CAPITULO XXXV
Descomponer un trinomio en factores hallando sus raices
Descomponer un trinomio en factores hallando sus raices
- Ejercicio 280
Descomponer en factores, hallando las raΓces:
SoluciΓ³n:
- Se iguala la ecuaciΓ³n a 0
- Se buscan las raΓces
- Se remmplazan los valores en la fΓ³rmula a x 2 +bx+c=a( xβ x 1 ) ( xβ x 2 )
- x 2 β16x+63 x 2 β16x+63 =0 ( xβ9 ) ( xβ7 ) =0 xβ9 =0 x 1 =9 xβ7 =0 x 2 =7 x 2 β16x+63 =( xβ9 ) ( xβ7 )
- x 2 +24x+143 x 2 +24x+143 =0 ( x+13 ) ( x+11 ) =0 x+13 =0 x 1 =β13 x+11 =0 x 2 =β11 x 2 +24x+143 =( x+13 ) ( x+11 )
- x 2 β26xβ155 x 2 β26xβ155 =0 ( xβ31 ) ( x+5 ) =0 xβ31 =0 x 1 =31 x+5 =0 x 2 =β5 x 2 β26xβ155 =( xβ31 ) ( x+5 )
- 2 x 2 +xβ6 2 x 2 +xβ6 =0 2 x 2 +4xβ3xβ6 =0 2x( x+2 ) β3( x+2 ) =0 ( x+2 ) ( 2xβ3 ) =0 x+2 =0 x 1 =β2 2xβ3 =0 2x =3 x 2 = 3 2 2 x 2 +xβ6 =2( x+2 ) ( xβ 3 2 ) 2 x 2 +xβ6 = 2 ( x+2 ) ( 2xβ3 2 ) 2 x 2 +xβ6 =( x+2 ) ( 2xβ3 )
- 12 x 2 +5xβ2 12 x 2 +5xβ2 =0 12 x 2 β3x+8xβ2 =0 3x( 4xβ1 ) +2( 4xβ1 ) =0 ( 4xβ1 ) ( 3x+2 ) =0 4xβ1 =0 4x =1 x 1 = 1 4 3x+2 =0 3x =β2 x 2 =β 2 3 12 x 2 +5xβ2 =12( xβ 1 4 ) ( x+ 2 3 ) 12 x 2 +5xβ2 = 3 ( 4xβ1 4 ) ( 3x+2 3 ) 12 x 2 +5xβ2 =( 4xβ1 ) ( 3x+2 )
- 5 x 2 +41x+8 5 x 2 +41x+8 =0 5 x 2 +40x+x+8 =0 5x( x+8 ) +( x+8 ) =0 ( x+8 ) ( 5x+1 ) =0 x+8 =0 x 1 =β8 5x+1 =0 5x =β1 x 2 =β 1 5 5 x 2 +41x+8 =5( x+8 ) ( x+ 1 5 ) 5 x 2 +41x+8 = 5 ( x+8 ) ( 5x+1 5 ) 5 x 2 +41x+8 =( x+8 ) ( 5x+1 )
- 6 x 2 +7xβ10 6 x 2 +7xβ10 =0 6 x 2 +12xβ5xβ10 =0 6x( x+2 ) β5( x+2 ) =0 ( x+2 ) ( 6xβ5 ) =0 x+2 =0 x 1 =β2 6xβ5 =0 6x =5 x 2 = 5 6 6 x 2 +7xβ10 =6( x+2 ) ( xβ 5 6 ) 6 x 2 +7xβ10 = 6 ( x+2 ) ( 6xβ5 6 ) 6 x 2 +7xβ10 =( x+2 ) ( 6xβ5 )
- 12 x 2 β25x+12 12 x 2 β25x+12 =0 12 x 2 β16xβ9x+12 =0 4x( 3xβ4 ) β3( 3xβ4 ) =0 ( 3xβ4 ) ( 4xβ3 ) =0 3xβ4 =0 3x =4 x 1 = 4 3 4xβ3 =0 4x =3 x 2 = 3 4 12 x 2 β25x+12 =12( xβ 4 3 ) ( xβ 3 4 ) 12 x 2 β25x+12 = 12 ( 3xβ4 3 ) ( 4xβ3 4 ) 12 x 2 β25x+12 =( 3xβ4 ) ( 4xβ3 )
- 8 x 2 +50x+63 8 x 2 +50x+63 =0 x = β50 Β± 5 0 2 β4( 8 ) ( 63 ) 2( 8 ) x = β50 Β± 2500β2016 16 x = β50 Β± 484 16 x = β50 Β± 22 16 x 1 = β50+22 16 =β =β 7 4 x 2 = β50β22 16 =β =β 9 2 8 x 2 +50x+63 =8( x+ 7 4 ) ( x+ 9 2 ) 8 x 2 +50x+63 = 8 ( 4x+7 4 ) ( 2x+9 2 ) 8 x 2 +50x+63 =( 4x+7 ) ( 2x+9 )
- 27 x 2 +30x+7 27 x 2 +30x+7 =0 27 x 2 +9x+21x+7 =0 9x( 3x+1 ) +7( 3x+1 ) =0 ( 3x+1 ) ( 9x+7 ) =0 3x+1 =0 3x =β1 x 1 =β 1 3 9x+7 =0 9x =β7 x 2 =β 7 9 27 x 2 +30x+7 =27( x+ 1 3 ) ( x+ 7 9 ) 27 x 2 +30x+7 = 27 ( 3x+1 3 ) ( 9x+7 9 ) 27 x 2 +30x+7 =( 3x+1 ) ( 9x+7 )
- 30 x 2 β61x+30 30 x 2 β61x+30 =0 30 x 2 β36xβ25x+30 =0 6x( 5xβ6 ) β5( 5xβ6 ) =0 ( 5xβ6 ) ( 6xβ5 ) =0 5xβ6 =0 5x =6 x 1 = 6 5 6xβ5 =0 6x =5 x 2 = 5 6 30 x 2 β61x+30 =30( xβ 6 5 ) ( xβ 5 6 ) 30 x 2 β61x+30 = 30 ( 5xβ6 5 ) ( 6xβ5 6 ) 30 x 2 β61x+30 =( 5xβ6 ) ( 6xβ5 )
- 11 x 2 β153xβ180 11 x 2 β153xβ180 =0 11 x 2 β165x+12xβ180 =0 11x( xβ15 ) +12( xβ15 ) =0 ( xβ15 ) ( 11x+12 ) =0 xβ15 =0 x 1 =15 11x+12 =0 11x =β12 x =β 12 11 11 x 2 β153xβ180 =11( xβ15 ) ( x+ 12 11 ) 11 x 2 β153xβ180 = 11 ( xβ15 ) ( 11x+12 11 ) 11 x 2 β153xβ180 =( xβ15 ) ( 11x+12 )
- 6βxβ x 2 x 2 +xβ6 =0 ( x+3 ) ( xβ2 ) =0 x+3 =0 x 1 =β3 xβ2 =0 x 2 =2 x 2 +xβ6 =( x+3 ) ( xβ2 )
- 5β9xβ2 x 2 2 x 2 +9xβ5 =0 2 x 2 βx+10xβ5 =0 x( 2xβ1 ) +5( 2xβ1 ) =0 ( 2xβ1 ) ( x+5 ) =0 2xβ1 =0 2x =1 x 1 = 1 2 x+5 =0 x 2 =β5 2 x 2 +9xβ5 =2( xβ 1 2 ) ( x+5 ) 2 x 2 +9xβ5 = 2 ( 2xβ1 2 ) ( x+5 ) 2 x 2 +9xβ5 =( 2xβ1 ) ( x+5 )
- 15+4xβ4 x 2 4 x 2 β4xβ15 =0 x = β( β4 ) Β± ( β4 ) 2 β4( 4 ) ( β15 ) 2( 4 ) x = 4 Β± 16+240 8 x = 4 Β± 256 8 x = 4 Β± 16 8 x 1 = 4+16 8 = = 5 2 x 2 = 4β16 8 =β =β 3 2 4 x 2 β4xβ15 =4( xβ 5 2 ) ( x+ 3 2 ) 4 x 2 β4xβ15 = 4 ( 2xβ5 2 ) ( 2x+3 2 ) 4 x 2 β4xβ15 =( 2xβ5 ) ( 2x+3 )
- 4β13xβ12 x 2 12 x 2 +13xβ4 =0 12 x 2 β3x+16xβ4 =0 3x( 4xβ1 ) +4( 4xβ1 ) =0 ( 4xβ1 ) ( 3x+4 ) =0 4xβ1 =0 4x =1 x 1 = 1 4 3x+4 =0 3x =β4 x 2 =β 4 3 12 x 2 +13xβ4 =12( xβ 1 4 ) ( x+ 4 3 ) 12 x 2 +13xβ4 = 12 ( 4xβ1 4 ) ( 3x+4 3 ) 12 x 2 +13xβ4 =( 4xβ1 ) ( 3x+4 )
- 72 x 2 β55xβ7 72 x 2 β55xβ7 =0 72 x 2 +8xβ63xβ7 =0 8x( 9x+1 ) β7( 9x+1 ) =0 ( 9x+1 ) ( 8xβ7 ) =0 9x+1 =0 9x =β1 x 1 =β 1 9 8xβ7 =0 8x =7 x = 7 8 72 x 2 β55xβ7 =72( x+ 1 9 ) ( xβ 7 8 ) 72 x 2 β55xβ7 = 72 ( 9x+1 9 ) ( 8xβ7 8 ) 72 x 2 β55xβ7 =( 9x+1 ) ( 8xβ7 )
- 6+31xβ30 x 2 30 x 2 β31xβ6 =0 30 x 2 +5xβ36xβ6 =0 5x( 6x+1 ) β6( 6x+1 ) =0 ( 6x+1 ) ( 5xβ6 ) =0 6x+1 =0 6x =β1 x 1 =β 1 6 5xβ6 =0 5x =6 x 2 = 6 5 30 x 2 β31xβ6 =30( x+ 1 6 ) ( xβ 6 5 ) 30 x 2 β31xβ6 = 30 ( 6x+1 6 ) ( 5xβ6 5 ) 30 x 2 β31xβ6 =( 6x+1 ) ( 5xβ6 )
- 10 x 2 +207xβ63 x = β207 Β± 20 7 2 β4( 10 ) ( β63 ) 2( 10 ) x = β207 Β± 42849+2520 20 x = β207 Β± 45369 20 x = β207 Β± 213 20 x 1 = β207+213 20 = = 3 10 x 2 = β207β213 20 =β 20 =β24 10 x 2 +207xβ63 =10( xβ 3 10 ) ( x+24 ) 10 x 2 +207xβ63 = 10 ( 10xβ3 10 ) ( xβ24 ) 10 x 2 +207xβ63 =( 10xβ3 ) ( xβ24 )
- 100β15xβ x 2 x 2 +15xβ100 =0 ( x+20 ) ( xβ5 ) =0 x+20 =0 x 1 =β20 xβ5 =0 x 2 =5 x 2 +15xβ100 =( x+20 ) ( xβ5 )
- 18 x 2 +31xβ49 18 x 2 +31xβ49 =0 18 x 2 β18x+49xβ49 =0 18x( xβ1 ) +49( xβ1 ) =0 ( xβ1 ) ( 18x+49 ) =0 xβ1 =0 x 1 =1 18x+49 =0 18x =β49 x 2 =β 49 18 18 x 2 +31xβ49 =18( xβ1 ) ( x+ 49 18 ) 18 x 2 +31xβ49 = 18 ( xβ1 ) ( 18x+49 18 ) 18 x 2 +31xβ49 =( xβ1 ) ( 18x+49 )
- 6 x 2 βaxβ2 a 2 6 x 2 βaxβ2 a 2 =0 6 x 2 +3axβ4axβ2 a 2 =0 3x( 2x+a ) β2a( 2x+a ) =0 ( 2x+a ) ( 3xβ2a ) =0 2x+a =0 2x =βa x 1 =β a 2 3xβ2a =0 3x =2a x 2 = 2a 3 6 x 2 βaxβ2 a 2 =6( x+ a 2 ) ( xβ 2a 3 ) 6 x 2 βaxβ2 a 2 = 6 ( 2x+a 2 ) ( 3xβ2a 3 ) 6 x 2 βaxβ2 a 2 =( 2x+a ) ( 3xβ2a )
- 5 x 2 +22xyβ15 y 2 5 x 2 +22xyβ15 y 2 =0 5 x 2 +25xyβ3xyβ15 y 2 =0 5x( x+5y ) β3y( x+5y ) =0 ( x+5y ) ( 5xβ3y ) =0 x+5y =0 x 1 =β5y 5xβ3y =0 5x =3y x 2 = 3y 5 5 x 2 +22xyβ15 y 2 =5( x+5y ) ( xβ 3y 5 ) 5 x 2 +22xyβ15 y 2 = 5 ( x+5y ) ( 5xβ3y 5 ) 5 x 2 +22xyβ15 y 2 =( x+5y ) ( 5xβ3y )
- 15 x 2 β32mxβ7 m 2 15 x 2 β32mxβ7 m 2 =0 15 x 2 +3mxβ35mxβ7 m 2 =0 3x( 5x+m ) β7m( 5x+m ) =0 ( 5x+m ) ( 3xβ7m ) =0 5x+m =0 5x =βm x 1 =β m 5 3xβ7m =0 3x =7m x 2 = 7m 3 15 x 2 β32mxβ7 m 2 =15( x+ m 5 ) ( xβ 7m 3 ) 15 x 2 β32mxβ7 m 2 = 15 ( 5x+m 5 ) ( 3xβ7m 3 ) 15 x 2 β32mxβ7 m 2 =( 5x+m ) ( 3xβ7m )