Problemas resueltos sobre Polinomios

DIEGO CORTEZ 1
1. Si 𝑃(𝑥) = 2 𝑥
− 2 𝑥−1
Calcule: 𝑃(1) + 𝑃(2) + 𝑃(3)
a) 7 b) 2 c) 3 d) 4 e) 5
Resolución:
𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 𝑙𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 1,2 𝑦 3 𝑒𝑛 𝑙𝑎 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑥:
𝑃(1) = 21
− 21−1
= 2 − 20
= 2 − 1 = 1
𝑃(2) = 22
− 22−1
= 22
− 21
= 4 − 2 = 2
𝑃(3) = 23
− 23−1
= 23
− 22
= 8 − 4 = 4
𝑁𝑜𝑠 𝑝𝑖𝑑𝑒𝑛:
𝑃(1) + 𝑃(2) + 𝑃(3) = 1 + 2 + 4 = 7
2. Sea 𝑃(𝑥) = 4𝑥 + 1
Halle: 𝐸 =
𝑃(1)+𝑃(2)
𝑃(3)+𝑃(0)
a) 1 b) 2 c) 3 d) 4 e) 5
Resolución:
𝑃(0) = 4(0) + 1 = 1
𝑃(1) = 4(1) + 1 = 5
𝑃(2) = 4(2) + 1 = 9
𝑃(3) = 4(3) + 1 = 13
𝐸 =
𝑃(1) + 𝑃(2)
𝑃(3) + 𝑃(0)
=
5 + 9
13 + 1
=
14
14
= 1

DIEGO CORTEZ 2
3. Siendo: 𝑃(𝑥) = 𝑥2
− 4𝑥 − 6, halle:
𝐸 = 𝑃(−2) + 𝑃(2) − 𝑃(1)
a) -5 b) 2 c) -4 d) -2 e) 5
Resolución:
𝑃(−2) = (−2)2
− 4(−2) − 6 = 4 + 8 − 6 = 6
𝑃(2) = 22
− 4(2) − 6 = 4 − 8 − 6 = −10
𝑃(1) = 12
− 4(1) − 6 = 1 − 4 − 6 = −9
𝐸 = 𝑃(−2) + 𝑃(2) − 𝑃(1) = 6 − 10 − (−9) = 6 − 10 + 9 = 5
4. Si 𝐹(𝑥) = 3𝑥2
− 2; calcular:
𝐸 = 𝐹(2) 𝐹(0) 𝐹(−1)
a) 100 b) 10 c) 1 d) 0.1 e) 0.01
Resolución:
𝐹(−1) = 3(−1)2
− 2 = 3 − 2 = 1
𝐹(0) = 3(0)2
− 2 = 0 − 2 = −2
𝐹(2) = 3(2)2
− 2 = 12 − 2 = 10
𝐸 = 𝐹(2) 𝐹(0) 𝐹(−1)
= 10−21
= 10−2
=
1
100
= 0.01

DIEGO CORTEZ 3
5. Siendo 𝑃(𝑥) = 𝑥2
+ 2𝑥, hallar:
𝑅 =
𝑃(0) 𝑃(1)
+ 𝑃(1) 𝑃(−1)
𝑃(2) 𝑃(0)
a) 3 b) 1/3 c) 2 d) 1/4 e) 5
Resolución:
𝑃(−1) = (−1)2
+ 2(−1) = 1 − 2 = −1
𝑃(0) = (0)2
+ 2(0) = 0
𝑃(1) = (1)2
+ 2(1) = 1 + 2 = 3
𝑃(2) = (2)2
+ 2(2) = 4 + 4 = 8
𝑅 =
𝑃(0) 𝑃(1)
+ 𝑃(1) 𝑃(−1)
𝑃(2) 𝑃(0)
=
03
+ 3−1
80
=
0 +
1
3
1
=
1
3
6. Si: 𝐹(𝑥) =
𝑥+1
2𝑥−1
Calcular el valor de:
𝑀 = (
𝐹(3) − 𝐹(1)
𝐹(2)
)
−1
a) 5/6 b) 5 c) 4 d) 6 e) -5/6
Resolución:
𝐹(1) =
1 + 1
2(1) − 1
=
2
1
= 2
𝐹(2) =
2 + 1
2(2) − 1
=
3
3
= 1
𝐹(3) =
3 + 1
2(3) − 1
=
4
5

DIEGO CORTEZ 4
𝑀 = (
𝐹(3) − 𝐹(1)
𝐹(2)
)
−1
= (
4
5
− 2
1
)
−1
= (
4 − 10
5
1
)
−1
= (
−6
5
1
)
−1
= (
−6
5
)
−1
=
−5
6
7. Si 𝑃(𝑥 + 1) = 3𝑥 − 2
Calcular 𝑃(2)
a) -5 b) -2 c) -4 d) 1 e) 5
Resolución:
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑣𝑎𝑙𝑜𝑟 𝑡𝑜𝑚𝑎𝑟í𝑎 𝑥 𝑝𝑎𝑟𝑎 𝑃(2):
𝑃(𝑥 + 1) = 𝑃(2)
𝑥 + 1 = 2
𝑥 = 1
𝑃𝑎𝑟𝑎 𝑃(2), 𝑥 = 1 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 𝑑𝑖𝑐ℎ𝑜 𝑣𝑎𝑙𝑜𝑟 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒𝑙 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎:
𝑃(2) = 3(1) − 2 = 3 − 2 = 1
8. Si 𝑃(𝑥 + 1) = 𝑥2
Halle: 𝑃(𝑃(𝑃(3)))
a) 3 b) 64 c) 49 d) 128 e) 25
Resolución:
𝐷𝑒𝑠𝑎𝑟𝑟𝑜𝑙𝑙𝑎𝑚𝑜𝑠 𝑑𝑒 𝑎𝑑𝑒𝑛𝑡𝑟𝑜 ℎ𝑎𝑐𝑖𝑎 𝑎𝑓𝑢𝑒𝑟𝑎. 𝐸𝑛 𝑝𝑟𝑖𝑚𝑒𝑟 𝑙𝑢𝑔𝑎𝑟 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑣𝑎𝑙𝑜𝑟
𝑡𝑜𝑚𝑎𝑟í𝑎 𝑥 𝑝𝑎𝑟𝑎 𝑃(3):
𝑃(𝑥 + 1) = 𝑃(3)
𝑥 + 1 = 3
𝑥 = 2

DIEGO CORTEZ 5
𝑃(3) = 22
= 4
𝐷𝑒𝑠𝑝𝑢é𝑠 𝑑𝑒 𝑒𝑙𝑙𝑜, 𝑙𝑜 𝑞𝑢𝑒 𝑛𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 ℎ𝑎𝑙𝑙𝑎𝑟 𝑞𝑢𝑒𝑑𝑎𝑟í𝑎:
𝑃(𝑃(4))
𝐴ℎ𝑜𝑟𝑎 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑣𝑎𝑙𝑜𝑟 𝑡𝑜𝑚𝑎𝑟í𝑎 𝑥 𝑝𝑎𝑟𝑎 𝑃(4):
𝑃(𝑥 + 1) = 𝑃(4)
𝑥 + 1 = 4
𝑥 = 3
𝑃(3) = 32
= 9
𝐿𝑢𝑒𝑔𝑜, 𝑙𝑜 𝑞𝑢𝑒 𝑛𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 ℎ𝑎𝑙𝑙𝑎𝑟 𝑞𝑢𝑒𝑑𝑎𝑟í𝑎:
𝑃(9)
𝑃(𝑥 + 1) = 𝑃(9)
𝑥 + 1 = 9
𝑥 = 8
𝑃𝑎𝑟𝑎 𝑃(9), 𝑥 = 8 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 𝑑𝑖𝑐ℎ𝑜 𝑣𝑎𝑙𝑜𝑟 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒𝑙 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎 𝑦 𝑐𝑜𝑛
𝑒𝑠𝑡𝑜 ℎ𝑎𝑙𝑙𝑎𝑟í𝑎𝑚𝑜𝑠 𝑒𝑙 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜:
𝑃(8) = 82
= 64
9. Sea 𝐹(3𝑥 − 1) = 2𝑥 + 3
𝑃(𝑥) = 4𝑥 − 1
Halle: 𝑃(𝐹(2))
a) 19 b) 20 c) 2 d) 12 e) 11

DIEGO CORTEZ 6
Resolución:
𝐷𝑒𝑠𝑎𝑟𝑟𝑜𝑙𝑙𝑎𝑚𝑜𝑠 𝑑𝑒 𝑎𝑑𝑒𝑛𝑡𝑟𝑜 ℎ𝑎𝑐𝑖𝑎 𝑎𝑓𝑢𝑒𝑟𝑎. 𝐸𝑛 𝑝𝑟𝑖𝑚𝑒𝑟 𝑙𝑢𝑔𝑎𝑟 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑣𝑎𝑙𝑜𝑟
𝑡𝑜𝑚𝑎𝑟í𝑎 𝑥 𝑝𝑎𝑟𝑎 𝐹(2):
𝐹(3𝑥 − 1) = 𝐹(2)
3𝑥 − 1 = 2
𝑥 = 1
𝑃𝑎𝑟𝑎 𝐹(2), 𝑥 = 1 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 𝑑𝑖𝑐ℎ𝑜 𝑣𝑎𝑙𝑜𝑟 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒𝑙 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎:
𝐹(2) = 2(1) + 3 = 2 + 3 = 5
𝐸𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑙𝑜 𝑞𝑢𝑒 𝑛𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 ℎ𝑎𝑙𝑙𝑎𝑟 𝑞𝑢𝑒𝑑𝑎𝑟í𝑎:
𝑃(5)
𝑃(𝑥) = 𝑃(5)
𝑥 = 5
𝑃𝑎𝑟𝑎 𝑃(5), 𝑥 = 5 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 𝑑𝑖𝑐ℎ𝑜 𝑣𝑎𝑙𝑜𝑟 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑒𝑙 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎 𝑦 𝑐𝑜𝑛
𝑒𝑙𝑙𝑜 𝑙𝑙𝑒𝑔𝑎𝑚𝑜𝑠 𝑎𝑙 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜:
𝑃(5) = 4(5) − 1 = 20 − 1 = 19
10. Si: 𝑃(2𝑥 − 1) = 𝑥3
− 𝑥 + 1, halle:
𝑅 = 𝑃(1) + 𝑃(3) − 𝑃(−1)
a) 5 b) 4 c) 6 d) 7 e) 9
Resolución:
𝐸𝑛 𝑝𝑟𝑖𝑚𝑒𝑟 𝑙𝑢𝑔𝑎𝑟 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑚𝑜𝑠 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑥 𝑝𝑎𝑟𝑎 𝑐𝑎𝑑𝑎 𝑣𝑎𝑙𝑜𝑟 𝑞𝑢𝑒 𝑡𝑜𝑚𝑎𝑟í𝑎 𝑃 𝑦 𝑒𝑛 𝑠𝑒𝑔𝑢𝑛𝑑𝑜
𝑙𝑢𝑔𝑎𝑟 𝑙𝑜 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 𝑒𝑛 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑑𝑎𝑑𝑎:
𝑃(2𝑥 − 1) = 𝑃(1)
2𝑥 − 1 = 1
𝑥 = 1

DIEGO CORTEZ 7
𝑃(1) = 13
− 1 + 1 = 1
𝑃(2𝑥 − 1) = 𝑃(3)
2𝑥 − 1 = 3
𝑥 = 2
𝑃(3) = 23
− 2 + 1 = 7
𝑃(2𝑥 − 1) = 𝑃(−1)
2𝑥 − 1 = −1
𝑥 = 0
𝑃(−1) = 03
− 0 + 1 = 1
𝐹𝑖𝑛𝑎𝑙𝑚𝑒𝑛𝑡𝑒 𝑛𝑜𝑠 𝑝𝑖𝑑𝑒𝑛:
𝑅 = 𝑃(1) + 𝑃(3) − 𝑃(−1) = 1 + 7 − (1) = 7
11. Si 𝑃(𝑥) = 3𝑥 + 2
Halle: 𝑃(5𝑥) − 5𝑃(𝑥)
a) -6 b) -8 c) -4 d) -10 e) -20
Resolución:
𝑆𝑖 𝑃(𝑥) = 3𝑥 + 2, 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝑝𝑎𝑟𝑎 𝑙𝑙𝑒𝑔𝑎𝑟 𝑎 𝑃(5𝑥), 𝑠𝑜𝑙𝑜 𝑡𝑒𝑛𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑟 5 𝑎𝑙
𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑞𝑢𝑒 𝑒𝑠𝑡á 𝑎𝑐𝑜𝑚𝑝𝑎ñ𝑎𝑛𝑑𝑜 𝑎 𝑙𝑎 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑥:
𝑃(5𝑥) = 5(3𝑥) + 2 = 15𝑥 + 2
𝑃𝑎𝑟𝑎 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 5𝑃(𝑥), 𝑡𝑒𝑛𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑟 5 𝑎 𝑡𝑜𝑑𝑜 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜:
5[𝑃(𝑥)] = 5(3𝑥 + 2) = 15𝑥 + 10

DIEGO CORTEZ 8
𝑁𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 ℎ𝑎𝑙𝑙𝑎𝑟:
𝑃(5𝑥) − 5𝑃(𝑥) = 15𝑥 + 2 − (15𝑥 + 10) = 2 − 10 = −8
12. Sea la expresión 𝑃(𝑥) = 1 +
1
𝑥
Calcular: 𝑃(1). 𝑃(2). 𝑃(3) … 𝑃(20)
a) 20 b) 21 c) 22 d) 24 e) 25
Resolución:
𝑃(1) = 1 +
1
1
= 1 + 1 = 2
𝑃(2) = 1 +
1
2
=
3
2
𝑃(3) = 1 +
1
3
=
4
3
𝑃(19) = 1 +
1
19
=
20
19
𝑃(20) = 1 +
1
20
=
21
20
𝑃(1). 𝑃(2). 𝑃(3) … 𝑃(20)
2 .
3
2
.
4
3
…
20
19
.
21
20
𝑁𝑜𝑠 𝑝𝑒𝑟𝑐𝑎𝑡𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑒𝑙 𝑛𝑢𝑚𝑒𝑟𝑎𝑑𝑜𝑟 𝑑𝑒 𝑢𝑛 𝑛ú𝑚𝑒𝑟𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑒𝑠 𝑖𝑔𝑢𝑎𝑙 𝑎𝑙 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑑𝑜𝑟 𝑑𝑒𝑙
𝑛ú𝑚𝑒𝑟𝑜 𝑞𝑢𝑒 𝑙𝑒 𝑠𝑖𝑔𝑢𝑒, 𝑝𝑜𝑟 𝑙𝑜 𝑡𝑎𝑛𝑡𝑜 𝑎𝑙 𝑠𝑒𝑟 𝑢𝑛𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑐𝑖ó𝑛, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑟:
2 .
3
2
.
4
3
…
20
19
.
21
20
𝐹𝑖𝑛𝑎𝑙𝑚𝑒𝑛𝑡𝑒 𝑛𝑜𝑠 𝑞𝑢𝑒𝑑𝑎 𝑒𝑙 𝑛ú𝑚𝑒𝑟𝑜 21

DIEGO CORTEZ 9
13. Si el polinomio es homogéneo:
𝑃(𝑥, 𝑦) = 3𝑥 𝑎+2
𝑦 𝑏+8
+ 𝑥 𝑑+3
𝑦7
+ 2𝑥8
𝑦5
Calcular a + b + d
a) 1 b) 5 c) 6 d) 8 e) 13
Resolución:
𝑈𝑛 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑒𝑠 ℎ𝑜𝑚𝑜𝑔é𝑛𝑒𝑜 𝑐𝑢𝑎𝑛𝑑𝑜 𝑠𝑢𝑠 𝑔𝑟𝑎𝑑𝑜𝑠 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜𝑠 𝑑𝑒 𝑐𝑎𝑑𝑎 𝑡é𝑟𝑚𝑖𝑛𝑜 𝑠𝑜𝑛 𝑖𝑔𝑢𝑎𝑙𝑒𝑠:
𝑎 + 2 + 𝑏 + 8 = 𝑑 + 3 + 7 = 8 + 5
𝑎 + 𝑏 + 10 = 𝑑 + 10 = 13
𝐼𝑔𝑢𝑎𝑙𝑎𝑚𝑜𝑠 𝑒𝑙 𝑐𝑜𝑙𝑜𝑟 𝑎𝑧𝑢𝑙:
𝑎 + 𝑏 + 10 = 13
𝑎 + 𝑏 = 3
𝐼𝑔𝑢𝑎𝑙𝑎𝑚𝑜𝑠 𝑒𝑙 𝑐𝑜𝑙𝑜𝑟 𝑛𝑎𝑟𝑎𝑛𝑗𝑎:
𝑑 + 10 = 13
𝑑 = 3
𝑁𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 𝑎 + 𝑏 + 𝑑 = 3 + 3 = 6
14. Dado el polinomio homogéneo:
𝑃(𝑥, 𝑦) = 𝑎𝑥 𝑎+2
𝑦4
+ 2𝑏𝑥 𝑏
𝑦7
− 𝑐𝑥6
𝑦8
+ 2𝑥 𝑐
Calcule la suma de coeficientes:
a) 8 b) 13 c) 12 d) 11 e) 10
Resolución:
𝐼𝑔𝑢𝑎𝑙𝑎𝑚𝑜𝑠 𝑙𝑜𝑠 𝑔𝑟𝑎𝑑𝑜𝑠 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜𝑠 𝑝𝑜𝑟 𝑠𝑒𝑟 𝑢𝑛 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 ℎ𝑜𝑚𝑜𝑔é𝑛𝑒𝑜:
𝑎 + 2 + 4 = 𝑏 + 7 = 6 + 8 = 𝑐

DIEGO CORTEZ 10
𝑎 + 6 = 𝑏 + 7 = 14 = 𝑐
𝑃𝑙𝑎𝑛𝑡𝑒𝑎𝑚𝑜𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑝𝑜𝑟 𝑠𝑒𝑝𝑎𝑟𝑎𝑑𝑜 𝑖𝑔𝑢𝑎𝑙𝑎𝑛𝑑𝑜 𝑐𝑎𝑑𝑎 𝑒𝑐𝑢𝑎𝑐𝑖ó𝑛 𝑎 14 𝑝𝑢𝑒𝑠 𝑒𝑠 𝑒𝑙 ú𝑛𝑖𝑐𝑜
𝑡é𝑟𝑚𝑖𝑛𝑜 𝑞𝑢𝑒 𝑛𝑜 𝑝𝑜𝑠𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑦 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑚𝑜𝑠 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑐𝑎𝑑𝑎 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒:
𝑎 + 6 = 14
𝑎 = 8
𝑏 + 7 = 14
𝑏 = 7
𝑐 = 14
𝐿𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑠𝑜𝑛: 𝑎, 2𝑏, −𝑐 𝑦 2
𝑁𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 𝑙𝑎 𝑠𝑢𝑚𝑎 𝑑𝑒 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠:
𝑎 + 2𝑏 − 𝑐 + 2
8 + 2(7) − 14 + 2
10
𝑃(𝑥, 𝑦) = 2𝑏𝑥 𝑏
𝑦 𝑐
+ 5𝑥7
𝑦2
+ 3𝑐𝑥 𝑏+7
𝑦
Calcule la suma de coeficientes:
a) 30 b) 31 c) 32 d) 33 e) 36
Resolución:
𝑏 + 𝑐 = 7 + 2 = 𝑏 + 7 + 1
𝑏 + 𝑐 = 9 = 𝑏 + 8

DIEGO CORTEZ 11
𝑏 + 8 = 9
𝑏 = 1
𝑏 + 𝑐 = 9
𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑚𝑜𝑠 𝑒𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑏 𝑜𝑏𝑡𝑒𝑛𝑖𝑑𝑜 𝑝𝑟𝑒𝑣𝑖𝑎𝑚𝑒𝑛𝑡𝑒:
1 + 𝑐 = 9
𝑐 = 8
𝐿𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑠𝑜𝑛: 2𝑏, 5 𝑦 3𝑐
𝑁𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 𝑙𝑎 𝑠𝑢𝑚𝑎 𝑑𝑒 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠:
2𝑏 + 5 + 3𝑐
2(1) + 5 + 3(8)
31
16. Si P(x) y Q(x) son idénticos donde:
𝑃(𝑥) = 𝑎𝑥5
+ 3𝑥2
− 4
𝑄(𝑥) = (2𝑎 − 3)𝑥5
+ (𝑐 + 2)𝑥2
+ 𝑏
Calcular a + b + c:
a) 0 b) 1 c) -1 d) 2 e) 8
Resolución:
𝐷𝑜𝑠 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜𝑠 𝑒𝑛 𝑢𝑛𝑎 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑦 𝑑𝑒𝑙 𝑚𝑖𝑠𝑚𝑜 𝑔𝑟𝑎𝑑𝑜 𝑠𝑜𝑛 𝑖𝑑é𝑛𝑡𝑖𝑐𝑜𝑠 𝑠𝑖 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒 𝑠𝑢𝑠
𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑜𝑠 𝑡é𝑟𝑚𝑖𝑛𝑜𝑠 𝑠𝑒𝑚𝑒𝑗𝑎𝑛𝑡𝑒𝑠 𝑠𝑜𝑛 𝑖𝑔𝑢𝑎𝑙𝑒𝑠:
𝑎𝑥5
+ 3𝑥2
− 4 ≡ (2𝑎 − 3)𝑥5
+ (𝑐 + 2)𝑥2
+ 𝑏

DIEGO CORTEZ 12
𝑃𝑙𝑎𝑛𝑡𝑒𝑎𝑚𝑜𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑖𝑔𝑢𝑎𝑙𝑎𝑛𝑑𝑜 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒 𝑃(𝑥) 𝑐𝑜𝑛 𝑙𝑜𝑠 𝑑𝑒 𝑄(𝑥)𝑠𝑒𝑔ú𝑛 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑎:
𝑎 = 2𝑎 − 3
𝑎 = 3
3 = 𝑐 + 2
𝑐 = 1
𝑏 = −4
𝑎 + 𝑏 + 𝑐 = 3 − 4 + 1 = 0
17. Si 𝑅(𝑥) = 2𝑥2
+ 5𝑥 − 3 Es idéntica con 𝑆(𝑥) = (𝑎2
− 2)𝑥2
+ (𝑏2
+ 1)𝑥 + 𝑐
Calcular a + b + c:
a) -1 b) 1 c) 0 d) 2 e) 3
Resolución:
2𝑥2
+ 5𝑥 − 3 ≡ (𝑎2
− 2)𝑥2
+ (𝑏2
+ 1)𝑥 + 𝑐
𝑃𝑙𝑎𝑛𝑡𝑒𝑎𝑚𝑜𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑖𝑔𝑢𝑎𝑙𝑎𝑛𝑑𝑜 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑑𝑒 𝑅(𝑥) 𝑐𝑜𝑛 𝑙𝑜𝑠 𝑑𝑒 𝑆(𝑥)𝑠𝑒𝑔ú𝑛 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑎:
2 = 𝑎2
− 2
𝑎 = 2
5 = 𝑏2
+ 1
𝑏 = 2
𝑐 = −3

DIEGO CORTEZ 13
𝑎 + 𝑏 + 𝑐 = 2 + 2 − 3 = 1
18. Sea el polinomio completo y ordenado descendentemente:
𝑃(𝑥) = 2𝑥 𝑚−2
+ 3𝑥 𝑚−𝑛+1
+ 5𝑥 𝑚−𝑝+7
− 𝑥 𝑝−𝑞−2
Calcular q:
a) 7 b) 8 c) 9 d) 5 e) 13
Resolución:
𝐸𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑒𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑜 𝑦 𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑜 𝑑𝑒 𝑓𝑜𝑟𝑚𝑎 𝑑𝑒𝑠𝑐𝑒𝑛𝑑𝑒𝑛𝑡𝑒:
𝑃(𝑥) ≡ 2𝑥3
+ 3𝑥2
+ 5𝑥1
− 𝑥0
𝐸𝑠𝑡𝑎𝑏𝑙𝑒𝑐𝑒𝑚𝑜𝑠 𝑙𝑜 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒:
2𝑥 𝑚−2
+ 3𝑥 𝑚−𝑛+1
+ 5𝑥 𝑚−𝑝+7
− 𝑥 𝑝−𝑞−2
≡ 2𝑥3
+ 3𝑥2
+ 5𝑥1
− 𝑥0
𝑃𝑙𝑎𝑛𝑡𝑒𝑎𝑚𝑜𝑠 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑖𝑔𝑢𝑎𝑙𝑎𝑛𝑑𝑜 𝑙𝑜𝑠 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒𝑠 𝑠𝑒𝑔ú𝑛 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑎:
𝑚 − 2 = 3
𝑚 = 5
𝑚 − 𝑝 + 7 = 1
𝑚 − 𝑝 = −6
5 − 𝑝 = −6
𝑝 = 11
𝑝 − 𝑞 − 2 = 0
𝑞 = 𝑝 − 2
𝑞 = 11 − 2
Término independiente

DIEGO CORTEZ 14
𝑞 = 9
𝑞 = 9
19. Dado: 𝑃(𝑥) = (4 + 𝑎)𝑥 + 5𝑐 + 𝑑 y 𝑄(𝑥) = 4𝑐 + 3 + (2𝑎 + 2)𝑥
Son idénticos. Calcular: a + c + d
a) 7 b) 8 c) 6 d) 5 e) 4
Resolución:
𝑂𝑟𝑑𝑒𝑛𝑎𝑚𝑜𝑠 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑄(𝑥):
𝑄(𝑥) = 4𝑐 + 3 + (2𝑎 + 2)𝑥
𝑄(𝑥) = (2𝑎 + 2)𝑥 + 4𝑐 + 3
𝑃(𝑥)𝑦 𝑄(𝑥) 𝑠𝑜𝑛 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜𝑠 𝑖𝑑é𝑛𝑡𝑖𝑐𝑜𝑠, 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠:
𝑃(𝑥) ≡ 𝑄(𝑥)
(4 + 𝑎)𝑥 + 5𝑐 + 𝑑 ≡ (2𝑎 + 2)𝑥 + 4𝑐 + 3
4 + 𝑎 = 2𝑎 + 2
𝑎 = 2
5𝑐 + 𝑑 = 4𝑐 + 3
𝑐 + 𝑑 = 3
𝑁𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑎 + 𝑐 + 𝑑:
2 + 3 = 5

DIEGO CORTEZ 15
20. Si los siguientes polinomios son idénticos
𝑃(𝑥) = 𝑚𝑥2
+ 𝑛𝑥 + 𝑝
𝑄(𝑥) = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐
Calcular:
𝐴 =
𝑚 + 𝑛 + 𝑝
𝑎 + 𝑏 + 𝑐
a) 1 b) 2 c) 3 d) 4 e) 5
Resolución:
𝑚𝑥2
+ 𝑛𝑥 + 𝑝 ≡ 𝑎𝑥2
+ 𝑏𝑥 + 𝑐
𝑚 = 𝑎
𝑛 = 𝑏
𝑝 = 𝑐
𝐴 =
𝑚 + 𝑛 + 𝑝
𝑎 + 𝑏 + 𝑐
=
𝑎 + 𝑏 + 𝑐
𝑎 + 𝑏 + 𝑐
= 1
𝐸𝑙 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝐴 𝑒𝑠 1. 𝑃𝑎𝑟𝑎 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝐴 ℎ𝑒𝑚𝑜𝑠 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑑𝑜 𝑙𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑚, 𝑛 𝑦 𝑝, sin 𝑒𝑚𝑏𝑎𝑟𝑔𝑜,
𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑡𝑎𝑚𝑏𝑖é𝑛 𝑟𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑟 𝑙𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑎, 𝑏 𝑦 𝑐 𝑝𝑎𝑟𝑎 𝑙𝑙𝑒𝑔𝑎𝑟 𝑎 𝑙𝑎 𝑟𝑒𝑠𝑝𝑢𝑒𝑠𝑡𝑎, 𝑙𝑎 𝑐𝑢𝑎𝑙
𝑠𝑒𝑔𝑢𝑖𝑟á 𝑠𝑖𝑒𝑛𝑑𝑜 𝑙𝑎 𝑚𝑖𝑠𝑚𝑎:
𝐴 =
𝑚 + 𝑛 + 𝑝
𝑎 + 𝑏 + 𝑐
=
𝑚 + 𝑛 + 𝑝
𝑚 + 𝑛 + 𝑝
= 1
21. Dado el polinomio idénticamente nulo:
𝑃(𝑥) = (𝑎 − 2)𝑥2
+ 𝑏𝑥 + 𝑐 + 3

DIEGO CORTEZ 16
Calcular a.b.c
a) -1 b) 0 c) 1 d) 2 e) 3
Resolución:
𝑈𝑛 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑒𝑠 𝑖𝑑é𝑛𝑡𝑖𝑐𝑎𝑚𝑒𝑛𝑡𝑒 𝑛𝑢𝑙𝑜, 𝑠𝑖 𝑠𝑢𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑛𝑢𝑚é𝑟𝑖𝑐𝑜𝑠 𝑝𝑎𝑟𝑎 𝑐𝑢𝑎𝑙𝑞𝑢𝑖𝑒𝑟 𝑣𝑎𝑙𝑜𝑟
𝑎𝑠𝑖𝑔𝑛𝑎𝑑𝑜 𝑎 𝑠𝑢 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎 𝑠𝑖𝑒𝑚𝑝𝑟𝑒 𝑐𝑒𝑟𝑜.
𝑃(𝑥) ≡ 0
(𝑎 − 2)𝑥2
+ 𝑏𝑥 + 𝑐 + 3 ≡ 0
𝐶𝑎𝑑𝑎 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑑𝑎𝑑𝑜 𝑠𝑒 𝑖𝑔𝑢𝑎𝑙𝑎 𝑎 𝑐𝑒𝑟𝑜:
𝑎 − 2 = 0
𝑎 = 2
𝑏 = 0
𝑐 + 3 = 0
𝑐 = −3
𝑎. 𝑏. 𝑐 = 2(0)(−3) = 0
22. Dado el polinomio idénticamente nulo:
𝑄(𝑥) = 3𝑥2
+ 5𝑥 − 3 + 𝑎𝑥2
+ 𝑏𝑥 − 𝑐
a) -10 b) -11 c) 7 d) -12 e) -13
Hallar a + b + c
Resolución:
𝑂𝑟𝑑𝑒𝑛𝑎𝑚𝑜𝑠 𝑦 𝑎𝑔𝑟𝑢𝑝𝑎𝑚𝑜𝑠 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜:

DIEGO CORTEZ 17
𝑄(𝑥) = 3𝑥2
+ 5𝑥 − 3 + 𝑎𝑥2
+ 𝑏𝑥 − 𝑐
𝑄(𝑥) = (3 + 𝑎)𝑥2
+ (5 + 𝑏)𝑥 − 𝑐 − 3
𝑄(𝑥) ≡ 0
(3 + 𝑎)𝑥2
+ (5 + 𝑏)𝑥 − 𝑐 − 3 ≡ 0
𝐶𝑎𝑑𝑎 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑑𝑎𝑑𝑜 𝑠𝑒 𝑖𝑔𝑢𝑎𝑙𝑎 𝑎 𝑐𝑒𝑟𝑜:
3 + 𝑎 = 0
𝑎 = −3
5 + 𝑏 = 0
𝑏 = −5
−𝑐 − 3 = 0
𝑐 = −3
𝑁𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 ℎ𝑎𝑙𝑙𝑎𝑟:
𝑎 + 𝑏 + 𝑐 = −3 − 5 − 3 = −11
23. El polinomio completo y ordenado:
𝐹(𝑥) = 8𝑥 𝑛−2
+ 9𝑥 𝑛−3
+. . . + 𝑥 𝑚−10
Tiene 20 términos, halle m + n
a) 30 b) 31 c) 32 d) 27 e) 29
Resolución:
𝑃𝑜𝑟 𝑙𝑎 𝑓𝑜𝑟𝑚𝑎 𝑑𝑒 𝑙𝑜𝑠 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑒𝑠 𝑑𝑒 𝐹(𝑥), 𝑒𝑛𝑡𝑒𝑛𝑑𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑒𝑠𝑡á 𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑜 𝑑𝑒
𝑓𝑜𝑟𝑚𝑎 𝑑𝑒𝑠𝑐𝑒𝑛𝑑𝑒𝑛𝑡𝑒:
𝐹(𝑥) = 8𝑥 𝑛−2
+ 9𝑥 𝑛−3
+. . . + 𝑥 𝑚−10
≡ 8𝑥19
+ 9𝑥18
+. . . + 𝑎𝑥1
+ 𝑥0
𝐷𝑜𝑛𝑑𝑒:

DIEGO CORTEZ 18
𝑥19
: 𝑡é𝑟𝑚𝑖𝑛𝑜 1
𝑥18
𝑎𝑥1
𝑥0
𝐼𝑔𝑢𝑎𝑙𝑎𝑚𝑜𝑠:
8𝑥 𝑛−2
+ 9𝑥 𝑛−3
+. . . + 𝑥 𝑚−10
≡ 8𝑥19
+ 9𝑥18
+. . . + 𝑎𝑥1
+ 𝑥0
𝑛 − 2 = 19
𝑛 = 21
𝑚 − 10 = 0
𝑚 = 10
𝑚 + 𝑛 = 10 + 21 = 31
𝑃(𝑥, 𝑦) = 2𝑥 𝑎
𝑦3
+ 3𝑥5
𝑦7
− 𝑥 𝑏
𝑦8
Calcular a + b
a) 13 b) 14 c) 15 d) 16 e) 17
Resolución:
𝑎 + 3 = 5 + 7 = 𝑏 + 8
𝑎 + 3 = 12 = 𝑏 + 8

DIEGO CORTEZ 19
𝑎 + 3 = 12
𝑎 = 9
𝑏 + 8 = 12
𝑏 = 4
𝑎 + 𝑏 = 9 + 4 = 13
25. Calcular a + b + c, si:
𝑎𝑥(𝑥 + 1) + 𝑏(𝑥 + 𝑐) + 𝑥2
≡ 3𝑥2
+ 8𝑥 − 12
a) 8 b) 6 c) 10 d) 11 e) 12
Resolución:
𝑂𝑟𝑑𝑒𝑛𝑎𝑚𝑜𝑠 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑑𝑒𝑙 𝑙𝑎𝑑𝑜 𝑖𝑧𝑞𝑢𝑖𝑒𝑟𝑑𝑜:
𝑎𝑥(𝑥 + 1) + 𝑏(𝑥 + 𝑐) + 𝑥2
𝑎𝑥2
+ 𝑎𝑥 + 𝑏𝑥 + 𝑏𝑐 + 𝑥2
(𝑎 + 1)𝑥2
+ (𝑎 + 𝑏)𝑥 + 𝑏𝑐
𝑇𝑒𝑛𝑖𝑒𝑛𝑑𝑜 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑜, 𝑝𝑟𝑜𝑐𝑒𝑑𝑒𝑚𝑜𝑠 𝑎 𝑖𝑔𝑢𝑎𝑙𝑎𝑟 𝑐𝑜𝑛 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑑𝑒 𝑙𝑎 𝑑𝑒𝑟𝑒𝑐ℎ𝑎
𝑑𝑒𝑙 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎:
(𝑎 + 1)𝑥2
+ (𝑎 + 𝑏)𝑥 + 𝑏𝑐 ≡ 3𝑥2
+ 8𝑥 − 12
𝑎 + 1 = 3
𝑎 = 2

DIEGO CORTEZ 20
𝑎 + 𝑏 = 8
2 + 𝑏 = 8
𝑏 = 6
𝑏𝑐 = −12
6𝑐 = −12
𝑐 = −2
𝑁𝑜𝑠 𝑝𝑖𝑑𝑒𝑛 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑎 + 𝑏 + 𝑐:
2 + 6 − 2 = 6

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