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CAPITULO VII

Corolario del Teorema del residuo

Ejercicio 76

Hallar, sin efectuar la división, si son exactas las divisiones siguientes:

$x^{2} - x - 6$ entre $x - 3$
$\begin{matrix} x - 3 & = 0 \\ x & = 3 \\ P (x) & = x^{2} - x - 6 \\ P (3) & = {(3)}^{2} - 3 - 6 \\ P (3) & = 9 - 9 \\ P (3) & = 0 \Rightarrow e x a c t a \end{matrix}$
$x^{3} + 4 x^{2} - x - 10$ entre $x + 2$
$\begin{matrix} x + 2 & = 0 \\ x & = - 2 \\ P (x) & = x^{3} + 4 x^{2} - x - 10 \\ P (- 2) & = {(- 2)}^{3} + 4 {(- 2)}^{2} - (- 2) - 10 \\ P (- 2) & = - 8 + 4 (4) + 2 - 10 \\ P (- 2) & = 16 - 16 \\ P (- 2) & = 0 \Rightarrow e x a c t a \end{matrix}$
$2 x^{4} - 5 x^{3} + 7 x^{2} - 9 x + 3$ entre $x - 1$
$\begin{matrix} x - 1 & = 0 \\ x & = 1 \\ P (x) & = 2 x^{4} - 5 x^{3} + 7 x^{2} - 9 x + 3 \\ P (1) & = 2 {(1)}^{4} - 5 {(1)}^{3} + 7 {(1)}^{2} - 9 (1) + 3 \\ P (1) & = 2 - 5 + 7 - 9 + 3 \\ P (1) & = - 2 \Rightarrow n o e s e x a c t a \end{matrix}$
$x^{5} + x^{4} - 5 x^{3} - 7 x + 8$ entre $x + 3$
$\begin{matrix} x + 3 & = 0 \\ x & = - 3 \\ P (x) & = x^{5} + x^{4} - 5 x^{3} - 7 x + 8 \\ P (- 3) & = {(- 3)}^{5} + {(- 3)}^{4} - 5 {(- 3)}^{3} - 7 (- 3) + 8 \\ P (- 3) & = - 243 + 81 - 5 (27) + 21 + 8 \\ P (- 3) & = - 243 + 81 - 135 + 21 + 8 \\ P (- 3) & = - 267 \Rightarrow n o e s e x a c t a \end{matrix}$
$4 x^{3} - 8 x^{2} + 11 x - 4$ entre $2 x - 1$
$\begin{matrix} 2 x - 1 & = 0 \\ 2 x & = 1 \\ x & = \frac{1}{2} \\ P (x) & = 4 x^{3} - 8 x^{2} + 11 x - 4 \\ P (\frac{1}{2}) & = 4 {(\frac{1}{2})}^{3} - 8 {(\frac{1}{2})}^{2} + 11 (\frac{1}{2}) - 4 \\ P (\frac{1}{2}) & = 4 (\frac{1}{}) - (\frac{1}{4}) + \frac{11}{2} - 4 \\ P (\frac{1}{2}) & = \frac{1}{2} - 2 + \frac{11}{2} - 4 \\ P (\frac{1}{2}) & = \frac{1 - 4 + 11 - 8}{2} \\ P (\frac{1}{2}) & = 0 \Rightarrow e x a c t a \end{matrix}$
$6 x^{5} + 2 x^{4} - 3 x^{3} - x^{2} + 3 x + 3$ entre $3 x + 1$
$\begin{matrix} 3 x + 1 & = 0 \\ 3 x & = - 1 \\ x & = - \frac{1}{3} \\ P (x) & = 6 x^{5} + 2 x^{4} - 3 x^{3} - x^{2} + 3 x + 3 \\ P (- \frac{1}{3}) & = 6 {(- \frac{1}{3})}^{5} + 2 {(- \frac{1}{3})}^{4} - 3 {(- \frac{1}{3})}^{3} - {(- \frac{1}{3})}^{2} + 3 (- \frac{1}{3}) + 3 \\ P (- \frac{1}{3}) & = (- \frac{1}{}) + 2 (\frac{1}{81}) - 3 (- \frac{1}{}) - \frac{1}{9} - 1 + 3 \\ P (- \frac{1}{3}) & = - \frac{2}{81} + \frac{2}{81} + \frac{1}{9} - \frac{1}{9} + 2 \\ P (- \frac{1}{3}) & = 2 \Rightarrow n o e s e x a c t a \end{matrix}$ Sin efectuar la división, probar que:
$a + 1$ es factor de $a^{3} - 2 a^{2} + 2 a + 5$
$\begin{matrix} a + 1 & = 0 \\ a & = - 1 \\ P (a) & = a^{3} - 2 a^{2} + 2 a + 5 \\ P (- 1) & = {(- 1)}^{3} - 2 {(- 1)}^{2} + 2 (- 1) + 5 \\ P (- 1) & = - 1 - 2 - 2 + 5 \\ P (- 1) & = 0 \\ a^{3} - 2 a^{2} + 2 a + 5 & e s d i v i s i b l e p a r a a + 1 \\ \Rightarrow & a + 1 e s f a c t o r d e a^{3} - 2 a^{2} + 2 a + 5 \end{matrix}$
$x - 5$ divide a $x^{5} - 6 x^{4} + 6 x^{3} - 5 x^{2} + 2 x - 10$
$\begin{matrix} x - 5 & = 0 \\ x & = 5 \\ P (x) & = x^{5} - 6 x^{4} + 6 x^{3} - 5 x^{2} + 2 x - 10 \\ P (5) & = {(5)}^{5} - 6 {(5)}^{4} + 6 {(5)}^{3} - 5 {(5)}^{2} + 2 (5) - 10 \\ P (5) & = 3125 - 6 (625) + 6 (125) - 5 (25) + 10 - 10 \\ P (5) & = 3125 - 3750 + 750 - 125 \\ P (5) & = 0 \Rightarrow x - 5 d i v i d e a x^{5} - 6 x^{4} + 6 x^{3} - 5 x^{2} + 2 x - 10 \end{matrix}$
$4 x - 3$ divide a $4 x^{4} - 7 x^{3} + 7 x^{2} - 7 x + 3$
$\begin{matrix} 4 x - 3 & = 0 \\ 4 x & = 3 \\ x & = \frac{3}{4} \\ P (x) & = 4 x^{4} - 7 x^{3} + 7 x^{2} - 7 x + 3 \\ P (\frac{3}{4}) & = 4 {(\frac{3}{4})}^{4} - 7 {(\frac{3}{4})}^{3} + 7 {(\frac{3}{4})}^{2} - 7 (\frac{3}{4}) + 3 \\ P (\frac{3}{4}) & = 4 (\frac{3^{4}}{4^{}}) - 7 (\frac{27}{64}) + 7 (\frac{9}{16}) - \frac{21}{4} + 3 \\ P (\frac{3}{4}) & = \frac{81}{64} - \frac{189}{64} + \frac{63}{16} - \frac{21}{4} + 3 \\ P (\frac{3}{4}) & = \frac{81 - 189 + 252 - 336 + 192}{64} \\ P (\frac{3}{4}) & = 0 \Rightarrow 4 x - 3 d i v i d e a 4 x^{4} - 7 x^{3} + 7 x^{2} - 7 x + 3 \end{matrix}$
$3 n + 2$ no es factor de $3 n^{5} + 2 n^{4} - 3 n^{3} - 2 n^{2} + 6 n + 7$
$\begin{matrix} 3 n + 2 & = 0 \\ 3 n & = - 2 \\ n & = - \frac{2}{3} \\ P (n) & = 3 n^{5} + 2 n^{4} - 3 n^{3} - 2 n^{2} + 6 n + 7 \\ P (- \frac{2}{3}) & = 3 {(- \frac{2}{3})}^{5} + 2 {(- \frac{2}{3})}^{4} - 3 {(- \frac{2}{3})}^{3} - 2 {(- \frac{2}{3})}^{2} + 6 (- \frac{2}{3}) + 7 \\ P (- \frac{2}{3}) & = 3 (- \frac{2}{3^{}}) + 2 (\frac{16}{81}) - 3 (- \frac{8}{3^{}}) - 2 (\frac{4}{9}) + (- \frac{2}{3}) + 7 \\ P (- \frac{2}{3}) & = - \frac{2}{81} + \frac{32}{81} + \frac{8}{9} - \frac{8}{9} - 4 + 7 \\ P (- \frac{2}{3}) & = \frac{- 2 + 32 + 243}{81} \\ P (- \frac{2}{3}) & = \frac{}{} \\ P (- \frac{2}{3}) & = \frac{91}{27} \neq 0 \Rightarrow 3 n + 2 n o e s f a c t o r d e 3 n^{5} + 2 n^{4} - 3 n^{3} - 2 n^{2} + 6 n + 7 \end{matrix}$ Sin efectuar la división, hallar si las divisiones siguientes son o no exactas y determinar el cociente en cada caso y el residuo, si lo hay:
$2 a^{3} - 2 a^{2} - 4 a + 16$ entre $a + 2$
$a^{4} - a^{2} + 2 a + 2$ entre $a + 1$
$x^{4} + 5 x - 6$ entre $x - 1$
$x^{6} - 39 x^{4} + 26 x^{3} - 52 x^{2} + 29 x - 30$ entre $x - 6$
$a^{6} - 4 a^{5} - a^{4} + 4 a^{3} + a^{2} - 8 a + 25$ entre $a - 4$
$16 x^{4} - 24 x^{3} + 37 x^{2} - 24 x + 4$ entre $4 x - 1$
$15 n^{5} + 25 n^{4} - 18 n^{3} - 18 n^{2} + 17 n - 11$ entre $3 n + 5$

En los ejemplos siguientes, hallar el valor de la constante K (término independiente del polinomio) para que:
$7 x^{2} - 5 x + K$ sea divisible por $x - 5$
$\begin{matrix} x - 5 & = 0 \\ x & = 5 \\ P (x) & = 7 x^{2} - 5 x + K \\ P (5) & = 0, p a r a q u e l a s e a u n a d i v i s i ó n e x a c t a \\ 7 x^{2} - 5 x + K & = 0 \\ 7 {(5)}^{2} - 5 (5) + K & = 0 \\ 7 (25) - 25 + K & = 0 \\ 175 - 25 + K & = 0 \\ 150 + K & = 0 \\ K & = - 150 \end{matrix}$
$x^{3} - 3 x^{2} + 4 x + K$ sea divisible por $x - 2$
$\begin{matrix} x - 2 & = 0 \\ x & = 2 \\ P (x) & = x^{3} - 3 x^{2} + 4 x + K \\ P (2) & = 0 \\ {(2)}^{3} - 3 {(2)}^{2} + 4 (2) + K & = 0 \\ 8 - 12 + 8 + K & = 0 \\ 4 + K & = 0 \\ K & = - 4 \end{matrix}$
$2 a^{4} + 25 a + K$ sea divisible por $a + 3$
$\begin{matrix} a + 3 & = 0 \\ a & = - 3 \\ P (a) & = 2 a^{4} + 25 a + K \\ P (- 3) & = 0 \\ 2 {(- 3)}^{4} + 25 (- 3) + K & = 0 \\ 2 (81) - 75 + K & = 0 \\ 162 - 75 + K & = 0 \\ 87 + K & = 0 \\ K & = - 87 \end{matrix}$
$20 x^{3} - 7 x^{2} + 29 x + K$ sea divisible por $4 x + 1$
$\begin{matrix} 4 x + 1 & = 0 \\ 4 x & = - 1 \\ x & = - \frac{1}{4} \\ P (x) & = 20 x^{3} - 7 x^{2} + 29 x + K \\ P (- \frac{1}{4}) & = 0 \\ 20 {(- \frac{1}{4})}^{3} - 7 {(- \frac{1}{4})}^{2} + 29 (- \frac{1}{4}) + K & = 0 \\ (- \frac{1}{}) - 7 (\frac{1}{16}) - \frac{29}{4} + K & = 0 \\ - \frac{5}{16} - \frac{7}{16} - \frac{29}{4} + K & = 0 \\ \frac{- 5 - 7 - 116}{16} + K & = 0 \\ - \frac{}{16} + K & = 0 \\ - 8 + K & = 0 \\ K & = 8 \end{matrix}$