Course: 9.5 多面体的特殊产品

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多元体特殊产品
Special Products of Polynomials
::多元体特殊产品

We saw that when we multiply two binomials we need to make sure to multiply each term in the first binomial with each term in the second binomial. Let’s look at another example.
::我们看到,当我们乘以两个二进制时,我们需要确保第一个二进制中每个词的乘数与第二个二进制中每个词的乘数。让我们看看另一个例子。

Multiply two linear binomials (binomials whose degree is 1):
::乘以两个线性二进制(二进制,其等级为 1) :

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

:2x+3)(x+4)

When we multiply, we obtain a quadratic polynomial (one with degree 2) with four terms :
::当我们乘以乘以时,我们获得一个四边多面体(一级加2级),有四个条件:

$\begin{aligned} 2 x^{2} + 8 x + 3 x + 12 \end{aligned}$

::2x2+8x+3x+12

The middle terms are like terms and we can combine them. We simplify and get $\begin{aligned} 2 x^{2} + 11 x + 12 \end{aligned}$ . This is a quadratic, or second-degree, trinomial (polynomial with three terms).
::中间术语相似于条件, 我们可以合并它们。我们简化并获得 2x2+11x+12。这是四边形, 或者二度, 三边形( 圆形, 共三个条件 ) 。

You can see that every time we multiply two linear binomials with one variable , we will obtain a quadratic polynomial. In this section we’ll talk about some special products of binomials.
::你可以看到,每次我们用一个变量乘以两个线性二进制时,我们都会获得一个四进制多元分子。在本节中,我们将讨论一些二进制的特殊产品。

Find the Square of a Binomial
::寻找二进制广场

One special binomial product is the square of a binomial . Consider the product $\begin{aligned} (x + 4) (x + 4) \end{aligned}$ .
::一种特殊的二进制产品是二进制的正方形。考虑一下该产品( x+4)( x+4)。

Since we are multiplying the same expression by itself, that means we are squaring the expression. $\begin{aligned} (x + 4) (x + 4) \end{aligned}$ is the same as $\begin{aligned} (x + 4)^{2} \end{aligned}$ .
::由于我们本身是乘以同一个表达式,这意味着我们是在对表达式进行对比。 (x+4)(x+4) 与(x+4) 2 相同。

When we multiply it out, we get $\begin{aligned} x^{2} + 4 x + 4 x + 16 \end{aligned}$ , which simplifies to $\begin{aligned} x^{2} + 8 x + 16 \end{aligned}$ .
::当我们将其乘出时,我们会得到 x2+4x+4x+16, 它会简化为 x2+8x+16 。

Notice that the two middle terms—the ones we added together to get $\begin{aligned} 8 x \end{aligned}$ —were the same. Is this a coincidence? In order to find that out, let’s square a general linear binomial.
::请注意,两个中间条件 — — 我们为了获得8x值而加在一起的那些条件 — — 是一样的。这是巧合吗? 为了找出答案,让我们来做个普通的线性二元论。

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

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

Sure enough, the middle terms are the same. How about if the expression we square is a difference instead of a sum?
::当然,中间条件是一样的。如果我们平方的表达方式是差异而不是总和呢?

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

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

It looks like the middle two terms are the same in general whenever we square a binomial. The general pattern is: to square a binomial, take the square of the first term, add or subtract twice the product of the terms, and add the square of the second term. You should remember these formulas:
::看起来中两个条件在我们平方一个二进制时是相同的。一般模式是:平方一个二进制,取第一个条件的平方,加或减两个条件的产物,加或减两个条件的产物,再加第二个条件的平方。您应该记住这些公式:

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

:a+b)2=a2+2ab+2ab+2b2and(a-b)2=a2-2ab+2b2

Remember! Raising a polynomial to a power means that we multiply the polynomial by itself however many times the exponent indicates. For instance, $\begin{aligned} (a + b)^{2} = (a + b) (a + b) \end{aligned}$ . Don’t make the common mistake of thinking that $\begin{aligned} (a + b)^{2} = a^{2} + b^{2} \end{aligned}$ ! To see why that’s not true, try substituting numbers for $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ into the equation (for example, $\begin{aligned} a = 4 \end{aligned}$ and $\begin{aligned} b = 3 \end{aligned}$ ), and you will see that it is not a true statement. The middle term, $\begin{aligned} 2 a b \end{aligned}$ , is needed to make the equation work.
::记住! 将一个多数值加到一个能量中意味着我们将多数值本身乘以多数值本身, 不管指数显示多少倍。例如, (a+b) 2=(a+b) (a+b) +b。不要做一个常见错误, 即认为 (a+b) 2=a2+b2! 要了解为什么这不是真的, 请尝试在方程中将数字和b值( 例如, a=4 和 b=3) 替换为数字, 你会看到它不是一个真实的语句。要使方程工作, 需要一个中值, 2ab = 2ab 。

We can apply the formulas for squaring binomials to any number of problems.
::我们可以对任何几个问题采用平衡二元论的公式。

Squaring Binomials
::平分分义

Square each binomial and simplify.
::每个平方平方平方平方平方平方平方平坦。

Let’s use the square of a binomial formula to multiply each expression.
::让我们使用二进制公式的平方来乘以每个表达式。

a) $\begin{aligned} (x + 10)^{2} \end{aligned}$
::a) (x+10)2

$\begin{aligned} (x + 10)^{2} \end{aligned}$
:x+10)2

If we let $\begin{aligned} a = x \end{aligned}$ and $\begin{aligned} b = 10 \end{aligned}$ , then our formula $\begin{aligned} (a + b)^{2} = a^{2} + 2 a b + b^{2} \end{aligned}$ becomes $\begin{aligned} (x + 10)^{2} = x^{2} + 2 (x) (10) + 10^{2} \end{aligned}$ , which simplifies to $\begin{aligned} x^{2} + 20 x + 100 \end{aligned}$ .
::如果我们让 a=x 和 b= 10, 那么我们的公式 (a+b) 2= a2+2ab+2b2 变成 (x+10)2=x2+2(x)(10)+102, 简化为 x2+20x+100 。

b) $\begin{aligned} (2 x - 3)^{2} \end{aligned}$
::b) (2x-3)2

$\begin{aligned} (2 x - 3)^{2} \end{aligned}$
:2x-3)2

If we let $\begin{aligned} a = 2 x \end{aligned}$ and $\begin{aligned} b = 3 \end{aligned}$ , then our formula $\begin{aligned} (a - b)^{2} = a^{2} - 2 a b + b^{2} \end{aligned}$ becomes $\begin{aligned} (2 x - 3)^{2} = (2 x^{2}) - 2 (2 x) (3) + (3)^{2} \end{aligned}$ , which simplifies to $\begin{aligned} 4 x^{2} - 12 x + 9 \end{aligned}$ .
::如果我们允许 a=2x和b=3,那么我们的公式(a-b)2=a2-2ab+b2变成(2x-3)2=(2x2)-2x(3)+(3)2,该公式简化为4x2-12x+9。

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

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

If we let $\begin{aligned} a = x^{2} \end{aligned}$ and $\begin{aligned} b = 4 \end{aligned}$ , then
::如果我们让 a=x2 和 b=4, 那么

$\begin{aligned} (x^{2} + 4)^{2} & = (x^{2})^{2} + 2 (x^{2}) (4) + (4)^{2} \\ = x^{4} + 8 x^{2} + 16 \end{aligned}$

:x2+4) 2=(x2) 2+2(x2) (4)+(4)2=x4+8x2+16)

Find the Product of Binomials Using Sum and Difference Patterns
::使用总和和差异模式查找二进制产品

Another special binomial product is the product of a sum and a difference of terms. For example, let’s multiply the following binomials.
::另一个特殊的二进制产品是总和和条件差异的产物。比如,让我们乘以以下二进制。

$\begin{aligned} (x + 4) (x - 4) & = x^{2} - 4 x + 4 x - 16 \\ = x^{2} - 16 \end{aligned}$

:x+4)(x-4)=x2-4x+4x-4x-16=x2-16)

Notice that the middle terms are opposites of each other, so they cancel out when we collect like terms. This is not a coincidence. This always happens when we multiply a sum and difference of the same terms. In general,
::注意中间条件是相互对立的, 所以当我们收集类似条件时, 它们就会取消。这不是巧合。这总是当我们乘以相同条件的总和和和差额时发生。一般来说,

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

::a+b(a-b)=a2-ab+ab-2=a2-bb2=a2-b2=a2-b2=a2-b2

When multiplying a sum and difference of the same two terms, the middle terms cancel out. We get the square of the first term minus the square of the second term. You should remember this formula.
::当乘以相同两个任期的总和和差额时, 中间条件取消。我们得到第一个任期的平方减去第二个任期的平方。您应该记住这个公式。

Sum and Difference Formula : $\begin{aligned} (a + b) (a - b) = a^{2} - b^{2} \end{aligned}$
::公式(a+b)(a-b)=a2-b2

Let’s apply this formula to a few examples.
::让我们把这个公式应用到几个例子中。

Multiplying Binomials
::乘数分义

Multiply the following binomials and simplify.
::乘以以下二进制和简化。

a) $\begin{aligned} (x + 3) (x - 3) \end{aligned}$
::a) (x+3)(x-3)

Let $\begin{aligned} a = x \end{aligned}$ and $\begin{aligned} b = 3 \end{aligned}$ , then:
::a=x和b=3,然后:

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

:a+b)(a-b)=a2-b2(x+3(x-3)=x2-3=x2-3=x2-32=x2-9)

b) $\begin{aligned} (5 x + 9) (5 x - 9) \end{aligned}$
:b) (5x+9) (5x-9)

Let $\begin{aligned} a = 5 x \end{aligned}$ and $\begin{aligned} b = 9 \end{aligned}$ , then:
::Let a=5x和b=9,然后:

$\begin{aligned} (a + b) (a - b) & = a^{2} - b^{2} \\ (5 x + 9) (5 x - 9) & = (5 x)^{2} - 9^{2} \\ = 25 x^{2} - 81 \end{aligned}$

:a+b)(a-b)=a2-b2(5x+9)(5x-9)=(5x-9)=(5x)2-92=25x2-81)

c) $\begin{aligned} (2 x^{3} + 7) (2 x^{3} - 7) \end{aligned}$
:c) (2x3+7)(2x3-7)

Let $\begin{aligned} a = 2 x^{3} \end{aligned}$ and $\begin{aligned} b = 7 \end{aligned}$ , then:
::a=2x3和b=7,然后:

$\begin{aligned} (2 x^{3} + 7) (2 x^{3} - 7) & = (2 x^{3})^{2} - (7)^{2} \\ = 4 x^{6} - 49 \end{aligned}$

:2x3+7)(2x3-7)=(2x3)2-(7)2=4x6-49

Solve Real-World Problems Using Special Products of Polynomials
::利用多元性特殊产品解决现实世界问题

Now let’s see how special products of polynomials apply to geometry problems and to mental arithmetic.
::现在让我们来看看多种族的特殊产品如何适用于几何学问题和精神算术。

Real-World Application: Finding Area
::真实世界应用程序:寻找地区

Find the area of the following square:
::查找以下方块的区域:

The length of each side is $\begin{aligned} (a + b) \end{aligned}$ , so the area is $\begin{aligned} (a + b) (a + b) \end{aligned}$ .
::每一边的长度是(a+b),因此区域是(a+b)(a+b)。

Notice that this gives a visual explanation of the square of a binomial. The blue square has area $\begin{aligned} a^{2} \end{aligned}$ , the red square has area $\begin{aligned} b^{2} \end{aligned}$ , and each rectangle has area $\begin{aligned} a b \end{aligned}$ , so added all together, the area $\begin{aligned} (a + b) (a + b) \end{aligned}$ is equal to $\begin{aligned} a^{2} + 2 a b + b^{2} \end{aligned}$ .
::注意此选项可以直观地解释二进制方形的正方形。蓝色方形的面积为a2, 红色方形的面积为b2, 每个矩形的面积为ab, 所以加在一起, 区域( a+b) (a+b) 等于a2+2ab+b2。

The next example shows how you can use the special products to do fast mental calculations.
::下一个示例显示了您如何使用特殊产品快速进行心理计算。

Using the Difference of Squares and the Binomial Square Formulas
::使用广场和二元广场公式的差别

Use the difference of squares and the binomial square formulas to find the products of the following numbers without using a calculator.
::使用方形和二进制方形的差值,在不使用计算器的情况下找到下列数字的产品。

The key to these mental “tricks” is to rewrite each number as a sum or difference of numbers you know how to square easily.
::这些心理“tricks”的关键是将每个数字重写为数字的总和或差额,你知道如何轻易地平方。

a) $\begin{aligned} 43 \times 57 \end{aligned}$
:a) 43×57

Rewrite 43 as $\begin{aligned} (50 - 7) \end{aligned}$ and 57 as $\begin{aligned} (50 + 7) \end{aligned}$ .
::将43改为(50-7),57改为(50+7)。

Then $\begin{aligned} 43 \times 57 = (50 - 7) (50 + 7) = (50)^{2} - (7)^{2} = 2500 - 49 = 2451 \end{aligned}$
::然后43×57=(50-7)(50+7)=(50)2-(7)2=2500-49=2451

b) $\begin{aligned} 45^{2} \end{aligned}$
:b) 452

$\begin{aligned} 45^{2} = (40 + 5)^{2} = (40)^{2} + 2 (40) (5) + (5)^{2} = 1600 + 400 + 25 = 2025 \end{aligned}$

c) $\begin{aligned} 481 \times 319 \end{aligned}$
:c) 481x319

Rewrite 481 as $\begin{aligned} (400 + 81) \end{aligned}$ and 319 as $\begin{aligned} (400 - 81) \end{aligned}$ .
::将481改为(400+81),319改为(400-81)。

Then $\begin{aligned} 481 \times 319 = (400 + 81) (400 - 81) = (400)^{2} - (81)^{2} \end{aligned}$
::然后481×319=(400+81)(400-81)=(400-81)2-(81)2

$\begin{aligned} (400)^{2} \end{aligned}$ is easy - it equals 160000.
:4000) 简单 - 等于160000。

$\begin{aligned} (81)^{2} \end{aligned}$ is not easy to do mentally, so let’s rewrite 81 as $\begin{aligned} 80 + 1 \end{aligned}$ .
:812) 精神上不容易做,所以让我们把81改成80+1。

$\begin{aligned} (81)^{2} = (80 + 1)^{2} = (80)^{2} + 2 (80) (1) + (1)^{2} = 6400 + 160 + 1 = 6561 \end{aligned}$

Then $\begin{aligned} 481 \times 319 = (400)^{2} - (81)^{2} = 160000 - 6561 = 153439 \end{aligned}$
::然后481x319=(40002)-(812)=160000-6561=153439

Examples
::实例

Example 1
::例1

Square the binomial and simplify: $\begin{aligned} (5 x - 2 y)^{2} \end{aligned}$ .
::二进制平方和简化( 5x- 2- 2y) 2 。

$\begin{aligned} (5 x - 2 y)^{2} \end{aligned}$
:5x-2y)2

If we let $\begin{aligned} a = 5 x \end{aligned}$ and $\begin{aligned} b = 2 y \end{aligned}$ , then
::如果我们让 a=5x 和 b=2y, 那么

$\begin{aligned} (5 x - 2 y)^{2} & = (5 x)^{2} - 2 (5 x) (2 y) + (2 y)^{2} \\ = 25 x^{2} - 20 x y + 4 y^{2} \end{aligned}$

:5x-2y)2=(5x)2-2-2-2(5x)(2y)+(2y)2=(2y)2=25x2-20xy+4y2)

Example 2
::例2

Multiply $\begin{aligned} (4 x + 5 y) (4 x - 5 y) \end{aligned}$ and simplify.
::乘数(4x+5y)(4x-5y)和简化。

Let $\begin{aligned} a = 4 x \end{aligned}$ and $\begin{aligned} b = 5 y \end{aligned}$ , then:
::a=4x和b=5y,然后:

$\begin{aligned} (4 x + 5 y) (4 x - 5 y) & = (4 x)^{2} - (5 y)^{2} \\ = 16 x^{2} - 25 y^{2} \end{aligned}$

:4x+5y (4x-5y) = (4x) 2 - (5y) 2= 16x2 - 25y2)

Example 3
::例3

Use the difference of squares and the binomial square formulas to find the product of $\begin{aligned} 112 \times 88 \end{aligned}$ without using a calculator.
::使用方形和二进制方形的差数,在不使用计算器的情况下找到112×88的产物。

The key to these mental “tricks” is to rewrite each number as a sum or difference of numbers you know how to square easily.
::这些心理“tricks”的关键是将每个数字重写为数字的总和或差额,你知道如何轻易地平方。

Rewrite 112 as $\begin{aligned} (100 + 12) \end{aligned}$ and 88 as $\begin{aligned} (100 - 12) \end{aligned}$ .
::将112改写为(100+12),88改写为(100-12)。

Then
::然后

$\begin{aligned} 112 \times 88 & = (100 + 12) (100 - 12) \\ = (100)^{2} - (12)^{2} \\ = 10000 - 144 \\ = 9856 \end{aligned}$

Review
::回顾

Use the special product rule for squaring binomials to multiply these expressions.
::使用特殊产品规则来对比二进制来乘以这些表达式。

$\begin{aligned} (x + 9)^{2} \end{aligned}$
:x+9)2

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

$\begin{aligned} (5 x - y)^{2} \end{aligned}$
:5x-y)2

$\begin{aligned} (2 x^{3} - 3)^{2} \end{aligned}$
:2x3-3)2

$\begin{aligned} (4 x^{2} + y^{2})^{2} \end{aligned}$
:4x2+y2)2

$\begin{aligned} (8 x - 3)^{2} \end{aligned}$
:8x-3)2

$\begin{aligned} (2 x + 5) (5 + 2 x) \end{aligned}$
:2x+5)(5+2x)

$\begin{aligned} (x y - y)^{2} \end{aligned}$
:xy-y)2

Use the special product of a sum and difference to multiply these expressions.
::使用总和和差数的特殊产品乘以这些表达式。

$\begin{aligned} (2 x - 1) (2 x + 1) \end{aligned}$
:2x-1)(2x+1)

$\begin{aligned} (x - 12) (x + 12) \end{aligned}$
:x-12)(x+12)

$\begin{aligned} (5 a - 2 b) (5 a + 2 b) \end{aligned}$
:5a-2b(5a+2b))

$\begin{aligned} (a b - 1) (a b + 1) \end{aligned}$
:ab-1(ab+1))

$\begin{aligned} (z^{2} + y) (z^{2} - y) \end{aligned}$
:z2+y)(z2-y)

$\begin{aligned} (2 q^{3} + r^{2}) (2 q^{3} - r^{2}) \end{aligned}$
:2q3+r2)(2q3-r2)

$\begin{aligned} (7 s - t) (t + 7 s) \end{aligned}$
:7-t(t+7s))

$\begin{aligned} (x^{2} y + x y^{2}) (x^{2} y - x y^{2}) \end{aligned}$
:x2y+xy2)(x2y-xy2)

Find the area of the lower right square in the following figure.
::在下图中查找右下方方的区域。

Multiply the following numbers using special products.
::使用特殊产品乘以下列数字。

$\begin{aligned} 45 \times 55 \end{aligned}$

$\begin{aligned} 56^{2} \end{aligned}$

$\begin{aligned} 1002 \times 998 \end{aligned}$

$\begin{aligned} 36 \times 44 \end{aligned}$

$\begin{aligned} 10.5 \times 9.5 \end{aligned}$

$\begin{aligned} 10.2 \times 9.8 \end{aligned}$

$\begin{aligned} - 95 \times - 105 \end{aligned}$

$\begin{aligned} 2 \times - 2 \end{aligned}$

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。

9.5 多面体的特殊产品

Section outline

General

多元体特殊产品

Special Products of Polynomials
::多元体特殊产品

Find the Square of a Binomial
::寻找二进制广场

Squaring Binomials
::平分分义

Find the Product of Binomials Using Sum and Difference Patterns
::使用总和和差异模式查找二进制产品

Multiplying Binomials
::乘数分义

Solve Real-World Problems Using Special Products of Polynomials
::利用多元性特殊产品解决现实世界问题

Real-World Application: Finding Area
::真实世界应用程序:寻找地区

Using the Difference of Squares and the Binomial Square Formulas
::使用广场和二元广场公式的差别

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)

Section outline

General

多元体特殊产品

Special Products of Polynomials ::多元体特殊产品

Find the Square of a Binomial ::寻找二进制广场

Squaring Binomials ::平分分义

Find the Product of Binomials Using Sum and Difference Patterns ::使用总和和差异模式查找二进制产品

Multiplying Binomials ::乘数分义

Solve Real-World Problems Using Special Products of Polynomials ::利用多元性特殊产品解决现实世界问题

Real-World Application: Finding Area ::真实世界应用程序:寻找地区

Using the Difference of Squares and the Binomial Square Formulas ::使用广场和二元广场公式的差别

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Special Products of Polynomials
::多元体特殊产品

Find the Square of a Binomial
::寻找二进制广场

Squaring Binomials
::平分分义

Find the Product of Binomials Using Sum and Difference Patterns
::使用总和和差异模式查找二进制产品

Multiplying Binomials
::乘数分义

Solve Real-World Problems Using Special Products of Polynomials
::利用多元性特殊产品解决现实世界问题

Real-World Application: Finding Area
::真实世界应用程序:寻找地区

Using the Difference of Squares and the Binomial Square Formulas
::使用广场和二元广场公式的差别

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)