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# 数学代写|现代代数代考Modern Algebra代写|MATH611 The polynomial ring R[x]

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## 数学代写|现代代数代考Modern Algebra代写|The polynomial ring R[x]

Let’s turn our attention now to polynomials with real coefficients. Much of what we can say about $\mathbf{R}[x]$ comes from the relation of $\mathbf{R}$ as a subfield $\mathbf{C}$, and consequently $=$ from the relation of $\mathbf{R}[x]$ as a subring of $\mathbf{C}[x]$. That is to say, we can interpret a polynomial $f(x)$ with real coefficients as a polynomial with complex coefficients.

Theorem 3.54. If a polynomial $f(x)$ with real coefficients has a complex root $z$, then its complex conjugate $\bar{z}$ is also a root.

Proof. Let $f(x)=a_n x^n+\cdots+a_1 x+a_0$ where each $a_i \in \mathbf{R}$. If $z$ is a root of $f$, then $f(z)=a_n z^n+\cdots+a_1 z+a_0=0$. Take the complex conjugate of the equation, and note that $\bar{a}_i=a_i$. Then $f(\bar{z})=a_n \bar{z}^n+\cdots+a_1 \bar{z}+a_0=0$. Thus, $\bar{z}$ is also a root.

Q.E.D.
This theorem tells us for a polynomial $f(x)$ with real coefficients, its roots either come in $k$ pairs of a complex number or singly as real numbers. We can name the $2 k$ complex roots as
$$z_1, \bar{z}1, z_2, \bar{z}_2, \ldots, z_k, \bar{z}_k .$$ Writing $z_1=x_1+y_1 i, \ldots, z_k=x_i+y_k i$, the complex roots are $$x_1+y_1 i, x_1-y_1 i, x_2+y_2 i, x_2-y_2 i, \ldots, x_k+y_k i, x_k-y_k i$$ and the $n-2 k$ real roots as $$r{2 k+1}, \ldots, r_n$$

## 数学代写|现代代数代考Modern Algebra代写|Roots of polynomials

The quadratic case. Consider a quadratic polynomial $f(x)=a x^2+b x+c$ with coefficients in $\mathbf{Q}$.

Its roots are given by the quadratic formula $\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$ which can be shown by the process known as completing the square. The discriminant of a quadratic polynomial is $\Delta=b^2-4 a c$. When $\Delta$ is positive, there are two real roots; when 0 , there is one double root; and when negative, the roots are a pair of complex conjugate numbers.

When $\Delta$ is a perfect rational square, that is, the square of a rational number, then $f(x)$ factors, that is, it’s reducible. Otherwise, it’s irreducible.
Thus, $f(x)$ is irreducible if and only if the discriminant is not a perfect square.

The cubic case. It is more difficult to determine when a cubic polynomial $f(x)=a x^3+$ $b x^2+c x+d$ with rational coefficients is irreducible, but not too difficult. Note that if $f(x)$ factors, then one of the factors has to be linear, so the question of reducibility reduces to the existence of a rational root of $f(x)$.

Various solutions of a cubic equation $a x^3+b x^2+c x+d=0$ have been developed. Here’s one. First, we may assume that $f$ is monic by dividing by the leading coefficient. Our equation now has the form $x^3+b x^2+c x+d=0$. Second, we can eliminate the quadratic term by replacing $x$ by $y-\frac{1}{3} b$. The new polynomial in $y$ will have different roots, but they’re only translations by $\frac{1}{3} b$. We now have the cubic equation
$$y^3+\left(c-\frac{1}{3} b^2\right) y+\left(\frac{2}{27} b^3-\frac{1}{3} b c+d\right)=0$$
which we’ll write as
$$y^3+p y+q=0$$

# 现代代数代写

## 数学代写|现代代数代考Modern Algebra代写|The polynomial ring $R[x]$

QED

$$z_1, \bar{z} 1, z_2, \bar{z}_2, \ldots, z_k, \bar{z}_k .$$

$$x_1+y_1 i, x_1-y_1 i, x_2+y_2 i, x_2-y_2 i, \ldots, x_k+y_k i, x_k-y_k i$$

$$r 2 k+1, \ldots, r_n$$

## 数学代写|现代代数代考Modern Algebra代写|Roots of polynomials

$$y^3+\left(c-\frac{1}{3} b^2\right) y+\left(\frac{2}{27} b^3-\frac{1}{3} b c+d\right)=0$$

$$y^3+p y+q=0$$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。