%% This document created by Scientific Notebook (R) Version 3.5 %% Starting shell: article \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amsmath} \usepackage{amssymb} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Wednesday, February 10, 1999 13:29:48} %TCIDATA{LastRevised=Monday, October 03, 2005 09:54:41} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{Counters=arabic,1} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za III online - Komplexn\U{e9} \U{10d}\U{ed}sla a funkcie komplexnej premennej - Komplexn\U{e9} \U{10d}\U{ed}sla a algebraick\U{e9} oper\U{e1}cie s nimi\dotfill \thepage }} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \author{A. U. Thor} \title{Lab Report} \date{The Date } \maketitle \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{Polyn\'{o}my a racion\'{a}lne funkcie} \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{aindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{aindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{A5.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{A5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{A52.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{A52.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{A54.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{A54.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{OA5.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{OA5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{CA5.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{CA5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{INDA.tex}}% %BeginExpansion \msihyperref{Index}{}{}{INDA.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \subsection{Korene polyn\'{o}mu} \begin{definition} Komplexn\'{e} \v{c}\'{\i}slo $c$ naz\'{y}vame \emph{kore\v{n}om polyn\'{o}mu} $f,$ ak $f(c)=0$. \end{definition} \begin{example} Zistite, \v{c}i niektor\'{e} z \v{c}\'{\i}sel $3,\ 1+i$ je kore\v{n}om polyn% \'{o}mu $h(x)=2x^{3}-5x^{2}+6x-2$. \end{example} \begin{solution} Sta\v{c}\'{\i} zisti\v{t}, \v{c}i hodnota polyn\'{o}mu $h$ v dan\'{y}ch \v{c}% \'{\i}slach je $0$. Po\v{c}\'{\i}tajme \begin{equation*} h(3)=2\cdot 3^{3}-5\cdot 3^{2}+6\cdot 3-2=54-45+18-2=25. \end{equation*}% Vid\'{\i}me,~\v{z}e~\v{c}\'{\i}slo~$3$~nie~je~kore\v{n}om~polyn\'{o}mu~$h$.% \newline Hodnotu~polyn\'{o}mu~$h$~v~\v{c}\'{\i}sle~$1+i$~vypo\v{c}\'{\i}% tame~pomocou~Hornerovej~sch\'{e}my.\medskip \newline \begin{equation*} \begin{tabular}{c|rrrrl} & $2$ & $-5$ & $6$ & $-2$ & \\ \cline{1-5} $1+i$ & & $2+2i$ & $-5-i$ & 2 & \\ \cline{1-5} & $2$ & $-3+2i$ & $1-i$ & $% \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array}% $ & $=h(1+i)$% \end{tabular}% \end{equation*}% \newline Zistili~sme,~\v{z}e~$h(1+i)=0$,~teda~$1+i$~je~kore\v{n}om~polyn\'{o}mu~$h$.\ $\square $ \end{solution} \begin{theorem} Komplexn\'{e} \v{c}\'{\i}slo $c$ je kore\v{n}om polyn\'{o}mu $f$ pr\'{a}ve vtedy, ke\v{d} $(x-c)\mid f$. \end{theorem} \begin{tabular}{|l|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA531.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA531.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{1} \begin{definition} Ak \v{c}\'{\i}slo $c$ je kore\v{n} polyn\'{o}mu $f$ stup\v{n}a $n\in \mathbf{% N}$, tak line\'{a}rny polyn\'{o}m $x-c$ naz\'{y}vame \emph{kore\v{n}ov\'{y} \v{c}inite\v{l}} polyn\'{o}mu $f$. \end{definition} \begin{definition} Komplexn\'{e} \v{c}\'{\i}slo $c$ sa naz\'{y}va $k$\emph{-n\'{a}sobn\'{y} kore% \v{n}} $(k\in \mathbf{N})$ polyn\'{o}mu $f,\ \text{st}(f)\geq 1,$ ak $% (x-c)^{k}\mid f$ a $(x-c)^{k+1}\NEG{\hspace{2pt}}\mid f$. \ Pre $k=1$ hovor% \'{\i}me, \v{z}e $c$ je\emph{\ jednoduch\'{y} kore\v{n}}. \end{definition} \begin{quote} Ak $c$ je $k$-n\'{a}sobn\'{y} kore\v{n} polyn\'{o}mu $f,$ tak $% f(x)=(x-c)^{k}g(x)$ pre vhodn\'{y} polyn\'{o}m $g,\,$ale pre ka\v{z}d\'{y} polyn\'{o}m $h$, je $f(x)\neq (x-c)^{k+1}h(x)$. \end{quote} \begin{example} Zistite, ko\v{l}kon\'{a}sobn\'{y}m kore\v{n}om polyn\'{o}mu $% f(x)=x^{5}-5x^{4}+7x^{3}-2x^{2}+4x-8$ je \v{c}\'{\i}slo $2$. \end{example} \begin{solution} Pomocou Hornerovej sch\'{e}my vydel\'{\i}me polyn\'{o}m $f$ polyn\'{o}mom $% x-c$. Ak vyjde nulov\'{y} zvy\v{s}ok, opakujeme delenie pre podiel, ktor\'{y} sme predt\'{y}m dostali. Toto delenie opakujeme dovtedy, k\'{y}m vyjde nenulov\'{y} zvy\v{s}ok. N\'{a}sobnos\v{t} kore\v{n}a je potom zrejme po\v{c}% et delen\'{\i}, pri ktor\'{y}ch vy\v{s}iel nulov\'{y} zvy\v{s}ok. \begin{equation*} \begin{array}{c|rrrrrr} & 1 & -5 & 7 & -2 & 4 & -8 \\ \hline 2 & & 2 & -6 & 2 & 0 & 8 \\ \hline & 1 & -3 & 1 & 0 & 4 & \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array} \\ \cline{1-6} 2 & & 2 & -2 & -2 & -4 & \\ \cline{1-6} & 1 & -1 & -1 & -2 & \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array} & \\ \cline{1-5} 2 & & 2 & 2 & 2 & & \\ \cline{1-5} & 1 & 1 & 1 & \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array} & & \\ \cline{1-4} 2 & & 2 & 6 & & & \\ \cline{1-4} & 1 & 3 & \begin{array}{|r} 7\hspace{-5pt} \\ \hline \end{array} & & & \end{array}% \end{equation*}% \v{C}\'{\i}slo~$2$~je~trojn\'{a}sobn\'{y}m~kore\v{n}om~polyn\'{o}mu~$f$,~ktor% \'{y}~m\^{o}\v{z}eme~p\'{\i}sa\v{t}~v~tvare \begin{equation*} f(x)=(x-2)^{3}(x^{2}+x+1).\;\square \end{equation*} \end{solution} \begin{theorem} (o racion\'{a}lnych kore\v{n}och polyn\'{o}mov s celo\v{c}\'{\i}seln\'{y}mi koeficientami) Nech racion\'{a}lne \v{c}\'{\i}slo $\dfrac{p}{q}$, kde $p,q$ s% \'{u} nes\'{u}delite\v{l}n\'{e} cel\'{e} \v{c}\'{\i}sla, je kore\v{n} polyn% \'{o}mu $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0},\ a_{n}\neq 0,\,n\geq 1,\,$s celo\v{c}\'{\i}seln\'{y}mi koeficientami. Potom $q\mid a_{n},\quad p\mid a_{0}.$ \end{theorem} \begin{tabular}{|l|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA532.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA532.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{2} \begin{example} N\'{a}jdite v\v{s}etky racion\'{a}lne korene polyn\'{o}mu $f(x)=\frac{1}{3}% x^{5}-\frac{1}{2}x^{4}+\frac{1}{3}x^{3}-\frac{1}{3}x^{2}+\frac{1}{6}$ \end{example} \begin{solution} Polyn\'{o}m $f$ nem\'{a} celo\v{c}\'{\i}seln\'{e} koeficienty, ale polyn\'{o}% m $g(x)=6f(x)=2x^{5}-3x^{4}+2x^{3}-2x^{2}+1$ \'{a}no. Navy\v{s}e pre ka\v{z}d% \'{e} komplexn\'{e} \v{c}\'{\i}slo $x$ plat\'{\i} $f(x)=0$ pr\'{a}ve vtedy, ke\v{d} $g(x)=0$, To znamen\'{a}, \v{z}e $g$ m\'{a} rovnak\'{e} korene ako $% f $. M\^{o}\v{z}eme teda h\v{l}ada\v{t} racion\'{a}lne korene polyn\'{o}mu $% g $ v tvare $\frac{p}{q}$, kde $p$ je cel\'{e} \v{c}\'{\i}slo, $q$ je prirodzen\'{e} \v{c}\'{\i}slo vyhovuj\'{u}ce podmienkam: $p$ del\'{\i} $% a_{0}=1$, $q$ del\'{\i} $a_{5}=2$. Do \'{u}vahy pripadaj\'{u} \v{c}\'{\i}% sla: \begin{equation*} p\in \{\pm 1\},\ q\in \{1,2\},\ \text{odkia\v{l} vypl\'{y}va}\ \frac{p}{q}% \in \left\{ \pm 1,\pm \frac{1}{2}\right\} \end{equation*}% Ak polyn\'{o}m $g$ m\'{a} racion\'{a}lny kore\v{n}, tak pod\v{l}a predch\'{a}% dzaj\'{u}cej vety to m\^{o}\v{z}e by\v{t} len niektor\'{e} z \v{c}\'{\i}sel $% \pm 1,\ \pm \frac{1}{2}$ a \v{z}iadne in\'{e}. Ktor\'{e} z nich je kore\v{n}% om polyn\'{o}mu $g$, m\^{o}\v{z}eme zisti\v{t} Hornerovou sch\'{e}mou. V~pr% \'{\i}pade, \v{z}e natraf\'{\i}me na kore\v{n}, zist\'{\i}me hne\v{d} jeho n% \'{a}sobnos\v{t}. \begin{equation*} \begin{array}{c|rrrrrr} & 2 & -3 & 2 & -2 & 0 & 1 \\ \hline 1 & & 2 & -1 & 1 & -1 & -1 \\ \hline & 2 & -1 & 1 & -1 & -1 & \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array} \\ \cline{1-6} 1 & & 2 & 1 & 2 & 1 & \\ \cline{1-6} & 2 & 1 & 2 & 1 & \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array} & \\ \cline{1-5} 1 & & 2 & 3 & 5 & & \\ \cline{1-5} & 2 & 3 & 5 & \begin{array}{|r} 6\hspace{-5pt} \\ \hline \end{array} & & \end{array}% \end{equation*}% Zistili sme, \v{z}e polyn\'{o}m $g$ je delite\v{l}n\'{y} polyn\'{o}mom $% (x-1)^{2}$, ale nie je delite\v{l}n\'{y} polyn\'{o}mom $(x-1)^{3}$, \v{c}o znamen\'{a}, \v{z}e $1$ je dvojn\'{a}sobn\'{y} kore\v{n} polyn\'{o}mu $g$ a tento polyn\'{o}m m\^{o}\v{z}eme nap\'{\i}sa\v{t} v tvare \begin{equation*} g(x)=(x-1)^{2}(2x^{3}+x^{2}+2x+1). \end{equation*} Tretie delenie polyn\'{o}mu $g$ polyn\'{o}mom $x-1$ v Hornerovej sch\'{e}me u% \v{z} nebolo nutn\'{e} vykona\v{t}. Sta\v{c}ilo si uvedomi\v{t}, \v{z}e re% \'{a}lny polyn\'{o}m $h(x)=2x^{3}+x^{2}+2x+1,\,$ktor\'{y} je nenulov\'{y} s nez\'{a}porn\'{y}mi koeficientami, nem\^{o}\v{z}e ma\v{t} kladn\'{e} korene, lebo hodnota tohto polyn\'{o}mu v kladnom \v{c}\'{\i}sle je kladn\'{e} \v{c}% \'{\i}slo, a teda nie nula. V\v{s}etky \v{d}al\v{s}ie korene polyn\'{o}mu $g$ u\v{z} musia by\v{t} kore\v{n}mi polyn\'{o}mu $h$. Preto sta\v{c}\'{\i} pokra% \v{c}ova\v{t} v Hornerovej sch\'{e}me pre polyn\'{o}m $h,\,$pri\v{c}om, kladn% \'{e} \v{c}\'{\i}sla u\v{z} nemus\'{\i}me overova\v{t}. \begin{equation*} \begin{array}{c|rrrr} & 2 & 1 & 2 & 1 \\ \hline -1 & & -2 & 1 & -3 \\ \hline & 2 & -1 & 3 & \begin{array}{|r} 2\hspace{-5pt} \\ \hline \end{array}% \end{array}% \end{equation*}% \v{C}islo $-1$ nie je kore\v{n}om polyn\'{o}mu $g$. \begin{equation*} \begin{array}{c|rrrr} & 2 & 1 & 2 & 1 \\ \hline -\frac{1}{2} & & -1 & 0 & -1 \\ \hline & 2 & 0 & 2 & \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array}% \end{array}% \end{equation*}% Polyn\'{o}m $2x^{2}+2$ u\v{z} nem\'{a} re\'{a}lne korene (teda ani racion% \'{a}lne), a preto \v{c}\'{\i}slo $-\frac{1}{2}$ je len jednoduch\'{y}m kore% \v{n}om~polyn\'{o}mu~$g$. Polyn\'{o}m $g,\,$a teda aj \ $f$ \ m\'{a} pr\'{a}% ve tieto racion\'{a}lne korene: $1$ - dvojn\'{a}sobn\'{y}, $-\frac{1}{2}$ - jednoduch\'{y} a polyn\'{o}m~$f$~m\^{o}\v{z}eme~p\'{\i}sa\v{t}~v~tvare~s\'{u}% \v{c}inu$~$% \begin{equation*} f(x)=\dfrac{1}{6}g(x)=\dfrac{1}{6}(x-1)^{2}\left( x+\dfrac{1}{2}\right) (2x^{2}+2).\blacksquare \end{equation*} \end{solution} \begin{theorem} Nech $f\in P(\mathbf{R}),\ \text{st}(f)\geq 1,\,$potom plat\'{\i}: Ak $c\in \mathbf{C}$ je kore\v{n} polyn\'{o}mu $f,\,$tak aj komplexne zdru\v{z}en\'{e} \v{c}\'{\i}slo $\overline{c}$ je kore\v{n} polyn\'{o}mu $f.$ Naviac, ak $c$ je $k$-n\'{a}sobn\'{y} kore\v{n} polyn\'{o}mu $f,\,$tak aj $\overline{c}$ je $k$-n\'{a}sobn\'{y} kore\v{n} polyn\'{o}mu $f.$ \end{theorem} \begin{tabular}{|l|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA533.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA533.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{3} Re\'{a}lny polyn\'{o}m stup\v{n}a aspo\v{n} $1$ nemus\'{\i} ma\v{t} \v{z}% iadny re\'{a}lny kore\v{n}. M\^{o}\v{z}e sa sta\v{t}, \v{z}eby komplexn\'{y} polyn\'{o}m stup\v{n}a aspo\v{n} $1$ nemal \v{z}iadny komplexn\'{y} kore\v{n}% ? Tak\'{a}to situ\'{a}cia nem\^{o}\v{z}e nasta\v{t}, preto\v{z}e plat\'{\i}: \begin{theorem} (Z\'{a}kladn\'{a} veta algebry) Ka\v{z}d\'{y} komplexn\'{y} polyn\'{o}m stup% \v{n}a aspo\v{n} jedna m\'{a} aspo\v{n} jeden komplexn\'{y}~kore\v{n}.% \newline \end{theorem} \begin{example} N\'{a}jdite v\v{s}etky korene polyn\'{o}mu $% f(x)=2x^{5}+5x^{4}+8x^{3}+7x^{2}+6x+2,\,$ak viete, \v{z}e jedn\'{y}m kore% \v{n}om je $-1+i$. \end{example} \begin{solution} $f$ je polyn\'{o}m s re\'{a}lnymi koeficientami, preto \v{d}al\v{s}\'{\i}m jeho kore\v{n}om je $\overline{-1+i}=-1-i$. Vyde\v{l}me polyn\'{o}m $f$ kore% \v{n}ov\'{y}mi \v{c}\'{\i}nite\v{l}mi $x+1-i,\,$ $x+1+i$: \begin{equation*} \begin{array}{c|rrrrrr} & 2 & 5 & 8 & 7 & 6 & 2 \\ \hline -1+i & & -2+2i & -5+i & -4+2i & -5+i & -2 \\ \hline & 2 & 3+2i & 3+i & 3+2i & 1+i & \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array} \\ \cline{1-6} -1-i & & -2-2i & -1-i & -2-2i & -1-i & \\ \cline{1-6} & 2 & 1 & 2 & 1 & \begin{array}{|r} 0\hspace{-5pt} \\ \hline \end{array} & \end{array}% \end{equation*}% Podielom je polyn\'{o}m $h(x)=2x^{3}+x^{2}+2x+1$. Zvy\v{s}n\'{e} korene polyn% \'{o}mu $f$ s\'{u} kore\v{n}mi polyn\'{o}mu $h$. Jeho jedin\'{y} racion\'{a}% lny kore\v{n} sme na\v{s}li v predch\'{a}dzaj\'{u}com pr\'{\i}klade, je n% \'{\i}m \v{c}\'{\i}slo $-\frac{1}{2}$. Tie\v{z} sme tam zistili, \v{z}e $% h(x)=\left( x+\frac{1}{2}\right) (2x^{2}+2)$. Zost\'{a}vaj\'{u}ce korene polyn\'{o}mu $f$ s\'{u} preto kore\v{n}mi polyn\'{o}mu $2x^{2}+2$. N\'{a}% jdeme ich rie\v{s}en\'{\i}m kvadratickej rovnice $2x^{2}+2=0$: \begin{equation*} 2x^{2}+2=0 \end{equation*}% \begin{equation*} x^{2}=-1 \end{equation*}% \begin{equation*} x=\pm i \end{equation*}% Polyn\'{o}m~$f$~m\'{a}~korene~$-1+i,~-1-i,~-\frac{1}{2},~i,-i$.~S\'{u}~to~v% \v{s}etko~jednoduch\'{e}~korene. $\square $\newline \end{solution} Nech $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}$, $a_{n}\neq 0,\ n\geq 1$. Pod\v{l}a z\'{a}kladnej vety algebry m\'{a} tento polyn\'{o}m aspo% \v{n} jeden kore\v{n}. Ozna\v{c}me ho $c_{1}$. Polyn\'{o}m $f$ je delite\v{l}% n\'{y} kore\v{n}ov\'{y}m \v{c}inite\v{l}om $x-c_{1}$, teda existuje polyn% \'{o}m \begin{equation*} f_{1}\text{ st}\left( f_{1}\right) =n-1\text{ \ tak, \v{z}e}\ f(x)=(x-c_{1})f_{1}(x). \end{equation*}% Ak $n-1\geq 1,\,$tak $f_{1}$ m\'{a} kore\v{n}, m\^{o}\v{z}eme ho ozna\v{c}i% \v{t} $c_{2}$ a existuje tak\'{y} polyn\'{o}m \begin{equation*} f_{2},\ \text{st}(f_{2})=n-2,\,\text{\v{z}e \ }f_{1}(x)=(x-c_{2})f_{2}(x)% \text{ \ a potom }f(x)=(x-c_{1})(x-c_{2})f_{2}(x). \end{equation*}% Takto m\^{o}\v{z}eme pokra\v{c}ova\v{t} \v{d}alej a po $n$-tom zopakovan% \'{\i} tohto kroku dostaneme \begin{equation*} f(x)=(x-c_{1})(x-c_{2})\ldots (x-c_{n})f_{0}(x),\,\text{st}(f_{0})=0 \end{equation*}% \newline $f_{0}$ je kon\v{s}tantn\'{y} polyn\'{o}m, preto pre v\v{s}etky komplexn\'{e} \v{c}\'{\i}sla $x$ je $f_{0}(x)=b$, kde $b$ je komplexn\'{e} \v{c}\'{\i}slo. Vzh\v{l}adom k tomu, \v{z}e najvy\v{s}\v{s}\'{\i} koeficient polyn\'{o}mu $f$ je $a_{n}$, mus\'{\i} plati\v{t}: $b=a_{n}$.Tak sme dospeli k tvaru polyn% \'{o}mu $f$ \begin{equation*} f(x)=a_{n}(x-c_{1})(x-c_{2})\ldots (x-c_{n}) \end{equation*}% ktor\'{y} naz\'{y}vame \emph{rozklad polyn\'{o}mu }$f$\emph{\ na s\'{u}\v{c}% in kore\v{n}ov\'{y}ch \v{c}inite\v{l}ov}. Z tohto tvaru polyn\'{o}mu $f$ vypl% \'{y}va, \v{z}e \v{c}\'{\i}sla $c_{1},c_{2},\ldots ,c_{n}$ s\'{u} jeho korene a okrem nich \v{z}iadne in\'{e} nem\'{a}. Preto plat\'{\i} \begin{theorem} Polyn\'{o}m~stup\v{n}a~$n,~n\geq 1,$~m\'{a}~najviac~$n$~r\^{o}znych~kore\v{n}% ov.\newline \end{theorem} \begin{quote} Polyn\'{o}m nult\'{e}ho stup\v{n}a nem\'{a} \v{z}iadny kore\v{n}. Kore\v{n}% om polyn\'{o}mu stup\v{n}a $-\infty $ (nulov\'{e}ho polyn\'{o}mu) je ka\v{z}d% \'{e} komplexn\'{e} \v{c}\'{\i}slo. \end{quote} \begin{theorem} Nech $f,g$ s\'{u} komplexn\'{e} polyn\'{o}my stup\v{n}a najviac $n,\ g\neq \mathbf{0},\,n\in \mathbf{N}\cup \{0\}$ a nech pre $n+1$ komplexn\'{y}ch \v{c}\'{\i}sel $x_{0},x_{1},\ldots ,x_{n}$ plat\'{\i} \begin{equation*} f(x_{0})=g(x_{0}),\text{\/}f(x_{1})=g(x_{1}),\ldots ,\text{\/}% f(x_{n})=g(x_{n}), \end{equation*}% potom $f=g.$ \end{theorem} \begin{tabular}{|l|} \hline \textbf{% %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DA534.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DA534.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{4} \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{aindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{aindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{A5.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{A5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{A52.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{A52.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{A54.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{A54.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{OA5.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{OA5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{CA5.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{CA5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{INDA.tex}}% %BeginExpansion \msihyperref{Index}{}{}{INDA.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematika 1} \section{Polyn\'{o}my a racion\'{a}lne funkcie} \end{document}