[A2/H3/E3] Solved

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+% Document MUST be built with XeLaTeX!
+\documentclass[11pt,a4paper]{article}
+
+\usepackage[margin=1.5cm]{geometry}
+
+% Поддръжка на български език
+% Thanks to https://tex.stackexchange.com/a/136926
+\usepackage{fontspec}
+\defaultfontfeatures{Ligatures=TeX}
+\setmainfont{CMU Serif}
+\setsansfont{CMU Sans Serif}
+
+\usepackage[bulgarian,english]{babel}
+
+\usepackage{amsmath}            % Математически символи
+\usepackage{amssymb}
+\usepackage{mathtools}
+\usepackage{amsfonts}           % Още математически символи
+\usepackage{cancel}             % Зачертаване
+\usepackage[dvipsnames]{xcolor} % Цветни кутийки + допълнителни цветове
+\usepackage{indentfirst}        % Индентация на параграф след секция
+
+\usepackage{enumitem}
+\setlist[enumerate]{label=\arabic*.}
+
+\renewcommand{\thesection}{}
+
+\let\oldforall\forall
+\renewcommand{\forall}{\; \oldforall}
+
+\let\oldexist\exists
+\renewcommand{\exists}{\; \oldexist}
+
+\usepackage[pdfauthor={Камен Димитров Младенов, КН Летен семестър, 2023},
+		    pdftitle={Домашна работа 3 по А2, Зад 3},
+		    pdfsubject={},
+		    pdfkeywords={},
+		    colorlinks=true,
+		    urlcolor=blue]{hyperref}
+
+\title{\vspace{-1.5cm}Домашно №3, Задача 3 по Алгебра 2 на Компютърни Науки}
+\author{Камен Димитров Младенов, Група 6}
+\date{31 Май 2023}
+
+\begin{document}
+
+\maketitle
+
+\section{Решение}
+
+Ще докажем, че $p(x)$ е неразложим (над $\mathbb{Q}$). $\gcd(1, -3, 6, 2) = \pm 1 \implies f(x)$ е примитивен, то тогава той е разложим над $\mathbb{Q} \iff$ е разложим над $\mathbb{Z}$.
+Разлгеждаме съществено разлагане $f(x) = f_1(x) f_2(x) \quad f_1, f_2 \in \mathbb{Z}[x]$:
+\begin{align*}
+	& 3 = \deg(f) = \deg(f_1) + \deg(f_2), \quad 0 < \deg (f_1), \quad \deg (f_2) < \deg (f) \quad \text{Б.О.О. } \deg (f_1) \leq \deg(f_2) \\
+	& \text{Тогава } \deg (f_1) = 1, \deg (f_2) = 2 \quad f_1(x) = x - r, \; r \in \mathbb{Z} \quad r \text{ е делител на свободния член} \\
+	& \implies r \in \{ -2, -1, 1, 2 \} \quad \text{Пресмятаме по правилото на Хорнер:} \quad \begin{array}{r|rrrr}
+		     & 1 & -3 & 6 & 2 \\
+		\hline
+		 -2 & 1 & -5 & 16 & -30 \\
+		 -1 & 1 & -4 & 10 & -8 \\
+		  1 & 1 & -2 & 4 & 6 \\
+		  2 & 1 & -1 & 4 & 10 \\
+	\end{array} \\
+	& \text{Следователно } f \text{ няма рационални корени и е неразложим над } \mathbb{Z} \text{, тоест е неразложим над } \mathbb{Q}.
+\end{align*}
+
+Следователно, $F = \mathbb{Q}[x] / \langle p \rangle$ е поле.
+От това следва, че $\mathbb{Q}[\alpha]^{\leq 2}$ е поле. Ще намерим $\lambda$:
+\begin{align*}
+	& \text{Нека } f(x) = x^2 + x - 1 \in \mathbb{Q}[x]^{\leq 2}, \quad g(x) = x^2 - x - 1 \in \mathbb{Q}[x]^{\leq 2} \implies f(x)g(x) = x^4 - 3x^2 + 1 \\
+	& \frac{f(x)g(x)}{p(x)} = \frac{x^4 - 3x^2 + 1}{x^3 - 3x^2 + 6x + 2} \qquad (x^4 - 3x^2 + 1) - x(x^3 - 3x^2 + 6x + 2) = 3x^3 - 9x^2 - 2x + 1 \\
+	& \qquad (3x^3 - 9x^2 - 2x + 1) - 3(x^3 - 3x^2 + 6x + 2) = -20x - 5 \quad f(x)g(x) = (x+3)p(x) - 20x - 5 \\
+	& \text{Понеже } p(\alpha) = 0 \text{, то следва че } \lambda = f(\alpha)g(\alpha) = (\alpha + 3) p(\alpha) - 20 \alpha -5 = -20 \alpha - 5
+\end{align*}
+
+Сега ще намерим $\mu$:
+\begin{align*}
+	& \text{Нека } h(x) = x^2 + x + 1 \in \mathbb{Q}[x] \quad \text{Ще намерим } u(x), v(x) \in \mathbb{Q}[x] \text{, такива че } \\
+	& \gcd (p(x), h(x)) = h(x)u(x) + p(x)v(x) \\
+	& \frac{p(x)}{h(x)} = \frac{x^3 - 3x^2 + 6x + 2}{x^2 + x + 1} \qquad (x^3 - 3x^2 + 6x + 2) - x(x^2 + x + 1) = -4x^2 + 5x + 2 \\
+	& \qquad (-4x^2 + 5x + 2) - (-4)(x^2 + x + 1) = 9x + 6 = r_1 \implies p(x) = (x-4)h(x) + r_1 \\
+	& \frac{h(x)}{r_1} = \frac{x^2 + x + 1}{9x + 6} \qquad (x^2 + x + 1) - \frac{1}{9}x(9x + 6) = \frac{1}{3}x + 1 \qquad \left(\frac{1}{3}x + 1\right) - \frac{1}{27}(9x + 6) = \frac{7}{9} \\
+	& \implies h(x) = \left(\frac{x}{9} + \frac{1}{27}\right)r_1 + \frac{7}{9} \implies h(x) \text{ и } p(x) \text{ са взаимно прости, тоест } \gcd(h(x), p(x)) = 1 \\
+	& p(x) = (x-4)h(x) + r_1 \iff r_1 = p(x) + (4 - x)h(x) \implies h(x) = \left(\frac{x}{9} + \frac{1}{27}\right)(p(x) + (4 - x)h(x)) + \frac{7}{9} \\
+	& \frac{9}{7} h(x) = \left(\frac{x}{7} + \frac{1}{21}\right)(p(x) + (4 - x)h(x)) + 1 \qquad 0 = \left(\frac{x}{7} + \frac{1}{21}\right)p(x) + \left(\frac{x}{7} + \frac{1}{21}\right)(4 - x)h(x) + 1 - \frac{9}{7} h(x) \\
+	& -1 = \left(\frac{x}{7} + \frac{1}{21}\right)p(x) + \left(-\frac{x^2}{7} + \frac{11x}{21} + \frac{4}{21} - \frac{9}{7} \right)h(x) \quad 1 = \left(-\frac{x}{7} - \frac{1}{21}\right)p(x) + \left(\frac{x^2}{7} - \frac{11x}{21} + \frac{23}{21} \right)h(x) \\
+	& \text{Заместваме с } \alpha \qquad 1 = \left(\frac{\alpha^2}{7} - \frac{11\alpha}{21} + \frac{23}{21} \right)h(\alpha) \quad 21 = \left(3\alpha^2 - 11\alpha + 23 \right)h(\alpha) \\
+	& \implies \mu = \frac{21}{h(\alpha)} = 3\alpha^2 - 11\alpha + 23
+\end{align*}
+
+\end{document}
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