课程： 5.3 特殊四方 | The Way To Learn

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选择节特殊四面形

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特殊四面形
The total time, in hours, it takes a rower to paddle upstream, turn around and come back to her starting point is $\begin{aligned} 18 x^{2} = 32 \end{aligned}$ . How long does it take her to make the round trip?
::整个时间,用小时,需要一位划船员在上游划船,转过身回到她的起点是18x2=32。她往返旅行需要多长时间?

Factoring Special Quadratics
::特殊四面形

There are a couple of special quadratics that, when factored, have a pattern .
::有几种特殊的象方体, 当计算系数时, 具有一种模式。

Multiplying the Square of a Binomial
::倍增二进制广场

Step 1: Rewrite $\begin{aligned} (a + b)^{2} \end{aligned}$ as the product of two factors. Expand $\begin{aligned} (a + b)^{2} \end{aligned}$ . $\begin{aligned} (a + b)^{2} = (a + b) (a + b) \end{aligned}$
::第1步:重写(a+b)2,作为两个因素的产物。扩展(a+b)2.(a+b)2=(a+b)(a+b)+b

Step 2: FOIL your answer from Step 1. This is a perfect square trinomial . $\begin{aligned} a^{2} + 2 a b + b^{2} \end{aligned}$
::步骤2:FIL 你从步骤1回答的答案。这是一个完美的平方三角。 a2+2ab+b2

Step 3: $\begin{aligned} (a - b)^{2} \end{aligned}$ also produces a perfect square trinomial. $\begin{aligned} (a - b)^{2} = a^{2} - 2 a b + b^{2} \end{aligned}$
::步骤3a-b)2还产生一个完全平方的三角。 (a-b)2=a2-2ab+b2

Step 4: Apply the formula above to factoring $\begin{aligned} 9 x^{2} - 12 x + 4 \end{aligned}$ . First, find $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ .
::第4步:将上面的公式应用到乘数 9x2 - 12x+4。首先,找到a和b。

$\begin{aligned} a^{2} & = 9 x^{2}, b^{2} = 4 \\ a & = 3 x, b = 2 \end{aligned}$

::a2=9x2, b2=4a=3x, b=2

Step 5: Now, plug $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ into the appropriate formula.
::第5步:现在,在适当的公式中插入a和b。

$\begin{aligned} (3 x - 2)^{2} & = (3 x)^{2} - 2 (3 x) (2) + 2^{2} \\ = 9 x^{2} - 12 x + 4 \end{aligned}$

:3x-2)2=(3x)2-2-2-2(3x)(2)+22=9x2-12x+4

Multiplying (a + b)(a - b)
::乘数 (a+ b) a-b)

Step 1: FOIL $\begin{aligned} (a - b) (a + b) \end{aligned}$ .
::步骤1:FOIL(a-b)(a+b)。

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

:a)-b(a+b)=a2+ab-ab-2=a2+ab-ab-2=a2-b2=a2-b2

Step 2: This is a difference of squares . The difference of squares will always factor to be $\begin{aligned} (a + b) (a - b) \end{aligned}$ .
::第2步:这是方形的差别,方形的差别总是以(a+b)(a-b)为因数。

Step 3: Apply the formula above to factoring $\begin{aligned} 25 x^{2} - 16 \end{aligned}$ . First, find $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ .
::第3步:将以上公式适用于乘数25x2-16。首先,找到a和b。

$\begin{aligned} a^{2} & = 25 x^{2}, b^{2} = 16 \\ a & = 5 x, b = 4 \end{aligned}$

::a2=25x2, b2=16a=5x, b=4

Step 4: Now, plug $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ into the appropriate formula. $\begin{aligned} (5 x - 4) (5 x + 4) = (5 x)^{2} - 4^{2} \end{aligned}$
::第4步:现在,将a和b插入适当的公式中。 (5x-4)(5x+4)=(5x)2-42

$\begin{aligned} ^{* *} \end{aligned}$ It is important to note that if you forget these formulas or do not want to use them, you can still factor all of these quadratics the same way you have already learned .
::必须指出,如果你忘记这些公式或不想使用这些公式,你仍然可以按照你已经学过的方法,将所有这些梯度都计算在内。

Now, let's factor the following quadratics.
::现在,让我们考虑以下的二次方位。

Factor $\begin{aligned} x^{2} - 81 \end{aligned}$ .
::系数x2-81。

Using the formula from the steps above, we need to first find the values of $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ .
::使用以上步骤的公式, 我们需要首先找到 a 和 b 的值。

$\begin{aligned} x^{2} - 81 = a^{2} - b^{2} \\ a^{2} = x^{2}, b^{2} = 81 \\ a = x, b = 9 \end{aligned}$

::x2-81=a2-b2a2=x2,b2=81 a=x,b=9

Now, plugging $\begin{aligned} x \end{aligned}$ and 9 into the formula, we have $\begin{aligned} x^{2} - 81 = (x - 9) (x + 9) \end{aligned}$ . To solve for $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ , we found the square root of each number. Recall that the square root is a number that, when multiplied by itself, produces another number. This other number is called a perfect square .
::现在,在公式中插入 x 和 9, 我们有 x2- 81 = (x- 9) (x+ 9) 。要解决 a 和 b , 我们找到了每个数字的平方根。提醒注意, 平方根是一个数字, 当自己乘以后, 产生另一个数字。另一个数字被称为完美的平方。

Alternate Method
::替代方法

Rewrite $\begin{aligned} x^{2} - 81 \end{aligned}$ so that the middle term is present. $\begin{aligned} x^{2} + 0 x - 81 \end{aligned}$
::重写 x2 - 81 , 以便中期内出现。 x2+0x- 81

Using a previously learned method, what are the two factors of -81 that add up to 0? 9 and -9
::使用以前学到的方法,81年加起来等于0的两种因素是什么? 9和 -9

Therefore , the factors are $\begin{aligned} (x - 9) (x + 9) \end{aligned}$ .
::因此,这些因素是(x-9)(x+9)。

Factor $\begin{aligned} 36 x^{2} + 120 x + 100 \end{aligned}$ .
::36x2+120x+100系数。

First, check for a GCF.
::首先,检查一个绿色气候基金。

$\begin{aligned} 4 (9 x^{2} + 30 x + 25) \end{aligned}$

::4(9x2+30x+25)

Now, double-check that the quadratic equation above fits into the perfect square trinomial formula.
::现在,再确认一下上面的二次方程符合完美的方形三角公式

$\begin{aligned} a^{2} & = 9 x^{2} b^{2} = 25 \\ \sqrt{a^{2}} & = \sqrt{9 x^{2}} \sqrt{b^{2}} = \sqrt{25} 2 a b = 30 x \\ a & = 3 x b = 5 2 (3 x) (5) = 30 x \end{aligned}$

::a2=9x2b2=25a2=9x2b2=252ab=30xx=3xb=52(3x)(5)=30x

Using $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ above, the equation factors to be $\begin{aligned} 4 (3 x + 5)^{2} \end{aligned}$ . If you did not factor out the 4 in the beginning, the formula will still work. $\begin{aligned} a \end{aligned}$ would equal $\begin{aligned} 6 x \end{aligned}$ and $\begin{aligned} b \end{aligned}$ would equal 10, so the factors would be $\begin{aligned} (6 x + 10)^{2} \end{aligned}$ . If you expand and find the GCF, you would have $\begin{aligned} (6 x + 10)^{2} = (6 x + 10) (6 x + 10) = 2 (3 x + 5) 2 (3 x + 5) = 4 (3 x + 5)^{2} \end{aligned}$ .
::使用 a 和 b 以上, 方程系数为 4( 3x+5) 2 。如果您在起始点未将 4( 3x+5) 计算在内, 公式仍然有效。等于 6x 和 b 等于 10, 所以系数为 6x+10 2 。如果您扩大并找到全球合作框架, 您将会拥有 6x+102 = ( 6x+10) (6x+10) = 2( 3x+5) 2( 3x+5) = 4( 3x+5) 。

Alternate Method
::替代方法

First, find the GCF. $\begin{aligned} 4 (9 x^{2} + 30 x + 25) \end{aligned}$
::首先,见全球合作框架4 (9x2+30+25)

Then, find $\begin{aligned} a c \end{aligned}$ and expand $\begin{aligned} b \end{aligned}$ accordingly. $\begin{aligned} 9 \cdot 25 = 225 \end{aligned}$ , the factors of 225 that add up to 30 are 15 and 15.
::然后找到 ac 并相应扩大 b. 925=225, 总计30的225因数是15和15。

$\begin{aligned} 4 (9 x^{2} + 30 x + 25) \\ 4 (9 x^{2} + 15 x + 15 x + 25) \\ 4 [(9 x^{2} + 15 x) + (15 x + 25)] \\ 4 [3 x (3 x + 5) + 5 (3 x + 5)] \\ 4 (3 x + 5) (3 x + 5) o r 4 (3 x^{2} + 5) \end{aligned}$

::4(9x2+30x+25)4(9x2+15x+15x+25)4[(9x2+15x+15x+25)+(15x+25)]4[3x(3x+5)+5(3x+5)]4(3x+5)(3x+5)(3x+5)或4(3x2+5)

Again, notice that if you do not use the formula discovered in this concept, you can still factor and get the correct answer.
::再次请注意,如果您不使用这个概念中发现的公式,您仍然可以计数并获得正确的答案。

Factor $\begin{aligned} 48 x^{2} - 147 \end{aligned}$ .
::48x2-147因数。

At first glance, this does not look like a difference of squares. 48 nor 147 are square numbers. But, if we take a 3 out of both, we have $\begin{aligned} 3 (16 x^{2} - 49) \end{aligned}$ . 16 and 49 are both square numbers, so now we can use the formula.
::乍一看,这看起来不像是平方之差。 48和147是平方之差。但是,如果我们同时选3个,我们就有3个(3,16x2-49 ) 。 16和49是平方之差,因此我们现在可以使用公式。

$\begin{aligned} 16 x^{2} & = a^{2} 49 = b^{2} \\ 4 x & = a 7 = b \end{aligned}$

::16x2=a249=b24x=a7=b

The factors are $\begin{aligned} 3 (4 x - 7) (4 x + 7) \end{aligned}$ .
::这些因素是3(3、4x-7)(4x+7)。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the time it takes to make the round trip.
::早些时候,你被要求寻找时间进行往返旅行。

$\begin{aligned} 18 x^{2} = 32 \end{aligned}$ can be rewritten as $\begin{aligned} 18 x^{2} - 32 = 0 \end{aligned}$ , so factor $\begin{aligned} 18 x^{2} - 32 \end{aligned}$ .
::18x2=32 可重写为 18x2-32=0, 所以系数 18x2- 32。

First, we must take greatest common factor of 2 out of both. We then have $\begin{aligned} 2 (9 x^{2} - 16) \end{aligned}$ . 9 and 16 are both square numbers, so now we can use the formula.
::首先,我们必须同时使用两个最大的共同系数。然后我们拥有2(9x2-16),9和16都是平方数字,所以我们现在可以使用这个公式。

$\begin{aligned} 9 x^{2} & = a^{2} 16 = b^{2} \\ 3 x & = a 4 = b \end{aligned}$

::9x2=a216=b23x=a4=b

The factors are $\begin{aligned} 2 (3 x - 4) (3 x + 4) \end{aligned}$ .
::这些因素是2(2、3x-4)(3x+4)。

Finally, to find the time, set these factors equal to zero and solve $\begin{aligned} 2 (3 x - 4) (3 x + 4) = 0 \end{aligned}$ .
::最后,为了找到时间,将这些因素设定为零,并解决2(2、3x-4)(3x+4)=0。

Because x represents the time, it must be positive. Only $\begin{aligned} (3 x - 4) = 0 \end{aligned}$ results in a positive value of x .
::由于 x 代表时间, 它必须是正数。只有( 3x- 4) = 0 得出正数 x 。

$\begin{aligned} x = \frac{4}{3} = 1.3333 \end{aligned}$ Therefore the round trip takes 1.3333 hours.
::因此,往返旅行需要1.33333小时。

Example 2
::例2

Factor the following quadratic equation.
::乘以以下的二次方程。

$\begin{aligned} x^{2} - 4 \end{aligned}$
::x2 - 4

$\begin{aligned} a = x \end{aligned}$ and $\begin{aligned} b = 2 \end{aligned}$ . Therefore, $\begin{aligned} x^{2} - 4 = (x - 2) (x + 2) \end{aligned}$ .
::a=x和b=2. 因此, x2-4=(x-2)(x+2)。

Example 3
::例3

Factor the following quadratic equation.
::乘以以下的二次方程。

$\begin{aligned} 2 x^{2} - 20 x + 50 \end{aligned}$
::2x2-20x+50

Factor out the GCF, 2. $\begin{aligned} 2 (x^{2} - 10 x + 25) \end{aligned}$ . This is now a perfect square trinomial with $\begin{aligned} a = x \end{aligned}$ and $\begin{aligned} b = 5 \end{aligned}$ .
::将绿色气候基金计算为2.2(x2-10x+25),现在这是一个完全平方的三角,A=x和b=5。

$\begin{aligned} 2 (x^{2} - 10 x + 25) = 2 (x - 5)^{2} . \end{aligned}$

::2(x2-10x+25)=2(x-5)2。

Example 4
::例4

Factor the following quadratic equation.
::乘以以下的二次方程。

$\begin{aligned} 81 x^{2} + 144 + 64 \end{aligned}$
::81x2+144+64

This is a perfect square trinomial and no common factors. Solve for $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ .
::这是一个完全平方的三角,没有共同的因素。

$\begin{aligned} 81 x^{2} & = a^{2} 64 = b^{2} \\ 9 x & = a 8 = b \end{aligned}$

::81x2=a264=b29x=a8=b

The factors are $\begin{aligned} (9 x + 8)^{2} \end{aligned}$ .
::这些因素是(9x+8)2。

Review
::回顾

List the perfect squares that are less than 200.
::列出低于200的完美方形。

Why do you think there is no sum of squares formula?
::为什么你认为没有平方公式的总和?

Factor the following quadratics, if possible.
::如果可能,则以下列二次方位因子为因数。

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

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

$\begin{aligned} 16 x^{2} - 24 x + 9 \end{aligned}$
::16x2 - 24x+9

$\begin{aligned} - 3 x^{2} + 36 x - 108 \end{aligned}$
::- 3x2+36x-108

$\begin{aligned} 144 x^{2} - 49 \end{aligned}$
::144x2-49

$\begin{aligned} 196 x^{2} + 140 x + 25 \end{aligned}$
::196x2+140x+25

$\begin{aligned} 100 x^{2} + 1 \end{aligned}$
::100x2+1

$\begin{aligned} 162 x^{2} + 72 x + 8 \end{aligned}$
::162x2+72x+8

$\begin{aligned} 225 - x^{2} \end{aligned}$
::225 - x2

$\begin{aligned} 121 - 132 x + 36 x^{2} \end{aligned}$
::121-132x+36x2

$\begin{aligned} 5 x^{2} + 100 x - 500 \end{aligned}$
::5x2+100x-500

$\begin{aligned} 256 x^{2} - 676 \end{aligned}$
::256x2-676

Error Analysis Spencer is given the following problem: Multiply $\begin{aligned} (2 x - 5)^{2} \end{aligned}$ . Here is his work:
::错误分析 Spencer 给出的问题如下: 乘数 (2x-5) 2。以下是他的工作 :

$\begin{aligned} (2 x - 5)^{2} = (2 x)^{2} - 5^{2} = 4 x^{2} - 25 \end{aligned}$

:2x-5)2=(2x)2-2-52-52=4x2-2-25

His teacher tells him the answer is $\begin{aligned} 4 x^{2} - 20 x + 25 \end{aligned}$ . What did Spencer do wrong? Describe his error and correct the problem.
::他的老师告诉他答案是4x2-20x+25 Spencer做错了什么?描述他的错误并纠正问题。

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。

章节大纲

常规

特殊四面形

Factoring Special Quadratics ::特殊四面形

Multiplying the Square of a Binomial ::倍增二进制广场

Multiplying (a + b)(a - b) ::乘数 (a+ b) a-b)

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)

Factoring Special Quadratics
::特殊四面形

Multiplying the Square of a Binomial
::倍增二进制广场

Multiplying (a + b)(a - b)
::乘数 (a+ b) a-b)

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)