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\usepackage[pdfauthor={Камен Димитров Младенов, КН Летен семестър, 2023},
		    pdftitle={Домашна работа 3 по А2, Зад 1},
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\title{\vspace{-1.5cm}Домашно №3, Задача 1 по Алгебра 2 на Компютърни Науки}
\author{Камен Димитров Младенов, Група 6}
\date{31 Май 2023}

\begin{document}

\maketitle

\section{Решение}

\begin{align*}
	\frac{f(x)}{g(x)} & = \frac{x^3+x^2-x+\bar{2}}{x^3-\bar{2}x^2-\bar{2}x+\bar{2}} \quad (x^3+x^2-x+\bar{2}) - (x^3-\bar{2}x^2-\bar{2}x+\bar{2}) = \bar{3}x^2+x = r_1 \quad f(x) = g(x) + r_1 \\
	\frac{g(x)}{r_1} & = \frac{x^3-\bar{2}x^2-\bar{2}x+\bar{2}}{\bar{3}x^2+x} \quad (x^3-\bar{2}x^2-\bar{2}x+\bar{2}) - \bar{2}x(\bar{3}x^2+x) = -\bar{4}x^2 - \bar{2}x + \bar{2} = x^2 + \bar{3}x + \bar{2} \\
	& \qquad\qquad\qquad\qquad\qquad\; (x^2 + \bar{3}x + \bar{2}) - \bar{2}(\bar{3}x^2+x) = x + \bar{2} = r_2 \quad g(x) = (\bar{2}x + \bar{2})r_1 + r_2 \\
	\frac{r_1}{r_2} & = \frac{\bar{3}x^2+x}{x + \bar{2}} \quad \qquad\qquad\;\; (\bar{3}x^2+x) - (\bar{3}x)(x + \bar{2}) = 0 \quad r_1 = (\bar{3}x) r_2 \\
\end{align*}
\begin{align*}
	& g(x) = \bar{2}(x + \bar{1})(\bar{3}x) r_2 + r_2 = (x^2 + x + \bar{1}) r_2 \qquad f(x) = g(x) + (\bar{3}x) r_2 = (x^2 + \bar{4} x + \bar{1})r_2 \implies (f, g) = r_2
\end{align*}
\begin{align*}
	& f(x) = g(x) + r_1 \iff r_1 = f(x) - g(x) \\
	& \implies g(x) = \bar{2}(x+\bar{1})r_1 + r_2 = \bar{2}(x+\bar{1})(f(x) - g(x)) + r_2 \\
	& \iff r_2 = \bar{2}(x+\bar{1})(g(x) - f(x)) + g(x) \\
	& \iff r_2 = (\bar{2}x + \bar{2} + \bar{1})g(x) + (-\bar{2})(x+\bar{1})f(x) \\
	& \iff r_2 = \bar{3}(x+\bar{1})f(x) + (\bar{2}x + \bar{3})g(x) \\
	& \implies u(x) = \bar{3}x + \bar{3} \quad v(x) = \bar{2}x + \bar{3}
\end{align*}


\end{document}