The document discusses algebraic division, including its definition as the operation that finds the quotient and remainder when dividing two polynomials. It covers the uniqueness of the quotient and remainder, exact and inexact division, properties of the division such as the degree of the quotient and maximum degree of the remainder, and criteria and methods for polynomial division.
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Methods of integration, integration of rational algebraic functions, integration of irrational algebraic functions, definite integrals, properties of definite integral, integration by parts, Bernoulli's theorem, reduction formula
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June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
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The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
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Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
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3. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
OBJETIVOS
✓ Reconocer las partes de una
división algebraica y sus
propiedades.
𝑫 𝒙 = 𝒅 𝒙 𝑸 𝒙 + 𝑹 𝒙
✓ Aplicar la regla de Ruffini.
✓ Aplicar el método de Horner.
✓ Aplicar el teorema del resto.
4. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
DIVISIÓN
ALGEBRAICA
Es aquella operación en la cual a partir de dos polinomios en la misma variable conocidos como
D(x): Dividendo y d(x): divisor, encontramos dos únicos polinomios para cada división, llamados
q(x): cociente y R(x): residuo
𝐷 𝑥 𝑑 𝑥
q 𝑥
R 𝑥
Polinomio dividendo Polinomio divisor
Polinomio residuo Polinomio cociente
Algoritmo de la división
𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥 + 𝑅 𝑥
Unicidad del cociente y residuo
𝐷𝑜𝑛𝑑𝑒: ° 𝐷 𝑥 ≥ ° 𝑑 𝑥 > ° 𝑅 𝑥
Suponiendo que existen otros polinomios 𝑞1 𝑥 𝑦 𝑅1 𝑥 donde
° 𝑅1 𝑥 < ° 𝑑 𝑥 , luego
𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥 + 𝑅 𝑥 … 𝐼
𝐷 𝑥 = 𝑑 𝑥 𝑞1 𝑥 + 𝑅1 𝑥 … 𝐼𝐼
Debemos demostrar que: 𝑞 𝑥 = 𝑞1 𝑥 ⋀ 𝑅 𝑥 = 𝑅1 𝑥
Tenemos: 𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥 + 𝑅 𝑥 = 𝑑 𝑥 𝑞1 𝑥 + 𝑅1 𝑥
𝑑 𝑥 𝑞 𝑥 − 𝑞1 𝑥 = 𝑅1 𝑥 − 𝑅 𝑥 ° 𝑑 𝑥 > ° 𝑅1 𝑥
El primer miembro tiene mayor grado que el segundo miembro,
lo cual es contradictorio. Esto implica que ambos son nulos.
𝑞 𝑥 − 𝑞1 𝑥 =0 𝑅1 𝑥 − 𝑅 𝑥 =0
𝑞 𝑥 = 𝑞1 𝑥 𝑅1 𝑥 = 𝑅 𝑥
⋀
⋀
; ∀𝑥
5. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Clases de división
I. División exacta
Toma esta denominación cuando el residuo es
un polinomio idénticamente nulo. 𝑅 𝑥 = 0
Luego, en el algoritmo de la división
𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥 + 𝑅 𝑥 𝑐𝑜𝑚𝑜 𝑅 𝑥 = 0
Nos queda:
𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥
Podemos decir que:
➢ 𝐷 𝑥 𝑒𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑒𝑛𝑡𝑟𝑒 𝑑 𝑥
➢ 𝐷 𝑥 𝑐𝑜𝑛𝑡𝑖𝑒𝑛𝑒 𝑑𝑒 𝑚𝑎𝑛𝑒𝑟𝑎 𝑒𝑥𝑎𝑐𝑡𝑎 𝑎 𝑑 𝑥
➢ d 𝑥 𝑑𝑖𝑣𝑖𝑑𝑒 𝑎 𝐷 𝑥
➢ 𝑑 𝑥 𝑒𝑠 𝑢𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝐷 𝑥
Ejemplos:
1. 𝐿𝑎 𝑑𝑖𝑣𝑖𝑠𝑖ó𝑛 (𝑥3−8) ÷ (𝑥 − 2 ) 𝑒𝑠 𝑒𝑥𝑎𝑐𝑡𝑎, 𝑙𝑢𝑒𝑔𝑜
(𝑥3
−8) = (𝑥 − 2 ) 𝑥2
+ 2𝑥 + 4
Entonces
𝑥 − 2 𝑒𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 (𝑥3 − 8)
2. 𝐿𝑎 𝑑𝑖𝑣𝑖𝑠𝑖ó𝑛 (𝑥4
+ 𝑥2
+1) ÷ (𝑥2
− 𝑥 + 1 ) 𝑒𝑠 𝑒𝑥𝑎𝑐𝑡𝑎, 𝑙𝑢𝑒𝑔𝑜
(𝑥4
+ 𝑥2
+1) = (𝑥2
− 𝑥 + 1 ) 𝑥2
+ 𝑥 + 1
Entonces
𝑥2
− 𝑥 + 1 𝑒𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 (𝑥4
+ 𝑥2
+ 1)
𝐷 𝑥
𝑑 𝑥
𝑑 𝑥
𝐷 𝑥 𝑞 𝑥
𝑞 𝑥
=
=
.
.
; ∀𝑥
6. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
II. División inexacta
Toma esta denominación cuando el residuo no
es un polinomio idénticamente nulo.
Luego, en el algoritmo de la división
𝑅 𝑥 ≠ 0
𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥 + 𝑅 𝑥 ; ∀𝑥;
Podemos decir que:
➢ 𝐷 𝑥 𝑛𝑜 𝑒𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑒𝑛𝑡𝑟𝑒 𝑑 𝑥
➢ 𝐷 𝑥 𝑛𝑜 𝑐𝑜𝑛𝑡𝑖𝑒𝑛𝑒 𝑑𝑒 𝑚𝑎𝑛𝑒𝑟𝑎 𝑒𝑥𝑎𝑐𝑡𝑎 𝑎 𝑑 𝑥
➢ d 𝑥 𝑛𝑜 𝑑𝑖𝑣𝑖𝑑𝑒 𝑎 𝐷 𝑥
➢ 𝑑 𝑥 𝑛𝑜 𝑒𝑠 𝑢𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑒 𝐷 𝑥
Ejemplo:
1. 𝐿𝑎 𝑑𝑖𝑣𝑖𝑠𝑖ó𝑛 (𝑥2−5𝑥 + 20) ÷ (𝑥 − 3 ) 𝑛𝑜 𝑒𝑠 𝑒𝑥𝑎𝑐𝑡𝑎
𝑥2 − 5𝑥 + 20 𝑥 − 3
𝑥 − 2
14
𝐸𝑛𝑡𝑜𝑛𝑐𝑒𝑠
(𝑥2
−5𝑥 + 20) = (𝑥 − 3)(𝑥 − 2) + 14
𝐿𝑢𝑒𝑔𝑜
𝑥2
− 5𝑥 + 20
𝑥 − 3
= 𝑥 − 2 +
14
𝑥 − 3
𝑁𝑜𝑡𝑎: 𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥 + 𝑅 𝑥
𝐷 𝑥
𝑑 𝑥
= 𝑞 𝑥 +
𝑅 𝑥
𝑑 𝑥
7. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
° 𝑞 𝑥 = ° 𝐷 𝑥 − ° 𝑑 𝑥
𝑀𝑎𝑥 ° 𝑅 𝑥 = ° 𝑑 𝑥 −1
Propiedades
Grado del cociente
Máximo grado del residuo
Ejemplo:
Ejemplo:
Como 𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥 + 𝑅 𝑥 , 𝑎𝑑𝑒𝑚á𝑠
° 𝑅 𝑥 < ° 𝑑 𝑥 → ° 𝑑 𝑥 𝑞 𝑥 = ° 𝐷 𝑥
→ ° 𝑑 𝑥 + ° 𝑞 𝑥 = ° 𝐷 𝑥
→ ° 𝑞 𝑥 = ° 𝐷 𝑥 − ° 𝑑 𝑥
En la división de 𝐴𝑥7
+ 𝐵𝑥5
+ 𝐶 ÷ 𝑀𝑥3
+ 𝑁𝑥 + 𝑃
El grado del cociente es ° 𝑞 𝑥 = ° 𝐷 𝑥 − ° 𝑑 𝑥
7 3
° 𝑞 𝑥 = −
° 𝑞 𝑥 = 4
𝐶𝑜𝑚𝑜: ° 𝐷 𝑥 ≥ ° 𝑑 𝑥 > ° 𝑅 𝑥
𝐷𝑒 ° 𝑑 𝑥 > ° 𝑅 𝑥 𝑠𝑜𝑛 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑒𝑛𝑡𝑒𝑟𝑜𝑠
𝑆𝑖 ° 𝑑 𝑥 = 𝑘 𝑎 𝑙𝑜 𝑚á𝑠 ° 𝑅 𝑥 = 𝑘 − 1
→ 𝑀𝑎𝑥 ° 𝑅 𝑥 = ° 𝑑 𝑥 −1
En la división de 𝐴𝑥7 + 𝐵𝑥5 + 𝐶 ÷ 𝑀𝑥3 + 𝑁𝑥 + 𝑃
𝑀𝑎𝑥 ° 𝑅 𝑥 = ° 𝑑 𝑥 −1 = 3 − 1 = 2
𝑅 𝑥 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐
𝑆𝑖 𝑎 ≠ 0 → 𝑅 𝑥 𝑒𝑠 𝑑𝑒 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑔𝑟𝑎𝑑𝑜
𝑆𝑖 𝑎 = 0 ∧ 𝑏 ≠ 0 → 𝑅 𝑥 𝑒𝑠 𝑑𝑒 𝑝𝑟𝑖𝑚𝑒𝑟 𝑔𝑟𝑎𝑑𝑜
𝑆𝑖 𝑎 = 0 ∧ 𝑏 = 0 ∧ 𝑐 ≠ 0 → 𝑅 𝑥 𝑒𝑠 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒
𝑆𝑖 𝑎 = 0 ∧ 𝑏 = 0 ∧ 𝑐 = 0 → 𝑅 𝑥 𝑒𝑠 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑛𝑢𝑙𝑜
8. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Aplicación
Se sabe que la división
𝑥 + 1 6 − 3𝑥4 + 𝑎𝑥 + 𝑏
𝑥2 + 𝑥 + 1
genera un cociente cuya suma de coeficientes es 21
Resolución
𝑆𝑎𝑏𝑒𝑚𝑜𝑠: 𝐷 𝑥 = 𝑑 𝑥 𝑞 𝑥 + 𝑅 𝑥
𝑥 + 1 6
− 3𝑥4
+ 𝑎𝑥 + 𝑏 = 𝑥2
+ 𝑥 + 1 𝑞 𝑥 + 4𝑥 + 5
La suma de coeficientes del cociente es 21 → 𝑞 1 = 21
𝑆𝑖 𝑥 = 1
1 + 1 6 − 3 1 4 + 𝑎(1) + 𝑏 = (1)2+1 + 1 𝑞 1 + 4 1 + 5
64 3 𝑎 3
𝑏 21 9
+ + +
= .
𝑎 + 𝑏 = 11
−
Determine el valor de 𝑎 + 𝑏.
y un residuo 𝑅 𝑥 = 4𝑥 + 5.
61 + 𝑎 + 𝑏 = 63 + 9
9. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Criterios y métodos para la división de polinomios
I. Criterio general
Si queremos efectuar la división de dos polinomios por
cualquier método, el dividendo y el divisor deben estar
completos y ordenados en forma descendente, donde
los exponentes de la variable se reduce de 1 en 1; si
faltan términos en forma práctica se completa con ceros
Ejemplo:
1) 𝑆𝑒𝑎 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝐷 𝑥 = 4𝑥3
− 3𝑥 + 5𝑥4
+ 8𝑥2
− 1
Ordenando en forma descendente, tenemos:
𝐷 𝑥 = 5𝑥4 + 4𝑥3 + 8𝑥2 − 3𝑥 − 1
2) 𝑆𝑒𝑎 𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝐷 𝑥 = 2𝑥5
− 6𝑥 + 9 − 3𝑥2
Ordenando en forma descendente, tenemos:
𝐷 𝑥 = 2𝑥5 − 3𝑥2 − 6𝑥 + 9
Completamos con ceros las potencias que faltan:
𝐷 𝑥 = 2𝑥5
+ 0𝑥4
+ 0𝑥3
− 3𝑥2
− 6𝑥 + 9
3) 𝑆𝑒𝑎𝑛 𝑙𝑜𝑠 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜𝑠 𝐷 𝑥 = 𝑥5 − 1; 𝑑 𝑥 = 𝑥2 − 1
𝐴𝑙 𝑒𝑥𝑝𝑟𝑒𝑠𝑎𝑟
𝐷 𝑥
𝑑 𝑥
𝑜𝑟𝑑𝑒𝑛𝑎𝑛𝑑𝑜 𝑒𝑛 𝑓𝑜𝑟𝑚𝑎 𝑑𝑒𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
𝑦 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑎𝑛𝑑𝑜 𝑐𝑒𝑟𝑜𝑠, 𝑠𝑒 𝑡𝑖𝑒𝑛𝑒
𝐷 𝑥
𝑑 𝑥
=
𝑥5
+ 0𝑥4
+ 0𝑥3
+ 0𝑥2
+ 0𝑥 − 1
𝑥2 + 0𝑥 − 1
10. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Método de William George Horner
Es un método general que permite la división de
polinomios de cualquier grado.
𝑆𝑒𝑎𝑛 𝐷 𝑥 = 𝑎0𝑥4 + 𝑎1𝑥3 + 𝑎2𝑥2 + 𝑎3𝑥 + 𝑎4
d 𝑥 = 𝑏0𝑥2 + 𝑏1𝑥 + 𝑏2
Esquema
𝑎0 𝑎1 𝑎2 𝑎3 𝑎4
𝑏0
𝑏1
𝑏2
𝑞0 𝑞1 𝑞2 𝑟0 𝑟1
* *
* *
* *
÷,×, +
Coeficientes
del cociente
Coeficientes
del residuo
Coeficientes del dividendo
Coef.
Divisor
×
−1
𝑞 𝑥 = 𝑞0𝑥2
+ 𝑞1𝑥 + 𝑞2 𝑅 𝑥 = 𝑟1𝑥 + 𝑟2
−
−
Ejemplo:
𝐸𝑓𝑒𝑐𝑡ú𝑒 𝑙𝑎 𝑠𝑖𝑔𝑢𝑖𝑒𝑛𝑡𝑒 𝑑𝑖𝑣𝑖𝑠𝑖ó𝑛
−23𝑥2 + 4𝑥4 + 16𝑥
5𝑥 + 2𝑥2 − 1
Ordenando y completando, tenemos
4𝑥4
+ 0𝑥3
− 23𝑥2
+ 16𝑥 + 0
2𝑥2 + 5𝑥 − 1
Aplicamos el método de Horner
4 0 −23 16 0
2
−5
1
2 -5 2 1 2
-10 2
25 -5
-10 2
-10
4
÷
÷
÷
𝑞 𝑥 = 2𝑥2 − 5𝑥 + 2 𝑅 𝑥 = 𝑥 + 2
∧
11. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Regla de Paolo Ruffini
Se aplica cuando el divisor es un polinomio lineal
Ejemplo:
𝑆𝑒𝑎𝑛 𝐷 𝑥 = 𝑎0𝑥4
+ 𝑎1𝑥3
+ 𝑎2𝑥2
+ 𝑎3𝑥 + 𝑎4
d 𝑥 = 𝑀𝑥 + 𝑁
Esquema
𝑀𝑥 + 𝑁 = 0
𝑥 = −
𝑁
𝑀
𝑎0 𝑎1 𝑎2 𝑎3 𝑎4
𝑎0 𝑏0 𝑏1 𝑏2 𝑅
𝑀 𝑀 𝑀 𝑀
𝑞0 𝑞1 𝑞2 𝑞3
𝑑 𝑥 = 0 Coeficientes del dividendo
Cociente falso
Coef. del cociente
Resto
𝑞 𝑥 = 𝑞0𝑥3
+ 𝑞1𝑥2
+ 𝑞2𝑥 + 𝑞3 R 𝑥 = 𝑅
∧
𝐸𝑓𝑒𝑐𝑡ú𝑒 𝑙𝑎 𝑑𝑖𝑣𝑖𝑠𝑖ó𝑛
3𝑥5
− 𝑥4
+ 4𝑥3
+ 4𝑥2
+ 9𝑥 + 1
3𝑥 + 2
3𝑥 + 2 = 0
𝑥 = −
2
3
3 −1 4 9 1
4
3
−2 2 −4 0 −6
−3 6 0 9 −5
÷ 3
Cociente falso
Coef. del cociente
3 3 3 3 3
1 −1 2 0 3
Entonces:
𝑞 𝑥 = 𝑥4 − 𝑥3 + 2𝑥2 +3 R 𝑥 = −5
∧
12. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Teorema del resto Calcula el resto de una división sin efectuarla
Teorema
𝐸𝑛 𝑡𝑜𝑑𝑎 𝑑𝑖𝑣𝑖𝑠𝑖ó𝑛
𝑃 𝑥
𝑎𝑥 + 𝑏
; 𝑠𝑢 𝑟𝑒𝑠𝑡𝑜 𝑒𝑠 𝑖𝑔𝑢𝑎𝑙 𝑎𝑙 𝑣𝑎𝑙𝑜𝑟 𝑛𝑢𝑚é𝑟𝑖𝑐𝑜
𝑑𝑒𝑙 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑜 𝑃 𝑥 𝑐𝑢𝑎𝑛𝑑𝑜 𝑥 = −
𝑏
𝑎
; 𝑒𝑠 𝑑𝑒𝑐𝑖𝑟 𝑅 𝑥 = 𝑃 −
𝑏
𝑎
Demostración:
𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒: 𝐷 𝑥 = 𝑑 𝑥 . 𝑞 𝑥 + 𝑅 𝑥
𝑃 𝑥 = (𝑎𝑥 + 𝑏). 𝑞 𝑥 + 𝑘
𝑇𝑒𝑛𝑒𝑚𝑜𝑠 𝑞𝑢𝑒: 𝑑 𝑥 = 𝑎𝑥 + 𝑏, 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 ° 𝑑 𝑥 = 1, 𝑐𝑜𝑚𝑜
𝑀𝑎𝑥 ° 𝑅 𝑥 = ° 𝑑 𝑥 − 1 → 𝑀𝑎𝑥 ° 𝑅 𝑥 = 0
1
s𝑖 𝑥 = −
𝑏
𝑎
0
𝑃 −
𝑏
𝑎
= 𝑞 −
𝑏
𝑎
. + 𝑘 𝑘 = 𝑃 −
𝑏
𝑎
𝐶𝑜𝑚𝑜 𝑅 𝑥 = 𝑘 𝑅 𝑥 = 𝑘 = 𝑃 −
𝑏
𝑎
∴ 𝑅 𝑥 = 𝑃 −
𝑏
𝑎
Ejemplos:
𝑃 𝑥
2𝑥 − 3
1) 𝑅 𝑥 = P
3
2
2) 𝑥5
+ 2𝑥3
+ 4
𝑥 + 1
Calcule el resto en cada caso
𝐼) 𝑥 + 1 = 0 → 𝑥 = −1
𝐼𝐼) 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 𝑒𝑙 𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑𝑜 𝑜𝑏𝑡𝑒𝑛𝑒𝑚𝑜𝑠
𝑒𝑙 𝑟𝑒𝑠𝑡𝑜
𝑅 = −1 5
+ 2 −1 3
+ 4 𝑅 = 1
𝑅 𝑥 = 𝑘
13. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Regla práctica para calcular el resto
I) Se iguala el divisor a cero
II) El resto se calcula reduciendo el dividendo, utilizando
la condición anterior.
NOTA
El grado del residuo es menor que el grado del divisor
𝐻𝑎𝑙𝑙𝑒 𝑒𝑙 𝑟𝑒𝑠𝑡𝑜 𝑑𝑒 𝑑𝑖𝑣𝑖𝑑𝑖𝑟
Ejemplo:
5𝑥12 + 12𝑥10 + 𝑥6 + 𝑥 + 3
𝑥2 + 1
𝐼) 𝑥2
+ 1 = 0 𝑥2
= −1
Aplicando el teorema del resto
𝐷 𝑥 = 5𝑥12 + 12𝑥10 + 𝑥6 + 𝑥 + 3
II) Para reducir el dividendo le damos forma
𝐷 𝑥 = 5 𝑥2 6 + 12 𝑥2 5 + 𝑥2 3 + 𝑥 + 3
𝑅 𝑥 = 5 −1 6
+ 12 −1 5
+ −1 3
+ 𝑥 + 3
𝑅 𝑥 = 5 1 + 12 −1 + −1 + 𝑥 + 3
𝑅 𝑥 = 5 − 12 − 1 + 𝑥 + 3
𝑅 𝑥 = 𝑥 − 5
Luego, el resto es
−𝟏 −𝟏 −𝟏
14. w w w . a c a d e m i a c e s a r v a l l e j o . e d u . p e