Matemáticas, pregunta formulada por Niky89, hace 1 año

factoriza la suma o diferencia de cubos, luego, factoriza la expresion completa

Adjuntos:

Respuestas a la pregunta

Contestado por judith0102

886

Factorizando la suma o diferencia de cubos y luego

factorizando la expresión completa se obtiene :

a ) ( x + 2 ) + ( x³ + 8 ) = ( x + 2 ) + ( x + 2 ) * ( x² - 2x + 4 )

= ( x + 2 ) * ( 1 + x² - 2x + 4 )

= ( x + 2 ) * ( x² - 2x + 5 )

b ) ( x - 4 ) + ( x³ - 64 ) = ( x - 4 ) + ( x - 4 ) * ( x² + 4x + 16 )

= ( x - 4 ) * ( 1 + x² + 4x + 16 )

= ( x - 4 ) * ( x² + 4x + 17)

c ) ( a + 5 ) + ( a³ + 125 ) = ( a + 5 ) + ( a + 5 ) * ( a² - 5a + 25 )

= ( a + 5 ) * ( 1 + a² - 5a + 25 )

= ( a + 5 ) * ( a² - 5a + 26 )

d) ( 2b +1 ) + ( 8b³ +1 ) = ( 2b + 1 ) + (2b +1 ) * ( 4b² - 2b + 1 )

= ( 2b + 1 ) * ( 1 + 4b² - 2b + 1 )

= ( 2b + 1 ) * ( 4b² - 2b + 2 )

e ) ( m + 1 ) + ( m³ + 1 )= ( m + 1 ) * ( m +1 ) * ( m² - m +1 )

= ( m + 1 ) * ( 1 + m² - m + 1 )

= ( m + 1 ) * ( m² - m + 2 )

f ) ( 3x + 2 ) + ( 27x³ + 8 ) = ( 3x +2 ) * ( 3x + 2 ) * ( 9x² - 6x + 4 )

= ( 3x + 2 ) * ( 1 + 9x² -6x +4 )

= ( 3x + 2 ) * ( 9x² - 6x + 5 )

Contestado por brayantom579

100

a ) ( x + 2 ) + ( x³ + 8 ) = ( x + 2 ) + ( x + 2 ) * ( x² - 2x + 4 )

= ( x + 2 ) * ( 1 + x² - 2x + 4 )

= ( x + 2 ) * ( x² - 2x + 5 )

b ) ( x - 4 ) + ( x³ - 64 ) = ( x - 4 ) + ( x - 4 ) * ( x² + 4x + 16 )

= ( x - 4 ) * ( 1 + x² + 4x + 16 )

= ( x - 4 ) * ( x² + 4x + 17)

c ) ( a + 5 ) + ( a³ + 125 ) = ( a + 5 ) + ( a + 5 ) * ( a² - 5a + 25 )

= ( a + 5 ) * ( 1 + a² - 5a + 25 )

= ( a + 5 ) * ( a² - 5a + 26 )

d) ( 2b +1 ) + ( 8b³ +1 ) = ( 2b + 1 ) + (2b +1 ) * ( 4b² - 2b + 1 )