{"id":1590,"date":"2022-01-16T21:48:15","date_gmt":"2022-01-17T00:48:15","guid":{"rendered":"http:\/\/localhost\/?page_id=1590"},"modified":"2023-01-06T14:51:16","modified_gmt":"2023-01-06T17:51:16","slug":"raizes-e-divisibilidade","status":"publish","type":"page","link":"http:\/\/localhost\/index.php\/ensino\/fundamentos-de-algebra\/raizes-e-divisibilidade\/","title":{"rendered":"Ra\u00edzes e divisibilidade"},"content":{"rendered":"
Na decomposi\u00e7\u00e3o dada pelo teorema anterior os escalares $\\alpha_1,\\ldots,\\alpha_k$ s\u00e3o as ra\u00edzes complexas de $f(x)$ contadas com multiplicidade. Isso quer dizer que uma raiz pode ocorrer v\u00e1rias vezes nesta lista.<\/p>\n
Lembre que se $z=a+bi\\in\\C$, ent\u00e3o o conjugado complexo<\/strong> $\\overline z$ de $z$ est\u00e1 definido como Seja $f(x)=x^2+\\alpha x+\\beta\\in\\R[x]$. Lembre que o discriminante<\/strong> $\\Delta$ de $f(x)$ \u00e9 definido como $\\Delta=\\alpha^2-4\\beta$. Ademais, $f(x)$ possui duas ra\u00edzes distintas em $\\R$ se $\\Delta > 0$, $f(x)$ possui uma \u00fanica raiz em $\\R$ se $\\Delta=0$ e $f(x)$ n\u00e3o possui ra\u00edzes em $\\R$ se $\\Delta < 0$.<\/p>\n Se $\\alpha\\in\\R$, ent\u00e3o $x-\\alpha\\mid f(x)$ e $f(x)=(x-\\alpha)q(x)$ com $q(x)\\in\\R[x]$ m\u00f4nico e $\\grau{q(x)}=k-1$. Pela hip\u00f3tese da indu\u00e7\u00e3o, Se $\\alpha\\not\\in\\R$, ent\u00e3o $f(\\overline\\alpha)=0$ e Na decomposi\u00e7\u00e3o dada pelo teorema anterior os escalares $\\alpha_1,\\ldots,\\alpha_s$ s\u00e3o as ra\u00edzes reais de $f(x)$ contadas com multiplicidade. Isso quer dizer que uma raiz pode ocorrer v\u00e1rias vezes nesta lista. No caso de polin\u00f4mios sobre $\\R$, o n\u00famero das ra\u00edzes de $f(x)$ pode ser menor que $\\grau{f(x)}$.<\/p>\n Seja $\\F$ um corpo. Lembre que $\\alpha\\in\\F$ \u00e9 raiz de um polin\u00f4mio $f(x)\\in\\F[x]$ se $f(\\alpha)=0$. Seja $f(x)\\in\\F[x]\\setminus\\{0\\}$ um polin\u00f4mio e $\\alpha\\in\\F$. Ent\u00e3o existe um polin\u00f4mio $q(x)\\in\\F[x]$ tal que \\[ f(x)=q(x)(x-\\alpha)+f(\\alpha). \\] Em particular, $x-\\alpha\\mid f(x)$ se e somente se $f(\\alpha)=0$ (ou seja, $\\alpha$ \u00e9 raiz do polin\u00f4mio $f(x)$). Use o Teorema de Divis\u00e3o de Euclides … Continue reading Ra\u00edzes e divisibilidade<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1193,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":[],"_links":{"self":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1590"}],"collection":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/comments?post=1590"}],"version-history":[{"count":9,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1590\/revisions"}],"predecessor-version":[{"id":2006,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1590\/revisions\/2006"}],"up":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/1193"}],"wp:attachment":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/media?parent=1590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\\[
\n\\overline z=a-bi.
\n\\]<\/p>\n\n
\n\\[
\nf(x)=\\alpha_nx^n+\\cdots+\\alpha_1x+\\alpha_0
\n\\]
\ne assuma que $\\alpha\\in\\C$ tal que $f(\\alpha)=0$. Ent\u00e3o, usando o exerc\u00edcio anterior, obtemos que
\n\\begin{align*}
\n0&=\\overline 0=\\overline{f(\\alpha)}=\\overline{\\alpha_n\\alpha^n+\\cdots+\\alpha_1\\alpha+\\alpha_0}\\\\&=
\n\\alpha_n\\overline \\alpha^n+\\cdots+\\alpha_1\\overline \\alpha+\\alpha_0.
\n\\end{align*}
\nLogo $f(\\overline\\alpha)=0$.<\/div>\n
\n\\[
\nf(x)=(x-\\alpha_1)\\cdots (x-\\alpha_s)(x^2+\\beta_1x+\\gamma_1)\\cdots (x^2+\\beta_rx+\\gamma_r)
\n\\]
\nonde $s+2r=k$, $\\alpha_i,\\beta_j,\\gamma_\\ell\\in\\R$ com $\\beta_i^2-4\\gamma_i < 0$ para todo $i\\in\\{1,\\ldots,r\\}$.<\/div>\n
\nAssuma que o teorema est\u00e1 verdadeiro para polin\u00f4mios $f(x)\\in\\F[x]$ com $\\grau{f(x)} < k$ e seja $f(x)\\in\\R[x]$ um polin\u00f4mio de grau $k$. Pelo Teorema Fundamental da \u00c1lgebra, existe $\\alpha\\in\\C$ tal que $f(\\alpha)=0$.<\/p>\n
\n\\[
\nq(x)=(x-\\alpha_1)\\cdots (x-\\alpha_s)(x^2+\\beta_1x+\\gamma_1)\\cdots (x^2+\\beta_rx+\\gamma_r)
\n\\]
\ncom $s+2r=k-1$ e $\\beta_i^2-4\\gamma_i < 0$ para todo $i$.
\nAssim
\n\\begin{align*}
\nf(x)&=(x-\\alpha)q(x)\\\\&=(x-\\alpha)(x-\\alpha_1)\\cdots(x-\\alpha_s)(x^2+\\beta_1x+\\gamma_1)\\cdots (x^2+\\beta_rx+\\gamma_r).
\n\\end{align*}
\ncom $\\alpha_i,\\beta_j,\\gamma_\\ell\\in\\R$ e $\\beta_i^2-4\\gamma_i < 0$ para todo $i$.<\/p>\n
\n\\[
\n(x-\\alpha)(x-\\overline\\alpha)\\mid f(x).
\n\\]
\nNote que
\n\\[
\n(x-\\alpha)(x-\\overline\\alpha)=x^2-(\\alpha+\\overline\\alpha)x+\\alpha\\overline\\alpha\\in\\R[x],
\n\\]
\npois se $\\alpha=a+bi$, ent\u00e3o $\\overline\\alpha=a-bi$ e assim
\n\\begin{align*}
\n\\alpha+\\overline\\alpha&=2a\\in\\R\\\\
\n\\alpha\\overline\\alpha&=a^2+b^2\\in\\R.
\n\\end{align*}
\nPortanto
\n\\[
\n(x-\\alpha)(x-\\overline\\alpha)=x^2+\\beta+\\gamma\\in\\R[x]
\n\\]
\ncom
\n\\begin{align*}
\n\\beta&=-\\alpha-\\overline{\\alpha}=-2a; \\\\
\n\\gamma&=\\alpha\\cdot\\overline{\\alpha}=a^2+b^2
\n\\end{align*}
\n\u00e9 um polin\u00f4mio sem ra\u00edzes em $\\R$ e assim $\\beta^2-4\\gamma < 0$.
\nEscreva $f(x)=(x^2+\\beta+\\gamma)q(x)$ com $q(x)\\in\\R[x]$ m\u00f4nico e $\\grau{g(x)}=k-2$. Pela hip\u00f3tese da indu\u00e7\u00e3o,
\n\\[
\nq(x)=(x-\\alpha_1)\\cdots (x-\\alpha_s)(x^2+\\beta_1x+\\gamma_1)\\cdots (x^2+\\beta_rx+\\gamma_r)
\n\\]
\ncom $\\alpha_i,\\beta_i,\\gamma_i\\in\\R$, $k-2=s+2r$ e $\\beta_i^2-4\\gamma_i < 0 $ para todo $i\\geq 1$. Ora,
\n\\[
\nf(x)=(x-\\alpha_1)\\cdots (x-\\alpha_s)(x^2+\\beta_1x+\\gamma_1)\\cdots (x^2+\\beta_rx+\\gamma_r)(x^2+\\beta x+\\gamma)
\n\\]
\ncomo foi afirmado.<\/p>\n<\/div>\n
\n\\[
\nf(x)=x^{n-1}+x^{n-2}+\\cdots+x+1\\in\\R[x]
\n\\]
\ncom $n\\geq 2$. No exemplo anterior determinamos as ra\u00edzes complexas de $f(x)$. Se $n$ for par, ent\u00e3o $f(x)$ possui uma \u00fanica raiz real, nomeadamente o $-1$, enquanto se $n$ for \u00edmpar, ent\u00e3o $f(x)$ n\u00e3o possui raiz real. Denotando por $\\alpha_k$ a $k$-\u00e9sima raiz complexa de $f(x)$ como no exemplo anterior, temos, para $k\\leq n\/2$ que
\n\\[
\n\\overline{\\alpha_k}=\\alpha_{n-k}.
\n\\]
\nPara $k\\leq n\/2$, seja
\n\\[
\n\\beta_k=-\\alpha_k-\\overline{\\alpha_k}=-2\\cos(2k\\pi \/n)
\n\\]
\ne note que
\n\\[
\n\\alpha_k\\cdot \\overline{\\alpha_k}=\\cos^2(2k\\pi \/n)+\\mbox{sen}^2(2k\\pi \/n)=1.
\n\\]
\nAssim obtemos a seguinte decomposi\u00e7\u00e3o. Se $n$ for par,
\n\\begin{align*}
\nf(x)&=(x+1)\\prod_{k=1}^{(n-2)\/2}(x^2+\\beta_kx+1)\\\\&=(x+1)\\prod_{i=1}^{(n-2)\/2}(x^2-2\\cos(2k\\pi\/n)x+1);
\n\\end{align*}
\nse $n$ for \u00edmpar, ent\u00e3o
\n\\begin{align*}
\nf(x)&=\\prod_{k=1}^{(n-1)\/2}(x^2+\\beta_kx+1)\\\\&=\\prod_{i=1}^{(n-1)\/2}(x^2-2\\cos(2k\\pi\/n)x+1).
\n\\end{align*}<\/div>\n