CAPITULO VII
Teorema del Residuo
- Ejercicio 74
Hallar sin efectuar la división, el residuo de dividir:
-
x
2
–2x+3
entre
x–1
x–1 =0 x =1 x 2 –2x+3 = ( 1 ) 2 –2( 1 ) +3 =1–2+3 =2 -
x
3
–3
x
2
+2x–2
entre
x+1
x+1 =0 x =–1 x 3 –3 x 2 +2x–2 = ( –1 ) 3 –3 ( –1 ) 2 +2( –1 ) –2 =–1–3–2–2 =–8 -
x
4
–
x
3
+5
entre
x–2
x–2 =0 x =2 x 4 – x 3 +5 = ( 2 ) 4 – ( 2 ) 3 +5 =16–8+5 =13 -
a
4
–5
a
3
+2
a
2
–6
entre
a+3
a+3 =0 a =–3 a 4 –5 a 3 +2 a 2 –6 = ( –3 ) 4 –5 ( –3 ) 3 +2 ( –3 ) 2 –6 =81–5( –27 ) +2( 9 ) –6 =81+135+18–6 =228 -
m
4
+
m
3
–
m
2
+5
entre
m–4
m–4 =0 m =4 m 4 + m 3 – m 2 +5 = ( 4 ) 4 + ( 4 ) 3 – ( 4 ) 2 +5 =256+64–16+5 =309 -
x
5
+3
x
4
–2
x
3
+4
x
2
–2x+2
entre
x+3
x+3 =0 x =–3 x 5 +3 x 4 –2 x 3 +4 x 2 –2x+2 = ( –3 ) 5 +3 ( –3 ) 4 –2 ( –3 ) 3 +4 ( –3 ) 2 –2( –3 ) +2 =– 243 + 243 –2( –27 ) +4( 9 ) +6+2 =54+36+8 =98 -
a
5
–2
a
3
+2a–4
entre
a–5
a–5 =0 a =5 a 5 –2 a 3 +2a–4 = ( 5 ) 5 –2 ( 5 ) 3 +2( 5 ) –4 =3125–2( 125 ) +10–4 =3131–250 =2881 -
6
x
3
+
x
2
+3x+5
entre
2x+1
2x+1 =0 2x =–1 x =– 1 2 6 x 3 + x 2 +3x+5 =6 ( – 1 2 ) 3 + ( – 1 2 ) 2 +3( – 1 2 ) +5 =( – 1 ) + 1 4 – 3 2 +5 =– 3 4 + 1 4 – 3 2 +5 = –3+1–6+20 4 = 4 =3 -
12
x
3
–21x+90
entre
3x–3
3x–3 =0 3x =3 x = 3 3 x =1 12 x 3 –21x+90 =12 ( 1 ) 3 –21( 1 ) +90 =12–21+90 =81 -
15
x
3
–11
x
2
+10x+18
entre
3x+2
3x+2 =0 3x =–2 x =– 2 3 15 x 3 –11 x 2 +10x+18 =15 ( – 2 3 ) 3 –11 ( – 2 3 ) 2 +10( – 2 3 ) +18 =( – 8 ) –11( 4 9 ) – 20 3 +18 =– 40 9 – 44 9 – 20 3 +18 = –40–44–60+162 9 = 9 =2 -
5
x
4
–12
x
3
+9
x
2
–22x+21
entre
5x–2
5x–2 =0 5x =2 x = 2 5 5 x 4 –12 x 3 +9 x 2 –22x+21 =5 ( 2 5 ) 4 –12 ( 2 5 ) 3 +9 ( 2 5 ) 2 –22( 2 5 ) +21 = 5 ( 16 5 ) –12( 8 125 ) +9( 4 25 ) – 44 5 +21 = 16 125 – 96 125 + 36 25 – 44 5 +21 = 16–96+180–1100+2625 125 = 125 =13 -
a
6
+
a
4
–8
a
2
+4a+1
entre
2a+3
2a+3 =0 2a =–3 a =– 3 2 a 6 + a 4 –8 a 2 +4a+1 = ( – 3 2 ) 6 + ( – 3 2 ) 4 –8 ( – 3 2 ) 2 +( – 3 2 ) +1 = 729 64 + 81 16 –( 9 4 ) –6+1 = 729 64 + 81 16 –18–5 = 729 64 + 81 16 –23 = 729+324–1472 64 =– 419 64