8.12 找到复合的多元职能零星

章节大纲

• 选择节常规

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• 选择节查找复数函数的复数零

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  查找复数函数的复数零

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  When we look at the night sky, stars are not always where they appear. Because the light from a star travels such a far distance , it is bent by the distribution of objects in between. This process is called "gravitational lensing," and was a prediction of Albert Einstein's General Theory of Relativity ^1,2 . 
  ::当我们观察夜空时,恒星并不总出现在它们出现的地方。 因为恒星的光线走得很远, 它被天体在天体之间的分布弯曲。 这个过程被称为“引力透镜”, 并且预测了阿尔伯特·爱因斯坦的“相对论通论”, 2。

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  Say you are looking at a particular star, but there is an object in the way. The light travels around the object, and instead of seeing the star at a particular point, you see two images of the star at two different points. To find the locations of lensed images, scientists use polynomial functions, like $\begin{array}{r}y=x^5+3x^3-4x\end{array}$ . The solutions of these polynomial functions tell us the locations of lensed images. 
  ::说您正在看着一个特定的恒星, 但路上有一个物体。 光线环绕着天体运行, 而不是在一个特定点看到恒星, 而是在两个不同点看到两个恒星的图像。 要找到有镜头的图像的位置, 科学家们会使用多面函数, 比如 y=x5+3x3- 4x。 这些多面函数的解决方案告诉我们有镜头的图像的位置 。

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  We continue to use the techniques from the previous section and discuss the nature of the solutions of polynomial equations below. 
  ::我们继续使用上一节的技术,并讨论下文多面方程式解决办法的性质。

  

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  Factoring a Polynomial 
  ::聚合系数乘数

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     Factorization
  ::定级化

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  A polynomial can be factored over as a product of linear and quadratic factors—that is, factors of the form  $\begin{array}{r}ax+b\end{array}$  or  $\begin{array}{r}a{x}^2+bx+c,\end{array}$  where  a,b, and  c  are real numbers and  $\begin{array}{r}a\ne 0.\end{array}$
  ::多元系数可以作为线性和二次系数的产物来计算,即以ax+b或ax2+bx+c为形式的系数,其中a、b和c为实际数字和a+%0。

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  However, a polynomial can be factored over the complex numbers as a product of linear factors . 
  ::然而,在复杂数字上可以将多数值作为线性因素的产物加以计算。

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  Let's see an example.
  ::让我们举个例子。

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  Example 1
  ::例1

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  Find the zeros or roots of  $\begin{array}{r}f(x)=4x^4+35x^2-9\end{array}$ .
  ::查找 f( x) =4x4+35x2- 9 的零或根。

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  Solution:   $\begin{array}{r}f(x)=4x^4+35x^2-9\end{array}$ is factorable since it is quadratic in form. $\begin{array}{r}ac=\text{-}36\end{array}$ .
  ::解析度: f(x) = 4x4+35x2- 9是可以考虑的,因为它是四方形。 ac=-36。

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  $$\begin{array}{r}4x^4+35x^2-9\\ 4x^4+36x^2-x^2-9\\ 4x^2(x^2+9)-1(x^2+9)\\ (x^2+9)(4x^2-1)\end{array}$$

  
  ::4x4+35x2-94x4+36x2-x2-94x2(x2+9)-1(x2+9)-(x2+9)-(x2+9)-(x2+9)-(4x2--1)

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  In the real numbers,  $\begin{array}{r}{x}^2+9\end{array}$  cannot be factored further.  $\begin{array}{r}4x^2-1\end{array}$  can be factored further over the real numbers. 
  ::按实际数字计算, x2+9 无法进一步计算。 4x2-1 可比实际数字进一步计算。

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  $$\begin{array}{rl}& 4x^2-1=0\\ & (2x+1)(2x-1)=0\\ & 2x+1=02x-1=0\\ & x=\pm \frac{1}{2}\end{array}$$

  
  ::4x2 - 1=0( 2x+1)( 2x- 1)=02x+1=0 2x-1=0x1=0x1=0x=0x1=0x=0x**12

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  So, there are two real zeros. In factored form ,  $\begin{array}{r}f(x)=4x^4+35x^2-9=(x^2+9)(2x+1)(2x-1)\end{array}$ . These are linear and quadratic factors. If we want to consider complex zeros, we can set  $\begin{array}{r}{x}^2+9\end{array}$ equal to 0.
  ::因此,有两个实际零。 以系数形式, f( x) = 4x4+35x2- 9= (x2+9) (2x+1)( 2x- 1) 。 这些是线性和二次系数。 如果我们想考虑复杂的零, 我们可以将x2+9 设为 0 。

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  $$\begin{array}{rl}& x^2+9=0\\ & x^2=\text{-}9\\ & x=\pm 3i\end{array}$$

  
  ::x2+9=0x2=-9x3i

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  In the complex numbers, $\begin{array}{r}f(x)=(x-3i)(x+3i)(2x+1)(2x-1)\end{array}$ . These are all linear factors.
  ::在复合数字f(x)=(x-3i)(x+3i)(2x+1)(2x-1)(2x-1)。)中,这些都是线性系数。

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  by Mathispower4u demonstrates how to find the zeros of a degree three polynomial function with the help of a graph of the function .  
  ::by Mathispower4u 演示如何在函数的图形帮助下找到三度多元函数的零。

   

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  Notice that however we factor this, we will get roots that are real numbers, which are also complex numbers; or complex numbers; or a combination of real and complex numbers. We can formalize this idea. 
  ::请注意,无论我们如何考虑这一点,我们都会找到真实数字的根,而实际数字也是复杂的数字;或复杂的数字;或真实数字和复杂数字的组合。 我们可以正式确定这个想法。

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  Fundamental Theorem of Algebra
  ::代数基本理论

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  Fundamental Theorem of Algebra
  ::代数基本理论

  Every non- constant polynomial with complex coefficients has a root in the complex number system. 

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  Example 2
  ::例2

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  Find the zeros of  $\begin{array}{r}f(x)=3x^3+6x^2+6x+12\end{array}$ . 
  ::查找 f(x)=3x3+6x2+6x+12的零。

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  Solution: When we set  $\begin{array}{r}f(x)=0\end{array}$ , we can factor out the GCF of 3.
  ::解决方案:当我们设定f(x)=0时,我们可以将全球合作框架为3。

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  $$\begin{array}{rl}f(x)=0& =3x^3+6x^2+6x+12\\ & =3(x^3+2x^2+2x+4)\\ & =x^3+2x^2+2x+4\end{array}$$

  
  ::f(x)=0=3x3+6x2+6x2+6x6x+12=3(x3+2x2+2x4)=x3+2x2+2x+4)=x3+2x2+2x4

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  Notice there are no sign changes in  $\begin{array}{r}f(x)\end{array}$ , so there are no positive real roots according to Descartes's Rule of Signs. We consider only the negative numbers from the list of possible roots.
  ::在 f( x) 中没有标记变化, 因此根据笛卡尔的标志规则, 没有正根。 我们只考虑可能根数列表中的负数 。

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  $$\begin{array}{r}\frac{p}{q}=\frac{\pm 1,\pm 2\pm 4}{\pm 1}=\pm 1,\pm 2,\pm 4\\ \\ f(\text{-}1)=(\text{-}1{)}^3+2(\text{-}1{)}^2+2(\text{-}1)+4=3\\ f(\text{-}2)=(\text{-}2{)}^3+2(\text{-}2{)}^2+2(\text{-}2)+4=0\end{array}$$

  
  ::pq1, 241, 2, 4f(-1) =(-1) 3+2(-1) 2(-1) +2(-1) +4= 3f(-2) = (-2) 3+2(-2) +2(-2)+4=0

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  This means that $\begin{array}{r}x+2\end{array}$  is a factor. In factored form,  $\begin{array}{r}f(x)=3(x+2)(x^2+2)\end{array}$ . If we want to find the complex zeros, we set $\begin{array}{r}{x}^2+2=0\end{array}$ .  
  ::这意味着 x+2 是一个系数。 在系数表格式中, f( x) =3( x+2)( x2+2) +2。 如果我们想要找到复数零, 我们设置 x2+2=0 。

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  $$\begin{array}{rl}& x^2+2=0\\ & x^2=\text{-}2\\ & x=\pm \sqrt{-2}=\pm i\sqrt{2}\end{array}$$

  
  ::x2+2=0x2=2xx2=2x%22i2

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  Note the number of roots in the complex numbers is the same as the degree of the polynomial .
  ::请注意,复杂数字中的根数与多元数值的程度相同。

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  by mrgibsonrhs provides a brief description of the Fundamental Theorem of Algebra and one example of an application. 
  ::由 mrgibsonrhs 提供对代数基本理论的简单描述,并举一个应用实例。

   

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  Example 3
  ::例3

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  To find the the locations of lensed images, scientists use polynomial functions, like $\begin{array}{r}y=x^5+3x^3-4x\end{array}$ . The solutions of these polynomial functions tell us the locations of lensed images ^1,2 .
  ::为了找到隐形图像的位置,科学家使用y=x5+3x3-4x等多元函数。这些多边函数的解决方案告诉我们隐形图像1、2的位置。

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  Solution:  We set  y  equal to 0 and solve.
  ::解答:我们设定y等于 0,然后解答。

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  $$\begin{array}{rl}y=0& =x^5+3x^3-4x\\ & =x(x^4+3x^2-4)\\ & =x(x^2-1)(x^2+4)\\ & =x(x+1)(x-1)(x^2+4)\\ \\ x& =0x+1=0x-1=0x^2+4=0\\ x& =0x=\text{-}1x=1x=\pm 2i\\ \\ y& =x(x+1)(x-1)(x-2i)(x+2i)\end{array}$$

   
  ::y==x5+3x3=3x3-4x=xxxxx4+3x2-4)=xx(x2--1)(x2+4)=x(x+1)(x2+1)(x-1)(x2+4)xx=0x1(x2+4)xx=0+1=0x1=0x1=0x2+4=0x=0x=0x=0x=0x=1xxxxxxxxxxx1x=1x*2iy=xx(x+1)(x-1)(x-1)(x-2i)(x-2)(x2i)(x+2i)

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  Complex Conjugate Theorem
  ::复杂共解理论

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     Complex Conjugate Theorem
  ::复杂共解理论

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  Complex zeros or roots of a polynomial function with real coefficients occur in conjugate pairs. That is, if  $\begin{array}{r}a+bi\end{array}$  is a zero , so is  $\begin{array}{r}a-bi\end{array}$  its complex conjugate.
  ::具有实际系数的多元函数的复杂零或根部在对等中发生。 也就是说,如果+Bi为零,那么A-Bi的复杂共产值也是零。

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  Example 4
  ::例4

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  Find the function that has the solution 3, -2, and $\begin{array}{r}4+i\end{array}$ .
  ::查找具有解决方案3、2-2和4+i的函数。

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  Solution: Notice that one of the given solutions i s a complex number. Thus,   $\begin{array}{r}4-i\end{array}$ is also a solution. Write  each solution as part of a factor, and multiply them all together.
  ::解决方案: 提醒注意给定的解决方案之一是一个复杂的数字。 因此, 4- i 也是个解决方案 。 将每个解决方案写成一个要素的一部分, 并一起乘以它们 。

  

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  Any multiple of this function would also have these roots. For example, $\begin{array}{r}2x^4-18x^3+38x^2+62x-204\end{array}$ would have these roots as well.
  ::此函数的任何多个函数也会有这些根。 例如, 2x4-18x3+38x2+62x-204也会有这些根。

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  by kevinmon68 explains the Complex Conjugate Pairs Theorem .  
  ::Kevinmon68解释复杂共产物对等理论。

   

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  Example 5
  ::例5

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  A function with  rational coefficients that has the following roots:  $\begin{array}{r}4,\sqrt{2},\text{and}1-i\end{array}$ .
  ::具有合理系数的函数,其根源如下:4,2,and1-i。

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  Solution:  Recall that if the coefficients are rational, the roots with radicals should be in conjugate pairs to eliminate the terms with radicals. Since rational coefficients are real numbers, the complex roots come in conjugate pairs as well. Therefore , all the roots are   $\begin{array}{r}4,\sqrt{2},\text{-}\sqrt{2},1+i,1-i\end{array}$ . Multiply the factors with all 5 roots together.
  ::解答:忆及如果系数是理性的,那么激进分子的根源应该是对立的,以消除与激进分子的术语。 由于理性系数是真实的数字,复杂的根源也是对立的。 因此,所有根源都是4,2,2,2,1+i,1-i。 将所有5个根源的因素加起来。

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  $$\begin{array}{r}f(x)=(x-4)(x-\sqrt{2})(x+\sqrt{2})(x-(1+i))(x-(1-i))\\ (x-4)(x^2-2)(x^2-2x+2)\\ (x^3-4x^2-2x+8)(x^2-2x+2)\\ x^5-6x^4+8x^3-4x^2-20x+16\end{array}$$

  
  :x) = (x) = (x) = (x-4) (x) = (x) = (x) = (x-4)(x) = (x) = (x) = (x) = (x) = (x) = (x) = (x-4)(x) = (x) = (x) = (x) = (x) = (x) = (x) = (x-4)(x) = (x) = (x-2) (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) = (x) (x-2) (x) (x) (x) (x) (x) (x) (x) = (x) (x) (x) = (x) (x) = (x) (x) (x- 4 (x) (x) (x) (x) (x) (x-2 = (x) (x) (x) (x) (x) (x) (x) (x) (x) =xxx-2-2 + (x)xxx)xxx =xxxxxxxxxxxxxxxxxxxxxx=xxxxxxxxxxxxxxxxxxxxx)xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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  Feature: Polynomial Art
  ::特色:多面艺术

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  by Deirdre Mundy
  ::由Deirdre Mundy 编辑

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  Can a polynomial be beautiful? A computer scientist from Rutgers University uses to create art. Dr. Bahman Kalantari has developed a computer program that visualizes solutions to complex polynomials. He calls his work "polynomiography," and he's shown his paintings and movies at museums and galleries in the United States and Europe.
  ::拉特格斯大学的计算机科学家用来创造艺术。巴赫曼·卡兰塔里博士开发了一个计算机程序,可以想象复杂的多面体的解决方案。他称自己的作品为“阴道学 ” , 他在美国和欧洲的博物馆和美术馆展示了自己的绘画和电影。

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  Solving Equations , Creating Patterns
  ::解决等号、创建模式

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  Both math and art use patterns to help people understand the world. Dr. Kalantari realized that patterns and images could help him see how computers solved complex polynomials. Computers use a guess-and-check method to find the roots to complex polynomials. Some programs are better at solving polynomials than others. Dr. Kalantari wrote a program that creates a picture based on a computer's attempts to solve polynomial equations. The user can choose a method for the computer to use, as well as a color scheme. After the computer has completed the image, the artist is free to modify the colors and highlights of the result to create a more impressive piece of art.
  ::数学和艺术使用模式帮助人们了解世界。 Kalantari博士意识到模式和图像可以帮助他了解计算机如何解决复杂的多面体问题。 计算机使用猜测和检查方法来找到复杂的多面体的根源。 有些程序比其他程序更擅长解决多面体问题。 Kalantari博士写了一个程序, 根据计算机试图解决多面体方程式来制作图片。 用户可以选择计算机使用的方法, 以及一个颜色方案。 在计算机完成图像后, 艺术家可以自由地修改结果的颜色和亮点, 从而创造出更令人印象深刻的艺术品 。

  

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  Dr. Kalantari uses polynomiography to teach high-school and college students about polynomials. He hopes to show others how beautiful mathematics can be through his art. He also uses polynomiography in his own research. The art form enables him to see new ways of approaching complex polynomials.
  
  ::卡兰塔里博士利用多民族学向高中和大学学生教授多民族学,他想向其他人展示他的艺术如何能带来美丽的数学。他也在自己的研究中使用多民族学。艺术形式使他能够看到新的方法来接近复杂的多民族学。

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  by Rutgers University shows Dr. Bahman Kalantari describing a piece of software he has developed through his research into polynomial root-finding. With this software, users can explore the world of polynomials and their natural and artistic applications. 
  ::Rutgers大学(Rutgers University)向Bahman Kalantari博士展示了他通过研究多分子根调查开发的软件。 使用这个软件,用户可以探索多民族世界及其自然和艺术应用。

   

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  Summary
  ::摘要

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  • We can factor polynomials over the complex numbers as linear factors.
    ::我们可以将复杂数字的多元系数作为线性系数。
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  • The Fundamental Theorem of Algebra guarantees a complex root.
    ::代数的基本理论保证了一个复杂的根。
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  • If the coefficients of a polynomial are real numbers and there is a complex root, then its complex conjugate is also a root.  
    ::如果多元数值的系数是真实数字,而且有复杂的根,那么其复杂的共产物也是根。

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  Review
  ::回顾

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  Find all zeros of  the following functions, using any method:
  ::使用任何方法查找下列函数的所有零:

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  1.  $\begin{array}{r}f(x)=x^4+x^3-12x^2-10x+20\end{array}$
  ::1. f(x)=x4+x3-12x2-10x+20

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  2.  $\begin{array}{r}f(x)=4x^3-20x^2-3x+15\end{array}$
  ::2. f(x)=4x3 - 20x2 - 3x+15

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  3.  $\begin{array}{r}f(x)=2x^4-7x^2-30\end{array}$
  ::3. f(x)=2x4-7x2-30

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  4.  $\begin{array}{r}f(x)=x^3+5x^2+12x+18\end{array}$
  ::4. f(x)=x3+5x2+12x18

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  5.  $\begin{array}{r}f(x)=4x^4+4x^3-22x^2-8x+40\end{array}$
  ::5. f(x)=4x4+4x3-22x2-8x+40

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  6.  $\begin{array}{r}f(x)=3x^4+4x^2-15\end{array}$
  ::6. f(x)=3x4+4x2-15

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  7.  $\begin{array}{r}f(x)=2x^3-6x^2+9x-27\end{array}$
  ::7. f(x)=2x3-6x2+9x-27

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  8.  $\begin{array}{r}f(x)=6x^4-7x^3-280x^2-419x+280\end{array}$
  ::8. f(x)=6x4-7x3-280x2-419x+280

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  9.  $\begin{array}{r}f(x)=9x^4+6x^3-28x^2+2x+11\end{array}$
  ::9. f(x)=9x4+6x3-28x2+2x+11

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  10.  $\begin{array}{r}f(x)=2x^5-19x^4+30x^3+97x^2-20x+150\end{array}$
  ::10. f(x)=2x5-194+30x3+97x2-20x+150

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  Write a function with real coefficients that has the following roots:
  ::以实际系数写出函数,其根源如下:

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  11.  $\begin{array}{r}4,i\end{array}$
  ::11. 4,i

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  12.  $\begin{array}{r}\text{-}3,\text{-}2i\end{array}$
  ::- 12.3,-2i

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  13.  $\begin{array}{r}\sqrt{5},\text{-}1+i\end{array}$
  ::13. 5,1+5,1+1

  14.  $\begin{array}{r}2,\frac{1}{3},4-\sqrt{2}\end{array}$

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  Explore More
  ::探索更多

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  1. a. Try to write a polynomial of degree 4 with no real zeros. 
  ::1. 尝试在无实际零位的情况下写出四级多位数。

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  b. Try to write a polynomial of degree 3 with no real zeros. 
  ::b. 尝试在无实际零位的情况下写出三度多位数。

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  c. What can you conclude about the degree and whether there is or is not a real zero? (Hint: Consider even and odd degree.)
  ::c. 关于程度以及是否存在实际零,你能得出什么结论? (提示:考虑偶数和奇数程度。 )

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  2. The nth roots of unity are the solutions to equations of the form  $\begin{array}{r}{x}^n-1=0\end{array}$ . Find the nth roots of unity for $\begin{array}{r}n=3,4,5\end{array}$ .
  ::2. 团结的 n 根是表xn-1=0的方程式的解决方案。 查找 n=3,4,5的n 源为团结的 nth根。

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  3. Find all the zeros of the polynomial function $\begin{array}{r}f(x)=x^4+5x^2-36\end{array}$ , given that $\begin{array}{r}3i\end{array}$  is a zero of $\begin{array}{r}f\end{array}$ .   
  ::3. 查找多元函数 f(x) =x4+5x2-36的所有零,因为 3i 是 f的零。

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  4.  Louis calculates that the area of a rectangle is represented by the equation $\begin{array}{r}3x^4+7x^2=2\end{array}$ . Did Louis calculate it right? Explain based on the degree and zeros of the function.
  ::4. 路易计算矩形区域代表方程式 3x4+7x2=2。路易计算正确吗?根据函数的度和零解释。

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  5. Consider  $\begin{array}{r}h(x)=\text{-}\frac{1}{4}{x}^4-2x^3-\frac{13}{4}{x}^2-8x-9\end{array}$ . The exact value for one of the zeros is $\begin{array}{r}\text{-}4-\sqrt{7}\end{array}$  . What is the exact value of the other root? Next, use this information to find the complex roots.
  ::5. 考虑 h(x) = 14x4-2x3- 134x2-8x- 9. 零的准确值为 - 4-7 。 其它根的准确值是多少? 下一步, 使用此信息查找复杂的根 。

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  Answers for Review and Explore More Problems
  ::回顾和探讨更多问题的答复

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  Please see the Appendix. 
  ::请参看附录。

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  PLIX
  ::PLIX

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  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:

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  References
  ::参考参考资料

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  1. "Gravitational Lens," last edited May 15, 2017,
  ::1. 2017年5月15日 上一次编辑的《万象镜头》

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  2. Khavinson, D., and G. Neumann. "From the Fundamental Theorem of Algebra to Astrophysics: A 'Harmonious' Path." Notices of the AMS 55, no. 6 (June/July 2008): 666-75. .
  ::2. Khavinson, D.和G. Neumann,“从代数的基本理论到天体物理学:一个`和谐'的路径”。 AMS 55通知,第6号(2008年6月/7月):666-75。