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- https://query.libretexts.org/%E7%AE%80%E4%BD%93%E4%B8%AD%E6%96%87/%E4%B8%AD%E7%BA%A7%E4%BB%A3%E6%95%B0_(OpenStax)/05%3A_%E5%A4%9A%E9%A1%B9%E5%BC%8F%E5%92%8C%E5%A4%9A%E9%A1%B9%E5%BC%8F%E5%87%BD%E6%95%B0/5.05%3A_%E9%99%A4%E4%BB%A5%E5%A4%9A%E9%A1%B9%E5%BC%8F\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=}...\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align} \)
- https://query.libretexts.org/Idioma_Portugues/Algebra_intermediaria_(OpenStax)/05%3A_Fun%C3%A7%C3%B5es_polinomiais_e_polinomiais/5.05%3A_Dividindo_polin%C3%B4mios\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=}...\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align} \) Quando o divisor é escrito como\(x−c\), o valor da função at\(c\),\(f(c)\), é o mesmo que o restante do problema de divisão.
- https://query.libretexts.org/%D8%A7%D9%84%D9%84%D8%BA%D8%A9_%D8%A7%D9%84%D8%B9%D8%B1%D8%A8%D9%8A%D8%A9/%D8%A7%D9%84%D8%AC%D8%A8%D8%B1_%D8%A7%D9%84%D9%85%D8%AA%D9%88%D8%B3%D8%B7_(OpenStax)/05%3A_%D8%AF%D9%88%D8%A7%D9%84_%D9%83%D8%AB%D9%8A%D8%B1%D8%A9_%D8%A7%D9%84%D8%AD%D8%AF%D9%88%D8%AF_%D9%88%D8%AF%D9%88%D8%A7%D9%84_%D9%83%D8%AB%D9%8A%D8%B1%D8%A9_%D8%A7%D9%84%D8%AD%D8%AF%D9%88%D8%AF/5.05%3A_%D9%82%D8%B3%D9%85%D8%A9_%D9%83%D8%AB%D9%8A%D8%B1%D8%A7%D8%AA_%D8%A7%D9%84%D8%AD%D8%AF%D9%88%D8%AF\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=}...\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align} \)
- https://query.libretexts.org/Francais/Alg%C3%A8bre_interm%C3%A9diaire_(OpenStax)/05%3A_Fonctions_polynomiales_et_polynomiales/5.05%3A_Diviser_des_polyn%C3%B4mes\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=}...\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align} \) En notation fonctionnelle, nous pourrions dire que pour obtenir le dividende\(f(x)\), nous multiplions le quotient,\(q(x)\) multiplions le diviseur et ajoutons le reste\(r\).\(x−c\)
- https://query.libretexts.org/Kiswahili/Algebra_ya_kati_(OpenStax)/05%3A_Kazi_za_Polynomial_na_Polynomial/5.05%3A_Kugawanya_Polynomials\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=}...\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align} \)