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CASOS DE FACTORIZACION (1,2,3,4,5)

Presentado al Docente: William Naranjo Gutiérrez

Universidad Uniminuto Fundamentos de la matemática NRC 43471 Ibagué, Mayo 2019

Introducción

En este documento encontrara ejercicios resueltos de casos de factorización con su debido procedimiento dependiendo de cada caso.

Caso I

𝟐. 𝑏 + 𝑏 2 = 𝑏(1 + 𝑏) 𝟒. 3𝑎2 − 𝑎2 = 𝑎2 (3 − 1) 6. 5m2 + 15m3 = 5m2 (1+ m) 8. x2 y +x2 z = x2 ( y+z ) 10. 8m2 – 12mn = 4m (2m – 3n ) 12. 15e3d2+ 60 c2d3 = 15(e3d2+4c4d3) 14. abc+abc2 = abc (1+c) 16. a3+a2+a = a (a2+a+1) = 3𝑎2 − 𝑎2 = 2𝑎2 𝟏𝟖. 15𝑦 3 + 20𝑦 2 − 5𝑦 = 5𝑦(3𝑦 2 + 4𝑦 − 1) 𝟐𝟎. 2𝑎2 𝑥 + 2𝑎𝑥 2 − 3𝑎𝑥 = 𝑎(2𝑎𝑥 + 2𝑥 2 − 3𝑥)

Caso II 2. a2+ab+ax+bx = (a2+ab)+ (ax+bx) = a(a+b) + x(a+b) =(a+b)(a+x) 4. a2x2-3bx2+a2y2-3by2 = (a2x2-3bx2) + (a2y2-3by2) = x2(a2-3b) + y2 (a2 – 3b) = ( x2+ y2)( a2 – 3b) 𝟔. 𝑥 2 − 𝑎2 + 𝑥 − 𝑎2 𝑥 = 𝑥 2 + 𝑥 − 𝑎2 𝑥 − 𝑎2 = (𝑥 2 + 𝑥)(−𝑎2 𝑥 − 𝑎2 ) = 𝑥(𝑥 + 1) − 𝑎2 (𝑥 + 1) = (𝑥 − 𝑎2 )(𝑥 + 1) 𝟖. 𝑥 + 𝑥 2 − 𝑥𝑦 2 − 𝑦 2 = 𝑥 2 + 𝑥 − 𝑥𝑦 2 − 𝑦 2 = (𝑥 2 + 𝑥)(−𝑥𝑦 2 − 𝑦 2 ) = 𝑥(𝑥 + 1) − 𝑦 2 (𝑥 + 1) = (𝑥 − 𝑦 2 )(𝑥 + 1)

𝟏𝟎. 3𝑎 − 𝑏 2 + 2𝑏 2 𝑥 − 6𝑎𝑥 = −𝑏 2 + 2𝑏 2 𝑥 + 3𝑎 − 6𝑎𝑥 = (−𝑏 2 + 2𝑏 2 𝑥)(+3𝑎 − 6𝑎𝑥) = −𝑏 2 (1 + 2𝑥) − 3𝑎(1 + 2𝑥) = (−𝑏 2 − 3𝑎)(1 − 2𝑥) 𝟏𝟐. 6𝑎𝑥 + 3𝑎 + 1 + 2𝑥 = 6𝑎𝑥 + 3𝑎 +2𝑥 + 1 = 3𝑎(2𝑥 + 1) + 1(2𝑥 + 1) = (3𝑎 + 1)(2𝑥 + 1) 𝟏𝟒. 2𝑎2 𝑥 − 5𝑎2 𝑦 + 15𝑏𝑦 − 6𝑏𝑥 = (2𝑎2 − 6𝑏𝑥)(−5𝑎2 + 15𝑏𝑦) = 2𝑥(𝑎2 − 𝑏 2 ) − 5𝑦(𝑎2 − 𝑏) = (2𝑥 − 5𝑦)(𝑎2 − 𝑏) 𝟏𝟔. 6𝑚 − 9𝑛 + 21𝑛𝑥 + 14𝑚𝑥 = 6𝑚 − 14𝑚𝑥 − 9𝑛 + 21𝑛𝑥 = (6𝑚 − 9𝑛)(−14𝑚𝑥 + 21𝑛𝑥) = 3(2𝑚 − 3𝑛) − 7𝑥(2𝑚 − 3𝑛) = (3 − 7𝑥)(2𝑚 − 3𝑛) 18. 1 + a + 3ab + 3b = (1 + a) + (3ab + 3b) = 1( 1 + a) + 3b( a + 1) = (1+ a) (1 + 3b) = ( a + 1) ( 3b + 1) 20. 20 ax – 5bx -2by + 8 ay = (20ax – 5by) – ( 2by + 8ay) = 5x (4a -b) – 2y (b + 4a) = 5x (4a – b) + 2y (4a – b) =(5x + 2y) ( 4a– b)

Caso III 𝟐. 𝑎2 + 2𝑎𝑏 + 𝑏 2 = (𝑎 + 𝑏)2 =. 𝑎2 + 2𝑎𝑏 + 𝑏 2 4. 𝑦 4 + 1 + 2𝑦 = 𝑦 4 + 2𝑦 + 1

(𝑦 2 + 1) = 𝑦 4 + 2𝑦 + 1 6. 9 - 6x+x2 = ( 3 – x )2 = (3)2 -2(3)(x)+(x)2 = 9 - 6x+x2 8. 1+ 49a2-14a2 = 1 - 14 a2 + 49a2 = (1-7a)2 = (1)2 -2(1)(7a)+(7a)2 = 1 - 14 a2 + 49a2 10. 1 – 2a3 + a6 = ( 1- a3 )2 = (1)2 -2(1)(a3)+(a3)2 = 1 – 2a3 + a6 12. a6 – 2a3 b3+b6 = (a3 – b3)2 = (a3)2 -2(a3)(b3)+(b3)2 = a6 – 2a3 b3+b6 14. 9 b2 -30 a2 b + 25 a4 = (3b – 5a2 )2 = (3b)2 -2(3b) (5a2) + (5a2)2 = 9 b2 -30 a2 b + 25 a4 = (3b – 5a2 ) 16. 1+ a10 – 2a5 = 1– 2a5 + a10 = (1 – a5)2 = (1)2 -2 (1) (a5) + (a5)2 = 1– 2a5 + a10 𝟏𝟖. 100𝑥10 − 60𝑎4 𝑥 5 𝑦 6 + 9𝑎8 𝑦12 (10𝑥 5 + 3𝑎4 𝑦 6 )2 = (10𝑥 5 )2 + 2(10𝑥 5 )(3𝑎4 𝑦 6 ) + (3𝑎4 𝑦 6 )2 = 100𝑥10 − 60𝑎4 𝑥 5 𝑦 6 + 9𝑎8 𝑦 12

𝟐𝟎. 𝑎2 − 24𝑎𝑚2 𝑥 2 + 144𝑚4 𝑥 4 = (𝑎 − 12𝑚2 𝑥 2 )2 = (𝑎)2 − 2(𝑎)(12𝑚2 𝑥 2 ) + (12𝑚2 𝑥 2 )2 = 𝑎2 − 24𝑎𝑚2 𝑥 2 + 144𝑚4 𝑥 4

Caso IV 2. a – 1 = (a +1) (a – 1) 4. 9 – b2 = ( 3 + b ) (3 – b 6. 16 – n2 = 4 − 𝑛 = (4 − 𝑛)2 = (4 + 𝑛)(4 − 𝑛) 𝟖. 1 − 𝑦 2 = 1 − 𝑦 = (1 − 𝑦)2 = (1 + 𝑦)(1 − 𝑦) 𝟏𝟎. 25 − 36𝑥 4 = 5 − 6𝑥 2 = ( 5 + 6𝑥 2 )(5 − 6𝑥 2 ) 12. 4𝑥 2 – 81𝑦 4

= 2𝑥

9𝑦 2 = (2𝑥 − 9𝑦 2 ) = ( 2𝑥 + 9𝑦 2 )(2𝑥 − 9𝑦 2 )

𝟏𝟒. 100 − 𝑥 2 𝑦 6 10

𝑥𝑦 3 = (10 − 𝑥𝑦 3 )2 = (10 + 𝑥𝑦 3 )(10 − 𝑥𝑦 3 )

𝟏𝟔. 25𝑥 2 𝑦 4 − 121 𝑦2

11 = (5𝑥𝑦 2 − 11)2 = (5𝑥𝑦 2 + 11)(5𝑥𝑦 2 − 11)

18. a2 m4 n6 – 144 = (a m4 n3 + 12) (a m4 n3 - 12) 20. 256 a12 – 289 b4 m10 = (16a6 + 17m5 ) ( 16a6 - 17m5 )

Caso V 𝟐. 𝑚4 + 𝑚2 𝑛2 + 𝑛4

𝑚2

𝑛4

= (𝑚2 + 𝑛2 )2 = 𝑚4 + 2𝑚2 𝑛2 + 𝑛4 𝑚 4 + 𝑚 2 𝑛2 + 𝑛4 +𝑚2 𝑛2

− 𝑚 2 𝑛2

_______________________ 𝑚4 + 2𝑚2 𝑛2 + 𝑛4 − 𝑚2 𝑛2 = (𝑚2 + 𝑛2 )2 − 𝑚𝑛 = (𝑚2 𝑛2 + 𝑚𝑛)(𝑚2 𝑛2 − 𝑚𝑛)

𝟒. 𝑎4 + 2𝑎2 + 9

𝑎2

3

= (𝑎2 + 3)2 = 𝑎4 + 6𝑎2 + 9 𝑎4 + 2𝑎2 + 9 +4𝑎2

− 4𝑎2

___________________________ 𝑎4 + 6𝑎2 + 9 − 4𝑎2 = (𝑎2 + 3)2 − 2𝑎 = (𝑎2 + 3 + 2𝑎)(𝑎2 + 3 − 2𝑎) 𝟔. 𝑥 4 − 6𝑥 2 + 1

𝑥2

1

(𝑥 2 − 1)2 = 𝑥 4 − 2𝑥 2 + 1 𝑥 4 − 6𝑥 2 + 1 +4𝑥 2

− 4𝑥 2

_________________________

𝑥 4 − 2𝑥 2 + 1 − 4𝑥 2 = (𝑥 2 − 1)2 − 4𝑥 2 (𝑥 2 − 1 + 2𝑥)(𝑥 2 − 1 + 2𝑥)

𝟖. 4𝑥 4 − 29𝑥 2 + 25 2𝑥 2

5

(2𝑥 2 − 5)2 = 4𝑥 4 − 20𝑥 2 + 25 4𝑥 4 − 29𝑥 2 + 25 +9𝑥 2

− 9𝑥 2

_______________________________ 4𝑥 4 − 20𝑥 2 + 25 − 9𝑥 2 (2𝑥 2 − 5)2 − 3𝑥 (2𝑥 2 − 5 + 3𝑥)(2𝑥 2 − 5 − 3𝑥)

𝟏𝟎. 16𝑚4 − 25𝑚2 𝑛2 + 9𝑛4 4𝑚2

3𝑛2

= (4𝑚2 − 3𝑛2 )2 = 16𝑚4 − 24𝑚2 𝑛2 + 9𝑛4 16𝑚4 − 25𝑚2 𝑛2 + 9𝑛4 +𝑚2 𝑛2

− 𝑚 2 𝑛2

_______________________________________ 16𝑚4 − 24𝑚2 𝑛2 + 9𝑛4 − 𝑚2 𝑛2 (4𝑚2 − 3𝑛2 )2 − 𝑚𝑛 (4𝑚2 − 3𝑛2 + 𝑚𝑛)(4𝑚2 − 3𝑛2 − 𝑚𝑛)

12. 36x4 - 109 x2 y2 +49 y4 = ( 6x2 - 7 y2 )2 = (6x2)2 - 2(6x2)(7y2)+ (7y2)2 = 36x4 - 84 x2 y2 +49 y4 = 36x4 - 109 x2 y2 +49 y4 =

+ 25 x2 y2

-25x2 y2

______________________________________________

(36x4 - 84 x2 y2 +49 y4) -25x2 y2 = (6x2 - 7 y2 )2 -25x2 y2 = (6x2 - 7 y2 + 5x y) (6x2 - 7 y2 - 5x y) = (6x2 + 5x y- 7 y2) (6x2 - 5x y- 7 y2)

14. c4 – 45 c2 +100 = (c2 - 10)2 = (c2)2 - 2(c2)(10)+ (10)2 = c4 – 20c2 + 100 = c4 – 45 c2 +100 =

+ 25 c2

-25c2

______________________________________________

(c4 – 20c2 + 100) -25c2 = (c2 - 10)2 -25c2 = (c2 – 10 +5c) (c2 – 10 -5c) = (5c + c2 - 10) (-5c + c2 - 10)

16. 49 + 76 n2 +64 n4 = ( 7 + 8 n2 )2 = (7)2 + 2(7) (8n2) + (8n2)2 = 49 + 108 n2 +64 n4 = 49 + 76 n2 +64 n4 =

+ 32 n2

-32n2

______________________________________________

(49 + 108 n2 +64 n4) -32n2 = (7 + 8 n2 )2 -32n2 = (7 + 8 n2 +16n) (7 + 8 n2 -16n)

18. 49x8 + 76 x4 y4 +100 y8 = ( 7x4 + 10 y4 )2 = (7x4)2 + 2(7x4)(10y4)+ (10y4)2 = 49x8 + 140 x4 y4 +100 y8

= 49x8 + 76 x4 y4 +100 y8 =

+ 64 x4 y4

-64x4 y4

______________________________________________

(49x8 + 140 x4 y4 +100 y8) -64x4 y4 = (7x4 + 10 y4 )2

-64x4 y4

= (7x4 + 10 y4 + 8x2 y2) (7x4 + 10 y4 - 8x2 y2) = (7x4 + 8x2 y2 + 10 y4) (7x4 - 8x2 y2+ 10 y4)

20. 121x4 - 133 x2 y4 +36 y8 = (11x2 - 6 y4 )2 = (11x2)2 - 2(11x2) (6y4) + (6y4)2 = 121x4 - 132 x2 y4 +36 y8 = 121x4 - 133 x2 y4 +36 y8 =

+ x2 y4

- x2 y4

______________________________________________

(121x4 - 132 x2 y4 +36 y8) - x2 y4 = (11x2 - 6 y4 )2 - x2 y4 = (11x2 - 6 y4+ x y2) (11x2 - 6 y4- x y2) = (11x2 + x y2- 6 y4) (11x2 - x y2- 6 y4)

Conclusiones Después de resolver los ejercicios nos dimos cuenta la importancia de identificar previamente cada caso de factorización, además que si no se identifica podríamos resolverlos de una forma inadecuada.

Bibliografía

Baldor, Aurelio. 1997. Álgebra. Pag-148-150-151-152