begin{aligned} a c = - 24 \end{aligned}$ . The factors of -24 that add up to -10 are -12 and 2.
::解决方案:24x2-30x-9还有3个全球合作框架,包括3个全球合作框架,我们有3(8x2-10x-3)、ac=24。

$\begin{aligned} 3 (8 x^{2} - 10 x - 3) \\ 3 [(8 x^{2} - 12 x) + (2 x - 3)] \\ 3 [4 x (2 x - 3) + 1 (2 x - 3)] \\ 3 (2 x - 3) (4 x + 1) \end{aligned}$

::3(3)8x2-10x-3)3[(8x2-12x)+(2x-3)]3[(4x(2x-3)+1(2x-3)]3(2x-3)(4x+1)

Factoring Quadratic Expressions By the Bottoms Up Method
::以自下游方法乘以二次曲线表达式

The "bottoms up" method starts out the same way as factoring by grouping, but then changes. Let's consider an example. (For a more extensive proof of this method, see the video in the Resources section.)
::“ 向上下” 方法与分组计数相同, 但会改变。让我们来举个例子。 (关于这一方法的更广泛证据, 请参见资源部分的视频。)

Example 4
::例4

Factor $\begin{aligned} 6 x^{2} - x - 2 \end{aligned}$ .
::因数 6x2-x-2。

Solution: We start the same by finding the two numbers whose product is $\begin{aligned} a c \end{aligned}$ and sum is $\begin{aligned} b \end{aligned}$ . Let's use the $\begin{aligned} X \end{aligned}$ to help us organize this.
::解决方案:我们从发现两个数字开始, 其产品为ac, 和总和是b。让我们用X来帮助我们组织这个。

The factors that work are 3 and -4. Next, we take these factors, divide each by the leading coefficient, 6, and simplify where possible.
::起作用的因素是3和4。接下来,我们采取这些因素,将每个因素除以主要系数6,并尽可能简化。

$\begin{aligned} \frac{3}{6} = \frac{1}{2} & \frac{- 4}{6} = \frac{- 2}{3} \end{aligned}$

Now, we write the factors as
::现在,我们把因素写成

$\begin{aligned} (x + \frac{1}{2}) (x - \frac{2}{3}) . \end{aligned}$

:x+12)(x-23)。

Next, we bring the "bottoms" or denominators up by multiplying each term in a factor through by the denominator in the fraction . We have
::接下来,我们把“底部”或分母通过一个因子乘以分母乘以一个因子,然后把“底部”或分母调高。我们有,我们有。

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

:2x+1)(3x-2)=6x2-x-2。

The factors of $\begin{aligned} 6 x^{2} - x - 2 \end{aligned}$ are $\begin{aligned} (2 x + 1) (3 x - 2) \end{aligned}$ . You can FOIL these to check your answer.
::6x2-x-2 的因数为 2x+1 (3x-2) 。您可查看这些因数以查看您的答案。

by MT Clinch demonstrates why the b ottoms up factoring method works.
::由MT Clinch 显示自下而上保理法为何有效。

Example 5
::例5

Factor $\begin{aligned} 12 x^{2} - 22 x - 20 \end{aligned}$ .
::系数 12x2-22x-20。

Solution: We will use the bottoms up method here, but 1st we need to factor out the GCF from all three terms.
::解决方案:在这里,我们将使用自下而上的方法,但首先,我们需要将全球合作框架从所有三个条件中考虑出来。

$\begin{aligned} 12 x^{2} - 22 x - 20 = 2 (6 x^{2} - 11 x - 10) \end{aligned}$

::12x2-22x-20=2(6x2-11x-10)

Using what is inside the " data-term="Parentheses" role="term" tabindex="0"> parentheses , find $\begin{aligned} a c \end{aligned}$ and determine the factors that add up to $\begin{aligned} b \end{aligned}$ .
::使用括号内的内容,查找 ac 并确定构成 b 的因素。

$\begin{aligned} 6 \cdot - 10 = - 60 \to - 15 \cdot 4 = - 60, - 15 + 4 = - 11 \end{aligned}$

The factors of -60 that add up to -11 are -15 and 4.
::总计为 -11的 -60 系数为 -15 和 4。

Next, we divide -15 and 4 by the leading coefficient, or 6.
::其次,我们将15和4除以主要系数,即6。

$\begin{aligned} \frac{- 15}{6} = \frac{- 5}{2} & \frac{4}{6} = \frac{2}{3} \end{aligned}$

Putting these into factors and bringing the "bottoms" up, we have
::将这些因素纳入因素,把“底底”加起来,我们有:

$\begin{aligned} (x - \frac{5}{2}) (x + \frac{2}{3}) \\ (2 x - 5) (3 x + 2) \end{aligned}$
So, factored completely, $\begin{aligned} 12 x^{2} - 22 x - 20 = 2 (2 x - 5) (3 x + 2) . \end{aligned}$
:x-52)(x+23)(2x-5)(3x+2)So,完全系数,12x2-22x-20=2(2x-5)(3x+2)。

by Mathispower4u demonstrates how to factor trinomials where the leading coefficient is not equal to 1 using the bottoms up method.
::由 Mathispower4u 演示如何用自下而上的方法将前位系数不等于 1 的三角系数乘以。

Example 6
::例6

The height achieved by a person when shot out of a cannon at the circus is $\begin{aligned} - 16 t^{2} + 64 t + 36 \end{aligned}$ , where t is time in seconds and height is measured in feet. Write this expression in factored form .
::一个人在马戏团从大炮中射出时达到的高度为 - 16t2+64t+36, 其中 t 以秒计时, 高度以脚计。以参数形式写入此表达式。

Solution: The 1st thing we should always look for when factoring is a GCF. A factor of each of the numbers in the quadratic expression is 4, so we can factor that out, and we will also factor out -1, so our leading coefficient is positive. This is for convenience.
::解决方案:当保理因素是一个全球合作框架时,我们总是需要寻找的第一件事。在二次表达中,每个数字的一个系数是 4, 所以我们可以考虑这个系数, 我们还可以考虑 -1, 所以我们的主要系数是正数。这是为方便起见。

This leaves us with $\begin{aligned} - 4 (4 t^{2} - 16 t - 9) \end{aligned}$ . To factor what remains in the parentheses, we need two numbers that multiply to -36 that add up to -16. The numbers 2 and -18 satisfy both conditions.
::这将给我们留下 -4(4t2-16t-9) 。要将括号中残留的值乘以 -36, 乘以 -16。数字2 和 -18 符合两个条件。

$\begin{aligned} \frac{2}{4} = \frac{1}{2} \frac{- 18}{4} = \frac{- 9}{2} \\ (x + \frac{1}{2}) (x - \frac{9}{2}) \\ (2 x + 1) (2 x - 9) \end{aligned}$
Altogether, we have $\begin{aligned} - 4 (2 x + 1) (2 x - 9) \end{aligned}$ .
::24=12-184=92(x+12)(x-92)(2x+1)(2x-9),我们共有 -4(2x+1)(2x-9)

Summary
::摘要

To factor quadratic expressions of the form $\begin{aligned} a x^{2} + b x + c \end{aligned}$ when $\begin{aligned} a \neq 1 \end{aligned}$ , you need two numbers whose product is ac and whose sum is b.

Then, you can separate the bx- term using those two numbers, and factor by grouping.
::然后,用这两个数字和因子分组来区分 bx-term。

Alternately, you can divide each of the numbers by a and put them as the 2nd term in a binomial. At least one will have a denominator, which becomes the leading coefficient of the binomial.
::或者,您可以将每个数字除以一个,并将它们作为二进制的第二个学期。至少一个会有一个分母,它将成为二进制的主要系数。

::当 a% 1 时, 您需要两个数字, 其产品为 ac , 其总和为 b 。然后, 您可以使用这两个数字来将 bx 条件和因子分组来分离。或者, 您可以将每个数字除以一个, 并把它们作为二进制的第二个条件。至少有一个分母, 这将成为二进制的主要系数。

Review
::回顾

Factor the quadratic expressions below , if possible. If they cannot be factored, write not factorable .
::如果可能的话, 以下面的二次表达式为因数。如果不能以二次表达式为因数, 请写不可以二次表达式为因数。

1. $\begin{aligned} 5 x^{2} + 18 x + 9 \end{aligned}$
::1. 5x2+18x+9

2. $\begin{aligned} 6 x^{2} - 21 x \end{aligned}$
::2. 6x2-21x

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

4. $\begin{aligned} 3 x^{2} + 2 x - 8 \end{aligned}$
::4. 3x2+2x-8

5. $\begin{aligned} 4 x^{2} + 8 x + 3 \end{aligned}$
::5. 4x2+8x+3

6. $\begin{aligned} 12 x^{2} - 12 x - 18 \end{aligned}$
::6. 12x2-12x-18

7. $\begin{aligned} 16 x^{2} - 6 x - 1 \end{aligned}$
::7. 16x2-6x-1

8. $\begin{aligned} 5 x^{2} - 35 x + 60 \end{aligned}$
::8. 5x2-35x+60

9. $\begin{aligned} 2 x^{2} + 7 x + 3 \end{aligned}$
::9. 2x2+7x+3

10. $\begin{aligned} 3 x^{2} + 3 x + 27 \end{aligned}$
::10. 3x2+3x+27

11. $\begin{aligned} 8 x^{2} - 14 x - 4 \end{aligned}$
::11. 8x2-14x-4

12. $\begin{aligned} 10 x^{2} + 27 x - 9 \end{aligned}$
::12. 10x2+27x-9

13. $\begin{aligned} 4 x^{2} + 12 x + 9 \end{aligned}$
::13. 4x2+12x9

14. $\begin{aligned} 15 x^{2} + 35 x \end{aligned}$
::14. 15x2+35x

15. $\begin{aligned} 6 x^{2} - 19 x + 15 \end{aligned}$
::15. 6x2-19x+15

Explore More
::探索更多

1. The area of a square is $\begin{aligned} 9 x^{2} + 24 x + 16 \end{aligned}$ . What are the dimensions of the square?
::1. 平方面积为9x2+24x+16。平方面积是多少?

by Mathispower4u explains how to determine when a projectile hits the ground.
::Mathispower4u解释如何确定弹丸何时撞击地面。

2. Using the zero p roduct property, determine how long the human cannonball noted above was in the air. Note the height of the human cannonball when he hits the ground is 0 feet.
::2. 使用零产品特性,确定上面提到的人类炮弹在空气中的长度,注意人体炮弹在撞击地面时的高度为0英尺。

3. Show why the bottoms up method actually works by factoring a out of $\begin{aligned} a x^{2} + b x + c, \end{aligned}$ and then factoring the remaining expression. One of the videos in the Resources section reviews this proof.
::3. 显示自下而上的方法为何实际上起作用,方法是乘以x2+bx+c,然后乘以剩余表达式。资源部分的视频之一审查了此证明。

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 these interactives that reinforce the concepts explored in this section:
::尝试这些强化本节所探讨概念的交互作用 :

章节大纲

常规

当领导系数不等于 1 时计数四方

Factoring Quadratic Expressions When the Leading Coefficient is Not Equal to 1 By Grouping ::组合计算,当前高系数不等于1时的二次曲线表达式乘数

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Factoring Quadratic Expressions By the Bottoms Up Method ::以自下游方法乘以二次曲线表达式

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

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

Factoring Quadratic Expressions When the Leading Coefficient is Not Equal to 1 By Grouping
::组合计算,当前高系数不等于1时的二次曲线表达式乘数

Example 1
::例1

Example 2
::例2

Example 3
::例3

Factoring Quadratic Expressions By the Bottoms Up Method
::以自下游方法乘以二次曲线表达式

Example 4
::例4

Example 5
::例5

Example 6
::例6

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