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
x–3 =0 x =3 P( x ) = x 2 –x–6 P( 3 ) = ( 3 ) 2 –3–6 P( 3 ) = 9 – 9 P( 3 ) =0 ⇒ exacta -
x
3
+4
x
2
–x–10
entre
x+2
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 ⇒ exacta -
2
x
4
–5
x
3
+7
x
2
–9x+3
entre
x–1
x–1 =0 x =1 P( x ) =2 x 4 –5 x 3 +7 x 2 –9x+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 ⇒ noesexacta -
x
5
+
x
4
–5
x
3
–7x+8
entre
x+3
x+3 =0 x =–3 P( x ) = x 5 + x 4 –5 x 3 –7x+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 ⇒ noesexacta -
4
x
3
–8
x
2
+11x–4
entre
2x–1
2x–1 =0 2x =1 x = 1 2 P( x ) =4 x 3 –8 x 2 +11x–4 P( 1 2 ) =4 ( 1 2 ) 3 –8 ( 1 2 ) 2 +11( 1 2 ) –4 P( 1 2 ) = 4 ( 1 ) –( 1 4 ) + 11 2 –4 P( 1 2 ) = 1 2 –2+ 11 2 –4 P( 1 2 ) = 1–4+11–8 2 P( 1 2 ) =0 ⇒ exacta -
6
x
5
+2
x
4
–3
x
3
–
x
2
+3x+3
entre
3x+1
3x+1 =0 3x =–1 x =– 1 3 P( x ) =6 x 5 +2 x 4 –3 x 3 – x 2 +3x+3 P( – 1 3 ) =6 ( – 1 3 ) 5 +2 ( – 1 3 ) 4 –3 ( – 1 3 ) 3 – ( – 1 3 ) 2 + 3 ( – 1 3 ) +3 P( – 1 3 ) =( – 1 ) +2( 1 81 ) – 3 ( – 1 ) – 1 9 –1+3 P( – 1 3 ) =– 2 81 + 2 81 + 1 9 – 1 9 +2 P( – 1 3 ) =2 ⇒ noesexacta Sin efectuar la división, probar que: -
a+1
es factor de
a
3
–2
a
2
+2a+5
a+1 =0 a =–1 P( a ) = a 3 –2 a 2 +2a+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 +2a+5 es divisible para a+1 ⇒ a+1 es factor de a 3 –2 a 2 +2a+5 -
x–5
divide a
x
5
–6
x
4
+6
x
3
–5
x
2
+2x–10
x–5 =0 x =5 P( x ) = x 5 –6 x 4 +6 x 3 –5 x 2 +2x–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 ⇒ x–5 divide a x 5 –6 x 4 +6 x 3 –5 x 2 +2x–10 -
4x–3
divide a
4
x
4
–7
x
3
+7
x
2
–7x+3
4x–3 =0 4x =3 x = 3 4 P( x ) =4 x 4 –7 x 3 +7 x 2 –7x+3 P( 3 4 ) =4 ( 3 4 ) 4 –7 ( 3 4 ) 3 +7 ( 3 4 ) 2 –7( 3 4 ) +3 P( 3 4 ) = 4 ( 3 4 4 ) –7( 27 64 ) +7( 9 16 ) – 21 4 +3 P( 3 4 ) = 81 64 – 189 64 + 63 16 – 21 4 +3 P( 3 4 ) = 81–189+252–336+192 64 P( 3 4 ) =0 ⇒ 4x–3 divide a 4 x 4 –7 x 3 +7 x 2 –7x+3 -
3n+2
no es factor de
3
n
5
+2
n
4
–3
n
3
–2
n
2
+6n+7
3n+2 =0 3n =–2 n =– 2 3 P( n ) =3 n 5 +2 n 4 –3 n 3 –2 n 2 +6n+7 P( – 2 3 ) =3 ( – 2 3 ) 5 +2 ( – 2 3 ) 4 –3 ( – 2 3 ) 3 –2 ( – 2 3 ) 2 +6( – 2 3 ) +7 P( – 2 3 ) = 3 ( – 2 3 ) +2( 16 81 ) – 3 ( – 8 3 ) –2( 4 9 ) +( – 2 3 ) +7 P( – 2 3 ) =– 2 81 + 32 81 + 8 9 – 8 9 –4+7 P( – 2 3 ) = –2+32+243 81 P( – 2 3 ) = P( – 2 3 ) = 91 27 ≠ 0 ⇒ 3n+2 no es factor de 3 n 5 +2 n 4 –3 n 3 –2 n 2 +6n+7 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
–4a+16
entre
a+2
-
a
4
–
a
2
+2a+2
entre
a+1
-
x
4
+5x–6
entre
x–1
-
x
6
–39
x
4
+26
x
3
–52
x
2
+29x–30
entre
x–6
-
a
6
–4
a
5
–
a
4
+4
a
3
+
a
2
–8a+25
entre
a–4
-
16
x
4
–24
x
3
+37
x
2
–24x+4
entre
4x–1
-
15
n
5
+25
n
4
–18
n
3
–18
n
2
+17n–11
entre
3n+5
En los ejemplos siguientes, hallar el valor de la constante K (término independiente del polinomio) para que: -
7
x
2
–5x+K
sea divisible por
x–5
x–5 =0 x =5 P( x ) =7 x 2 –5x+K P( 5 ) =0, para que la sea una división exacta 7 x 2 –5x+K =0 7 ( 5 ) 2 –5( 5 ) +K =0 7( 25 ) –25+K =0 175–25+K =0 150+K =0 K =–150 -
x
3
–3
x
2
+4x+K
sea divisible por
x–2
x–2 =0 x =2 P( x ) = x 3 –3 x 2 +4x+K P( 2 ) =0 ( 2 ) 3 –3 ( 2 ) 2 +4( 2 ) +K =0 8–12+8+K =0 4+K =0 K =–4 -
2
a
4
+25a+K
sea divisible por
a+3
a+3 =0 a =–3 P( a ) =2 a 4 +25a+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 -
20
x
3
–7
x
2
+29x+K
sea divisible por
4x+1
4x+1 =0 4x =–1 x =– 1 4 P( x ) =20 x 3 –7 x 2 +29x+K P( – 1 4 ) =0 20 ( – 1 4 ) 3 –7 ( – 1 4 ) 2 +29( – 1 4 ) +K =0 ( – 1 ) –7( 1 16 ) – 29 4 +K =0 – 5 16 – 7 16 – 29 4 +K =0 –5–7–116 16 +K =0 – 16 +K =0 –8+K =0 K =8
