Course: 7.3 特殊四方 | The Way To Learn

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特殊四面形
You work for a business that makes metal washers. Because the sizes of the washer and the hole inside can vary, to model the area we use the expression $\begin{aligned} π R^{2} - π r^{2} \end{aligned}$ , where R is the radius of the washer and r is the radius of the hole. We can write this expression in factored form , and we will discuss how in this section.
::您为一个制造金属洗衣机的企业工作。因为洗衣机的大小和里面的洞的大小可能不同, 用来模拟我们使用的表达式 R2r2, R是洗衣机的半径, r是洞的半径。我们可以以系数形式写下这个表达式, 我们将在本节中讨论这个表达式。

Perfect Square Trinomials
::完美的三边广场

There are a couple of special quadratics that, when factored, have a pattern . One is a perfect square trinomial , which is a trinomial that, when factored, results in two of the same factor .
::有一些特殊的二次方程式,当考虑到它们时,就有一个模式。一个是完美的正方形三角形,一个是三边形,三边形,当考虑到时,产生两个相同的因素。

Perfect Square Trinomials
::完美的三边广场

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

Perfect square trinomials are not necessarily easy to recognize. First, notice that the 1st and last terms are perfect squares. Then, note that the middle term is two times the square root of the 1st and last terms.
::完美的正方形三角词不一定容易识别。首先,请注意,第一和最后一个条件是完美的正方形。然后,请注意,中期是第一和最后一个条件的平方根的2倍。

Example 1
::例1

Factor $\begin{aligned} 9 x^{2} - 12 x + 4 \end{aligned}$ .
::9x2 - 12x+4系数。

Solution: Since there is no GCF, we then check to see if this is a perfect square trinomial.
::解决方案:由于没有全球合作框架,我们然后检查一下这是否是一个完美的三重正方形。

$\begin{aligned} a^{2} & = 9 x^{2} b^{2} = 4 \\ \sqrt{a^{2}} & = \sqrt{9 x^{2}} \sqrt{b^{2}} = \sqrt{4} 2 a b = 2 \cdot 3 x \cdot 2 \\ a & = 3 x b = 2 = 12 x \end{aligned}$

::a2=9x2b2=4a2=9x2b2=42ab=2}=2=3x2a=3xb=2=12x

The middle term is two times the square root of the 1st and last terms, so this is a perfect square trinomial. Using the 2nd formula above, since the middle term is subtracted from the expression, we have
::中期是第一个和最后一个条件的平方根的2倍, 所以这是一个完美的平方三角。使用上面的第二个公式, 因为从表达式中减去中值, 我们从表达式中减去了中值,

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

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

This video by CK-12 demonstrates how to .
::这段影片由CK-12拍摄,

Example 2
::例2

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

Solution: First, check for a GCF. Each of the numbers can be divided by 4, so we can factor that out.
::解决办法:首先,检查一个全球合作框架。每个数字可以除以4,这样我们就可以将这一点考虑在内。

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

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

Now we can check to see if what remains in the " data-term="Parentheses" role="term" tabindex="0"> parentheses is a perfect square trinomial.
::现在我们可以检查一下括号里还有什么是完美的三重正方形

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

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

Using the 1st formula above, the expression factors as
::使用上面第一个公式的表达式因子

$\begin{aligned} 4 (9 x^{2} + 30 x + 25) & = 4 ((3 x)^{2} + 2 (3 x) (5) + 5^{2}) \\ = 4 (3 x + 5)^{2} . \end{aligned}$

::4(9x2+30x+25)=4(3(3x)2+2(3x)(5)+52)=4(3x+5)2。

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 $\begin{aligned} (6 x + 10)^{2} = (6 x + 10) (6 x + 10) = 2 (3 x + 5) 2 (3 x + 5) = 4 (3 x + 5)^{2} . \end{aligned}$
::如果您没有在起始点中将4点计算出来, 公式仍然有效: 等于 6x 和 b 等于 10, 所以系数为 6x+102. 如果您扩大并找到绿色气候基金, 你将会拥有 (6x+10)2= (6x+10) (6x+10)= (23x+5)2(3x+5)= 4(3x+5) 。

Example 3
::例3

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

Solution: Factor out the GCF, 2: $\begin{aligned} 2 (x^{2} - 10 x + 25) \end{aligned}$ .
::解决办法:考虑到全球合作框架,2:2(x2-10x+25)。

What remains is now a perfect square trinomial with $\begin{aligned} a = x \end{aligned}$ and $\begin{aligned} b = 5 \end{aligned}$ .
::现在的遗体是A=x和b=5的完美的三重正方形

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

::2(x2-10x+25)=2(x-5)2

This video by CK-12 demonstrates how to factor by removing common terms 1st.
::CK-12的这段影片展示了如何通过删除共同术语1来考虑因素。

Difference of Two Squares
::两平方之差

Another special quadratic form is a difference of two squares, which as the name implies is a difference of two perfect squares.
::另一个特殊的二次形形式是两个方形的区别,因为名称意味着两个完美的方形的区别。

Difference of Two Squares
::两平方之差

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

Example 4
::例4

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

Solution: In this example, there is no GCF that can be factored out. Since $\begin{aligned} x^{2} \end{aligned}$ and 81 are perfect squares and the operation between them is subtraction , this is a difference of two squares. We need to first find the values of $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ .
::解决方案 : 在此示例中, 没有可以算出全球合作框架。由于 x2 和 81 是完美的方形, 而它们之间的操作是减去的, 这是两个方形的区别。我们需要首先找到 a 和 b 的值。

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

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

Now, plugging $\begin{aligned} x \end{aligned}$ and 9 into the formula, we have $\begin{aligned} x^{2} - 81 = (x - 9) (x + 9) \end{aligned}$ .
::现在,在公式中插入 x 和 9, 我们有 x2- 81 = (x- 9) (x+ 9) (x+ 9) 。

by Mathispower4u demonstrates how to factor a difference of two squares.
::由 Mathispower4u 演示如何乘以两个方形的差值。

Example 5
::例5

Factor $\begin{aligned} 25 x^{2} - 16 y^{2} . \end{aligned}$
::25x2 -16y2系数。

Solution: Here there is no GCF. Next, we notice that $\begin{aligned} 25 x^{2} \end{aligned}$ and $\begin{aligned} 16 y^{2} \end{aligned}$ are perfect squares, and determine their square roots, $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ .
::解决方案:这里没有全球合作框架。接下来,我们注意到25x2和16y2是完美的正方形,决定了它们的正方根,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

Since they are, we can now, plug $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ into the formula: $\begin{aligned} (5 x)^{2} - 4^{2} = (5 x - 4) (5 x + 4) . \end{aligned}$
::既然它们是,我们现在可以将a和b插入公式5x)2-42=(5x-4)(5x+4)。

by CK-12 provides a difference of two squares.
::CK-12提供两个方形的差数。

Example 6
::例6

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

Solution: At 1st glance, this does not look like a difference of squares . Neither 48 nor 147 are perfect squares. But remember that our 1st step when factoring is always to factor out the GCF (if one exists). In this example, we can factor a 3 out of both, which leaves us with $\begin{aligned} 3 (16 x^{2} - 49) \end{aligned}$ . Since 16 and 49 are both perfect squares, we can now use the formula.
::解决方案 : 乍一看, 这看起来不像是方形的差别。 48 和 147 都不是完美的方形。但记住, 我们的保理第一步始终是考虑全球合作框架( 如果存在的话 ) 。在这个例子中, 我们可以将其中的3个乘以其中的3个, 剩下的是 3( 16x2-49 ) 。由于16 和 49 都属于完美的方形, 我们现在可以使用公式了。

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

::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)。

Example 7
::例7

We can use an expression to model the area of a metal washer, $\begin{aligned} π R^{2} - π r^{2} \end{aligned}$ , where R is the radius of the washer, and r is the radius of the hole. Factor this expression.
::我们可以使用一个表达式来模拟金属洗衣机的区域, 也就是 R2r2, R是洗衣机的半径, R是洞的半径。乘此表达式。

Solution: Since $\begin{aligned} π \end{aligned}$ appears in both terms, it is a GCF, and we can factor it out: $\begin{aligned} π (R^{2} - r^{2}) \end{aligned}$ . What remains inside the parentheses is a difference of two squares. To factor it completely, we have $\begin{aligned} π (R + r) (R - r) \end{aligned}$ .
::解决办法:既然以两种语言出现,它就是一个全球合作框架,我们可以将其考虑在内: (R2-r2),括号内所保留的东西是两个方形的区别。为了完全考虑,我们已经有了 (R+r) (R-r) (R-r) (R-r) 。

WARNING
::警告

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

Summary
::摘要

To recognize a perfect square trinomial, note that the 1st and last terms must be perfect squares, and the middle term must be two times the square roots of each of the perfect squares.
::要承认一个完美的平方三角, 请注意,第一个和最后一个条件必须是完美的平方, 中期必须是每个完美的平方的平方根的2倍。

To factor a perfect square trinomial, we have two forms: either $\begin{aligned} (a + b)^{2} = a^{2} + 2 a b + b^{2}, \end{aligned}$ or $\begin{aligned} (a - b)^{2} = a^{2} - 2 a b + b^{2} . \end{aligned}$
::要计算一个完美的平方三角,我们有两种形式a+b)2=a2+2ab+2b2,或(a-b)2=a2-2ab+b2。

To recognize a difference of two squares, note that the two terms need to be perfect squares, and the operation between them needs to be subtraction.
::要承认两个方形的差别,请注意这两个词必须是完美的方形,它们之间的操作需要减去。

To factor a difference of two squares, we use the form $\begin{aligned} a^{2} - b^{2} = (a + b) (a - b) \end{aligned}$ .
::乘以两个方形的差,我们使用表A2-b2=(a+b)(a-b)-b。

Review
::回顾

Factor the following quadratics, if possible:
::如果可能,下列二次方位乘以:

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

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

3. $\begin{aligned} 16 x^{2} - 24 x + 9 \end{aligned}$
::3. 16x2-24x+9

4. $\begin{aligned} - 3 x^{2} + 36 x - 108 \end{aligned}$
::4.-3x2+36x-108

5. $\begin{aligned} 144 x^{2} - 49 \end{aligned}$
::5. 144x2-49

6. $\begin{aligned} 196 x^{2} + 140 x + 25 \end{aligned}$
::6. 196x2+140x+25

7. $\begin{aligned} 100 x^{2} + 1 \end{aligned}$
::7. 100x2+1

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

9. $\begin{aligned} 225 - x^{2} \end{aligned}$
::9. 225 - x2

10. $\begin{aligned} 121 - 132 x + 36 x^{2} \end{aligned}$
::10. 121-132x+36x2

11. $\begin{aligned} 5 x^{2} + 100 x - 500 \end{aligned}$
::11. 5x2+100x-500

12. $\begin{aligned} 256 x^{2} - 676 \end{aligned}$
::12. 256x2-676

Explore More
::探索更多

1. Spencer is given the following problem: Multiply $\begin{aligned} (2 x - 5)^{2} \end{aligned}$ . Here is his work:
::1. Spencer面临以下问题:乘数(2x-5),他的工作如下:

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

:2x-5)2=(2x)2-5-2-52-52=4x2-25。

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.
::答案是 4x2 - 20x+25。 Spencer 做错了什么? 描述他的错误并纠正问题。

2. Quadratic functions of the form $\begin{aligned} f (x) = x^{2} - a^{2} \end{aligned}$ are symmetric about the y -axis—that is, if you put a number and its opposite in the function, you will get the same result. Show this is true by substituting b and - b for x in the factored form of $\begin{aligned} x^{2} - a^{2} \end{aligned}$ .
::2. F(x)=x2-a2表的二次曲线函数是y轴的对称函数,也就是说,如果在函数中加上数字和相反的数字,结果就会相同。如果以 x2-a2 的系数形式替换 x x 的 b 和 -b , 则显示这是对的。

3. If the radius of the washer is 3 cm, how big can the radius of the hole be? Set the area equal to 0, and use the zero p roduct property to determine the bounds on the radius of the hole. Does your answer make practical sense?
::3. 如果洗衣机半径为3厘米,洞的半径可以有多大?将面积设为0,并使用零产品属性确定洞半径的界限。您的答复是否具有实际意义?

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:

Section outline

General

特殊四面形

Perfect Square Trinomials ::完美的三边广场

Perfect Square Trinomials ::完美的三边广场

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Difference of Two Squares ::两平方之差

Difference of Two Squares ::两平方之差

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

WARNING ::警告

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

Perfect Square Trinomials
::完美的三边广场

Perfect Square Trinomials
::完美的三边广场

Example 1
::例1

Example 2
::例2

Example 3
::例3

Difference of Two Squares
::两平方之差

Difference of Two Squares
::两平方之差

Example 4
::例4

Example 5
::例5

Example 6
::例6

Example 7
::例7

WARNING
::警告

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