课程： 9.8 赤道表达式的量化

章节大纲

选择节常规

折叠展开
常规

全部折叠展开全部
- 选择活动新闻通告
  
  新闻通告讨论区
选择节二次曲线表达式的量化

折叠展开
二次曲线表达式的量化
Factorization of Quadratic Expressions
::二次曲线表达式的量化

Quadratic polynomials are polynomials of the $\begin{aligned} 2^{n d} \end{aligned}$ degree . The standard form of a quadratic polynomial is written as
::二次二次多面性是第二度的多面性。二次多面性的标准形式是:

$\begin{aligned} a x^{2} + b x + c \end{aligned}$

::ax2+bx+c 轴x2+bx+c

where $\begin{aligned} a, b, \end{aligned}$ and $\begin{aligned} c \end{aligned}$ stand for constant numbers. Factoring these polynomials depends on the values of these constants. In this section we’ll learn how to factor quadratic polynomials for different values of $\begin{aligned} a, b, \end{aligned}$ and $\begin{aligned} c \end{aligned}$ . (When none of the coefficients are zero, these expressions are also called quadratic trinomials , since they are polynomials with three terms .)
::a, b, c 代表不变数值。计算这些多数值取决于这些常数的值。在本节中,我们将学习如何将a, b 和 c 等值的不同值乘以二次多数值。 (当没有一个系数为零时,这些表达也被称为二次三数值,因为它们是三术语的多元数。 )

You’ve already learned how to factor quadratic polynomials where $\begin{aligned} c = 0 \end{aligned}$ . For example, for the quadratic $\begin{aligned} a x^{2} + b x \end{aligned}$ , the common factor is $\begin{aligned} x \end{aligned}$ and this expression is factored as $\begin{aligned} x (a x + b) \end{aligned}$ . Now we’ll see how to factor quadratics where $\begin{aligned} c \end{aligned}$ is nonzero.
::您已经学会了如何在 c=0 的地方将四边形多面体乘以。例如,对于四方轴2+bx, 共同系数是 x, 这个表达式被乘以 x( 轴+b ) 。现在我们可以看到如何在 c 不为零的情况下将二次形因素乘以。

Factor when a = 1, b is Positive, and c is Positive
::当 a = 1, b = 阳, c = 阳时的乘数

First, let’s consider the case where $\begin{aligned} a = 1, b \end{aligned}$ is positive and $\begin{aligned} c \end{aligned}$ is positive. The quadratic trinomials will take the form
::首先,让我们考虑a=1,b是正数,c是正数。

$\begin{aligned} x^{2} + b x + c \end{aligned}$

::x2+bx+c x2+bx+c

You know from multiplying binomials that when you multiply two factors $\begin{aligned} (x + m) (x + n) \end{aligned}$ , you get a quadratic polynomial. Let’s look at this process in more detail. First we use distribution:
::从乘以二进制中可以知道,当乘以两个因数( x+m ( x+n)) 时,你就会得到一个四边形的多面体。让我们更详细地看一下这个过程。我们首先使用分布法 :

$\begin{aligned} (x + m) (x + n) = x^{2} + n x + m x + m n \end{aligned}$

:x+m(x+n)=x2+nx+mx+mn)

Then we simplify by combining the like terms in the middle. We get:
::然后我们把类似条件合并到中间来简化。我们得到:

$\begin{aligned} (x + m) (x + n) = x^{2} + (n + m) x + m n \end{aligned}$

:x+m)(x+n)=x2+(n+m)x+mn

So to factor a quadratic, we just need to do this process in reverse.
::因此,为了将二次方位乘法,我们只需要将这一过程倒转。

$\begin{aligned} We see that x^{2} + (n + m) x + m n \\ is the same form as x^{2} + b x + c \end{aligned}$

::我们看到 x2+(n+m) x+mnis 和 x2+bx+c 一样的窗体

This means that we need to find two numbers $\begin{aligned} m \end{aligned}$ and $\begin{aligned} n \end{aligned}$ where
::这意味着我们需要找到两个数字m和n

$\begin{aligned} n + m = b and m n = c \end{aligned}$

::n+m=bandmn=c = bandmn=c = n+m=bandmn=c

The factors of $\begin{aligned} x^{2} + b x + c \end{aligned}$ are always two binomials
::x2+bxx+c的因数总是两个二进制

$\begin{aligned} (x + m) (x + n) \end{aligned}$

:x+m(x+n))

such that $\begin{aligned} n + m = b \end{aligned}$ and $\begin{aligned} m n = c \end{aligned}$ .
::n+m=b和mn=c。

Factoring
::保理

1. Factor $\begin{aligned} x^{2} + 5 x + 6 \end{aligned}$ .
::1. 系数x2+5x+6。

We are looking for an answer that is a product of two binomials in " data-term="Parentheses" role="term" tabindex="0"> parentheses :
::我们正在寻找一个答案,这是括号中两个二项概念的产物:

$\begin{aligned} (x) (x) \end{aligned}$

:x)(x)

We want two numbers $\begin{aligned} m \end{aligned}$ and $\begin{aligned} n \end{aligned}$ that multiply to 6 and add up to 5. A good strategy is to list the possible ways we can multiply two numbers to get 6 and then see which of these numbers add up to 5:
::我们要两个数字 m 和 n , 乘以 6 和 5 。一个好的策略是列出我们乘以两个数字可能的方法来达到 6 , 然后看看其中哪个数字加到 5 :

$\begin{aligned} 6 = 1 \cdot 6 & and & 1 + 6 = 7 \\ 6 = 2 \cdot 3 & and & 2 + 3 = 5 T h i s i s t h e c o r r e c t c h o i c e . \end{aligned}$

::6=16和1+6=76=23和2+3=5 这是正确的选择。

So the answer is $\begin{aligned} (x + 2) (x + 3) \end{aligned}$ .
::答案是 (x+2)(x+3) 。

We can check to see if this is correct by multiplying $\begin{aligned} (x + 2) (x + 3) \end{aligned}$ :
::我们可以通过乘( x+2)( x+3) 来检查是否正确 :

$\begin{aligned} x + 2 \\ \underline{x + 3} \\ 3 x + 6 \\ \underline{x^{2} + 2 x} \\ x^{2} + 5 x + 6 \end{aligned}$

::x+2x+3_ 3x+6x2+2xx2+5x+6

The answer checks out.
::答案检查出来。

2. Factor $\begin{aligned} x^{2} + 7 x + 12 \end{aligned}$ .
::2. 系数x2+7x+12。

We are looking for an answer that is a product of two binomials in parentheses: $\begin{aligned} (x) (x) \end{aligned}$
::我们正在寻找一个答案,这是括号中两个二项概念的产物x)(x)

The number 12 can be written as the product of the following numbers:
::12号可以写成以下数字的产物:

$\begin{aligned} 12 = 1 \cdot 12 & and & 1 + 12 = 13 \\ 12 = 2 \cdot 6 & and & 2 + 6 = 8 \\ 12 = 3 \cdot 4 & and & 3 + 4 = 7 T h i s i s t h e c o r r e c t c h o i c e . \end{aligned}$

::12=112和1+12=1312=26和2+6=812=34和3+4=7 这是正确的选择。

The answer is $\begin{aligned} (x + 3) (x + 4) \end{aligned}$ .
::答案是 (x+3)(x+4) 。

3. Factor $\begin{aligned} x^{2} + 8 x + 12 \end{aligned}$ .
::3. 系数x2+8x+12。

We are looking for an answer that is a product of two binomials in parentheses: $\begin{aligned} (x) (x) \end{aligned}$
::我们正在寻找一个答案,这是括号中两个二项概念的产物x)(x)

The number 12 can be written as the product of the following numbers:
::12号可以写成以下数字的产物:

$\begin{aligned} 12 = 1 \cdot 12 & and & 1 + 12 = 13 \\ 12 = 2 \cdot 6 & and & 2 + 6 = 8 T h i s i s t h e c o r r e c t c h o i c e . \\ 12 = 3 \cdot 4 & and & 3 + 4 = 7 \end{aligned}$

::12=112和1+12=1312=26和2+6=8 这是正确的选择 12=34和3+4=7

The answer is $\begin{aligned} (x + 2) (x + 6) \end{aligned}$ .
::答案是 (x+2)(x+6) 。

Example
::示例示例示例示例

Example 1
::例1

Factor $\begin{aligned} x^{2} + 12 x + 36 \end{aligned}$ .
::系数 x2+12x+36。

We are looking for an answer that is a product of two binomials in parentheses: $\begin{aligned} (x) (x) \end{aligned}$
::我们正在寻找一个答案,这是括号中两个二项概念的产物x)(x)

The number 36 can be written as the product of the following numbers:
::36号可以写成以下数字的产物:

$\begin{aligned} 36 = 1 \cdot 36 & and & 1 + 36 = 37 \\ 36 = 2 \cdot 18 & and & 2 + 18 = 20 \\ 36 = 3 \cdot 12 & and & 3 + 12 = 15 \\ 36 = 4 \cdot 9 & and & 4 + 9 = 13 \\ 36 = 6 \cdot 6 & and & 6 + 6 = 12 T h i s i s t h e c o r r e c t c h o i c e . \end{aligned}$

::36=1*36和1+36=3736=2*18和2+18=2036=3*12和3+12=1536=4*9和4+9=1336=6*6和6+6=12 这是正确的选择。

The answer is $\begin{aligned} (x + 6) (x + 6) \end{aligned}$ .
::答案是 (x+6)(x+6) 。

Review
::回顾

Factor the following quadratic polynomials.
::乘以以下四边形多面体。

$\begin{aligned} x^{2} + 10 x + 9 \end{aligned}$
::x2+10x+9

$\begin{aligned} x^{2} + 15 x + 50 \end{aligned}$
::x2+15x+50

$\begin{aligned} x^{2} + 10 x + 21 \end{aligned}$
::x2+10x+21

$\begin{aligned} x^{2} + 16 x + 48 \end{aligned}$
::x2+16x+48

$\begin{aligned} x^{2} + 14 x + 45 \end{aligned}$
::x2+14x+45

$\begin{aligned} x^{2} + 15 x + 50 \end{aligned}$
::x2+15x+50

$\begin{aligned} x^{2} + 22 x + 40 \end{aligned}$
::x2+22x+40

$\begin{aligned} x^{2} + 15 x + 56 \end{aligned}$
::x2+15x+56

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

$\begin{aligned} x^{2} + 10 x + 24 \end{aligned}$
::x2+10x+24

$\begin{aligned} x^{2} + 17 x + 72 \end{aligned}$
::x2+17x+72

$\begin{aligned} x^{2} + 25 x + 150 \end{aligned}$
::x2+25x+150

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

二次曲线表达式的量化

Factorization of Quadratic Expressions ::二次曲线表达式的量化

Factoring ::保理

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Factorization of Quadratic Expressions
::二次曲线表达式的量化

Factoring
::保理

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)