9.9 具有负系数的赤道表达物的量化

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• 选择节负系数对具有负系数的赤道表达式进行分化

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  负系数对具有负系数的赤道表达式进行分化

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  Factorization of Quadratic Expressions with Negative Coefficients
  ::负系数对具有负系数的赤道表达式进行分化

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  Factor a quadratic trinomial when a = 1, b is negative and c is positive
  ::a = 1,b为负,c为正的二次三角数乘数

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  Now let’s see how this method works if the middle coefficient is negative.
  ::现在让我们看看如果中位系数为负数,

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  Factor $\begin{array}{r}{x}^2-6x+8\end{array}$ .
  ::系数 x2 - 6x+8

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  We are looking for an answer that is a product of two binomials in " data-term="Parentheses" role="term" tabindex="0"> parentheses : $\begin{array}{r}(x)(x)\end{array}$
  ::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)

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  When negative coefficients are involved, we have to remember that negative factors may be involved also. The number 8 can be written as the product of the following numbers:
  ::当涉及负系数时,我们必须记住也可能涉及消极因素。

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  $$\begin{array}{r}8=1\cdot 8\text{and}1+8=9\end{array}$$

  
  ::8=18和1+8=9

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  but also
  ::并同时

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  $$\begin{array}{r}8=(-1)\cdot (-8)\text{and}-1+(-8)=-9\end{array}$$

  
  ::8=(-1)__(-8)和-1+(-8)_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  and
  ::和

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  $$\begin{array}{r}8=2\cdot 4\text{and}2+4=6\end{array}$$

  
  ::8=24和2+4=6

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  but also
  ::并同时

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  $$\begin{array}{r}8=(-2)\cdot (-4)\text{and}-2+(-4)=-6.\end{array}$$

  
  ::8=(-2)-(-4)和-2+(-4)-6。

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  The last option is the correct choice. The answer is $\begin{array}{r}(x-2)(x-4)\end{array}$ . We can check to see if this is correct by multiplying $\begin{array}{r}(x-2)(x-4)\end{array}$ :
  ::最后一个选项是正确选择。 答案是 (x-2)(x- 4) , 我们可以检查乘( x-2)(x- 4) 是否正确 :

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  $$\begin{array}{rl}& x-2\\ & \underset{\_}{x-4}\\ & -4x+8\\ & \underset{\_}{x^2-2x}\\ & x^2-6x+8\end{array}$$

  
  ::x-2x-4_ - 4x+8x2 - 2x_x2 - 6x+8

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  The answer checks out.
  ::答案检查出来。

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  Factoring
  ::保理

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  1. Factor $\begin{array}{r}{x}^2-17x+16\end{array}$ .
  ::1. 系数x2-17x+16。

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  We are looking for an answer that is a product of two binomials in parentheses: $\begin{array}{r}(x)(x)\end{array}$
  ::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)

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  The number 16 can be written as the product of the following numbers:
  ::数字16可以写成以下数字的产物:

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  $$\begin{array}{rlrlrl}& 16=1\cdot 16& & \text{and}& & 1+16=17\\ & 16=(-1)\cdot (-16)& & \text{and}& & -1+(-16)=-17(Correctchoice)\\ & 16=2\cdot 8& & \text{and}& & 2+8=10\\ & 16=(-2)\cdot (-8)& & \text{and}& & -2+(-8)=-10\\ & 16=4\cdot 4& & \text{and}& & 4+4=8\\ & 16=(-4)\cdot (-4)& & \text{and}& & -4+(-4)=-8\end{array}$$

  
  ::16=116和1+1+16=1716=(-1)_(-16)和-1+(-16)_17(更正选择)16=28和2+8=1016=(-2)_(-8)和-2+(-8)_(1016)=44和4+4=816=(-4)_(-4)_(-4)和-4+(-4)_8

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  The answer is $\begin{array}{r}(x-1)(x-16)\end{array}$ .
  ::答案是(x-1)(x-16)。

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  In general, whenever $\begin{array}{r}b\end{array}$ is negative and $\begin{array}{r}a\end{array}$ and $\begin{array}{r}c\end{array}$ are positive, the two binomial factors will have minus signs instead of plus signs.
  ::一般而言,当b为负数,a和c为正数时,两个二元因素将具有减号而不是加号。

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  Factor when a = 1 and c is Negative
  ::=1和c为负时的因数

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  Now let’s see how this method works if the constant term is negative.
  ::现在让我们来看看如果常数为负值时该方法如何运作。

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  2. Factor $\begin{array}{r}{x}^2+2x-15\end{array}$ .
  ::2. 系数x2+2x-15。

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  We are looking for an answer that is a product of two binomials in parentheses: $\begin{array}{r}(x)(x)\end{array}$
  ::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)

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  Once again, we must take the negative sign into account. The number -15 can be written as the product of the following numbers:
  ::再次,我们必须考虑到负号。数字 -15可以写成为下列数字的产物:

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  $$\begin{array}{rlrlrl}& -15=-1\cdot 15& & \text{and}& & -1+15=14\\ & -15=1\cdot (-15)& & \text{and}& & 1+(-15)=-14\\ & -15=-3\cdot 5& & \text{and}& & -3+5=2(Correctchoice)\\ & -15=3\cdot (-5)& & \text{and}& & 3+(-5)=-2\end{array}$$

  
  ::-15115和-1+1+15=14-15=1(-15)和1+(-15)14-151535和-3+5=2(更正选择)-15=3(-5)和3+(-5)2

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  The answer is $\begin{array}{r}(x-3)(x+5)\end{array}$ .
  ::答案是(x-3)(x+5)。

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  We can check to see if this is correct by multiplying:
  ::我们可以通过乘法检查是否正确:

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  $$\begin{array}{rl}& x-3\\ & \underset{\_}{x+5}\\ & 5x-15\\ & \underset{\_}{x^2-3x}\\ & x^2+2x-15\end{array}$$

  
  ::- 3x+5_5x-15x2-3xxx2+2x-15

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  The answer checks out.
  ::答案检查出来。

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  3. Factor $\begin{array}{r}{x}^2-10x-24\end{array}$ .
  ::3. 系数x2-10x-24。

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  We are looking for an answer that is a product of two binomials in parentheses: $\begin{array}{r}(x)(x)\end{array}$
  ::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)

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  The number -24 can be written as the product of the following numbers:
  ::数字 - 24 可以写成为下列数字的产物:

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  $$\begin{array}{rlrlrl}& -24=-1\cdot 24& & \text{and}& & -1+24=23\\ & -24=1\cdot (-24)& & \text{and}& & 1+(-24)=-23\\ & -24=-2\cdot 12& & \text{and}& & -2+12=10\\ & -24=2\cdot (-12)& & \text{and}& & 2+(-12)=-10(Correctchoice)\\ & -24=-3\cdot 8& & \text{and}& & -3+8=5\\ & -24=3\cdot (-8)& & \text{and}& & 3+(-8)=-5\\ & -24=-4\cdot 6& & \text{and}& & -4+6=2\\ & -24=4\cdot (-6)& & \text{and}& & 4+(-6)=-2\end{array}$$

  
  ::- 24124和-1+24=23-24=1(24)和1+(24) +(24) 23-2212和-2-2+12=10-24=2(12)和2+(12)和2+(12) 10(Correct choose) -24__38和3+8=5-24=3__(8)和3+3+(8) 524 _46和4+6=2-24=4}(-6)和4+(-6)

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  The answer is $\begin{array}{r}(x-12)(x+2)\end{array}$ .
  ::答案是 (x-12)(x+2) 。

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  Factor when a = - 1
  ::a = - 1 时的因数

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  When $\begin{array}{r}a=-1\end{array}$ , the best strategy is to factor the common factor of -1 from all the terms in the quadratic polynomial and then apply the methods you learned so far in this section
  ::a1 1 时,最好的战略是从四边多元度中的所有术语中乘以 -1 的共同系数,然后运用本节中迄今学到的方法。

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  4. Factor $\begin{array}{r}-x^2+x+6\end{array}$ .
  ::4. 系数-x2+x+6。

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  First factor the common factor of -1 from each term in the trinomial. Factoring -1 just changes the signs of each term in the expression :
  ::第一个系数是三进制中每个术语的 - 1 的共同系数。乘数 - 1 只是改变表达式中每个术语的符号 :

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  $$\begin{array}{r}-x^2+x+6=-(x^2-x-6)\end{array}$$

  
  ::-x2+x+6(x2-x-6)

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  We’re looking for a product of two binomials in parentheses: $\begin{array}{r}-(x)(x)\end{array}$
  ::我们正在寻找括号中两个二进制的产物x)(x)

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  Now our job is to factor $\begin{array}{r}{x}^2-x-6\end{array}$ .
  ::现在我们的工作是乘以乘数 x2 - x - 6。

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  The number -6 can be written as the product of the following numbers:
  ::数字 - 6 可以写成以下数字的产物:

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  $$\begin{array}{rlrlrl}& -6=-1\cdot 6& & \text{and}& & -1+6=5\\ & -6=1\cdot (-6)& & \text{and}& & 1+(-6)=-5\\ & -6=-2\cdot 3& & \text{and}& & -2+3=1\\ & -6=2\cdot (-3)& & \text{and}& & 2+(-3)=-1(Correctchoice)\end{array}$$

  
  ::-616和-1+6+6=5-6=1(-6)和1+(-6)5-623和-2+3=1-6=2(-3)和2+2+(-3)+1(更正选择)

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  The answer is $\begin{array}{r}-(x-3)(x+2)\end{array}$ .
  ::答案是-(x-3)(x+2)。

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  Example
  ::示例示例示例示例

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  Example 1
  ::例1

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  Factor $\begin{array}{r}{x}^2+34x-35\end{array}$ .
  ::系数 x2+34x-35。

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  We are looking for an answer that is a product of two binomials in parentheses: $\begin{array}{r}(x)(x)\end{array}$
  ::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)

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  The number -35 can be written as the product of the following numbers:
  ::编号 - 35 可以是以下数字的产物:

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  $$\begin{array}{rlrlrl}& -35=-1\cdot 35& & \text{and}& & -1+35=34(Correctchoice)\\ & -35=1\cdot (-35)& & \text{and}& & 1+(-35)=-34\\ & -35=-5\cdot 7& & \text{and}& & -5+7=2\\ & -35=5\cdot (-7)& & \text{and}& & 5+(-7)=-2\end{array}$$

  
  ::-=35=5=5=5=5=5(-7)和5+(-7)+5+(-7)=2

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  The answer is $\begin{array}{r}(x-1)(x+35)\end{array}$ .
  ::答案是(x-1)(x+35)。

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  Review 
  ::回顾

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

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  1. $\begin{array}{r}{x}^2-11x+24\end{array}$
     ::x2 - 11x+24
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  2. $\begin{array}{r}{x}^2-13x+42\end{array}$
     ::x2 - 13x+42
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  3. $\begin{array}{r}{x}^2-14x+33\end{array}$
     ::x2 - 14x+33
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  4. $\begin{array}{r}{x}^2-9x+20\end{array}$
     ::x2-9x+20 x2 - 9x+20
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  5. $\begin{array}{r}{x}^2+5x-14\end{array}$
     ::x2+5x-14
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  6. $\begin{array}{r}{x}^2+6x-27\end{array}$
     ::x2+6x- 27 x2+6x- 27
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  7. $\begin{array}{r}{x}^2+7x-78\end{array}$
     ::x2+7x-78
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  8. $\begin{array}{r}{x}^2+4x-32\end{array}$
     ::x2+4x-32
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  9. $\begin{array}{r}{x}^2-12x-45\end{array}$
     ::x2 - 12x- 45
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  10. $\begin{array}{r}{x}^2-5x-50\end{array}$
      ::x2 - 5x- 50
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  11. $\begin{array}{r}{x}^2-3x-40\end{array}$
      ::x2 - 3x- 40
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  12. $\begin{array}{r}{x}^2-x-56\end{array}$
      ::x2 - x - 56 级
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  13. $\begin{array}{r}-x^2-2x-1\end{array}$
      ::-x2-2-2x-1
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  14. $\begin{array}{r}-x^2-5x+24\end{array}$
      ::−x2-5x+24
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  15. $\begin{array}{r}-x^2+18x-72\end{array}$
      ::-x2+18x-72
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  16. $\begin{array}{r}-x^2+25x-150\end{array}$
      ::-x2+25x-150
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  17. $\begin{array}{r}{x}^2+21x+108\end{array}$
      ::x2+21x+108
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  18. $\begin{array}{r}-x^2+11x-30\end{array}$
      ::-x2+11x-30
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  19. $\begin{array}{r}{x}^2+12x-64\end{array}$
      ::x2+12x-64
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  20. $\begin{array}{r}{x}^2-17x-60\end{array}$
      ::x2 - 17x- 60
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  21. $\begin{array}{r}{x}^2+5x-36\end{array}$
      ::x2+5x-36 x2+5x-36

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  Review (Answers)
  ::回顾(答复)

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  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 。