Icono del sitio Solucionario Baldor

Ejercicio 280

Comparte esto πŸ‘πŸ‘
CAPITULO XXXV

Descomponer un trinomio en factores hallando sus raices
Ejercicio 280
Descomponer en factores, hallando las raΓ­ces:
SoluciΓ³n:
  1. x 2 –16x+63 x 2 –16x+63 =0 ( x–9 ) ( x–7 ) =0 x–9 =0 x 1 =9 x–7 =0 x 2 =7 x 2 –16x+63 =( x–9 ) ( x–7 )
  2. x 2 +24x+143 x 2 +24x+143 =0 ( x+13 ) ( x+11 ) =0 x+13 =0 x 1 =–13 x+11 =0 x 2 =–11 x 2 +24x+143 =( x+13 ) ( x+11 )
  3. x 2 –26x–155 x 2 –26x–155 =0 ( x–31 ) ( x+5 ) =0 x–31 =0 x 1 =31 x+5 =0 x 2 =–5 x 2 –26x–155 =( x–31 ) ( x+5 )
  4. 2 x 2 +x–6 2 x 2 +x–6 =0 2 x 2 +4x–3x–6 =0 2x( x+2 ) –3( x+2 ) =0 ( x+2 ) ( 2x–3 ) =0 x+2 =0 x 1 =–2 2x–3 =0 2x =3 x 2 = 3 2 2 x 2 +x–6 =2( x+2 ) ( x– 3 2 ) 2 x 2 +x–6 = 2 ( x+2 ) ( 2x–3 2 ) 2 x 2 +x–6 =( x+2 ) ( 2x–3 )
  5. 12 x 2 +5x–2 12 x 2 +5x–2 =0 12 x 2 –3x+8x–2 =0 3x( 4x–1 ) +2( 4x–1 ) =0 ( 4x–1 ) ( 3x+2 ) =0 4x–1 =0 4x =1 x 1 = 1 4 3x+2 =0 3x =–2 x 2 =– 2 3 12 x 2 +5x–2 =12( x– 1 4 ) ( x+ 2 3 ) 12 x 2 +5x–2 = 3 ( 4x–1 4 ) ( 3x+2 3 ) 12 x 2 +5x–2 =( 4x–1 ) ( 3x+2 )
  6. 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 =–8 5x+1 =0 5x =–1 x 2 =– 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 )
  7. 6 x 2 +7x–10 6 x 2 +7x–10 =0 6 x 2 +12x–5x–10 =0 6x( x+2 ) –5( x+2 ) =0 ( x+2 ) ( 6x–5 ) =0 x+2 =0 x 1 =–2 6x–5 =0 6x =5 x 2 = 5 6 6 x 2 +7x–10 =6( x+2 ) ( x– 5 6 ) 6 x 2 +7x–10 = 6 ( x+2 ) ( 6x–5 6 ) 6 x 2 +7x–10 =( x+2 ) ( 6x–5 )
  8. 12 x 2 –25x+12 12 x 2 –25x+12 =0 12 x 2 –16x–9x+12 =0 4x( 3x–4 ) –3( 3x–4 ) =0 ( 3x–4 ) ( 4x–3 ) =0 3x–4 =0 3x =4 x 1 = 4 3 4x–3 =0 4x =3 x 2 = 3 4 12 x 2 –25x+12 =12( x– 4 3 ) ( x– 3 4 ) 12 x 2 –25x+12 = 12 ( 3x–4 3 ) ( 4x–3 4 ) 12 x 2 –25x+12 =( 3x–4 ) ( 4x–3 )
  9. 8 x 2 +50x+63 8 x 2 +50x+63 =0 x = –50 Β± 5 0 2 –4( 8 ) ( 63 ) 2( 8 ) x = –50 Β± 2500–2016 16 x = –50 Β± 484 16 x = –50 Β± 22 16 x 1 = –50+22 16 =– =– 7 4 x 2 = –50–22 16 =– =– 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 )
  10. 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 =–1 x 1 =– 1 3 9x+7 =0 9x =–7 x 2 =– 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 )
  11. 30 x 2 –61x+30 30 x 2 –61x+30 =0 30 x 2 –36x–25x+30 =0 6x( 5x–6 ) –5( 5x–6 ) =0 ( 5x–6 ) ( 6x–5 ) =0 5x–6 =0 5x =6 x 1 = 6 5 6x–5 =0 6x =5 x 2 = 5 6 30 x 2 –61x+30 =30( x– 6 5 ) ( x– 5 6 ) 30 x 2 –61x+30 = 30 ( 5x–6 5 ) ( 6x–5 6 ) 30 x 2 –61x+30 =( 5x–6 ) ( 6x–5 )
  12. 11 x 2 –153x–180 11 x 2 –153x–180 =0 11 x 2 –165x+12x–180 =0 11x( x–15 ) +12( x–15 ) =0 ( x–15 ) ( 11x+12 ) =0 x–15 =0 x 1 =15 11x+12 =0 11x =–12 x =– 12 11 11 x 2 –153x–180 =11( x–15 ) ( x+ 12 11 ) 11 x 2 –153x–180 = 11 ( x–15 ) ( 11x+12 11 ) 11 x 2 –153x–180 =( x–15 ) ( 11x+12 )
  13. 6–x– x 2 x 2 +x–6 =0 ( x+3 ) ( x–2 ) =0 x+3 =0 x 1 =–3 x–2 =0 x 2 =2 x 2 +x–6 =( x+3 ) ( x–2 )
  14. 5–9x–2 x 2 2 x 2 +9x–5 =0 2 x 2 –x+10x–5 =0 x( 2x–1 ) +5( 2x–1 ) =0 ( 2x–1 ) ( x+5 ) =0 2x–1 =0 2x =1 x 1 = 1 2 x+5 =0 x 2 =–5 2 x 2 +9x–5 =2( x– 1 2 ) ( x+5 ) 2 x 2 +9x–5 = 2 ( 2x–1 2 ) ( x+5 ) 2 x 2 +9x–5 =( 2x–1 ) ( x+5 )
  15. 15+4x–4 x 2 4 x 2 –4x–15 =0 x = –( –4 ) Β± ( –4 ) 2 –4( 4 ) ( –15 ) 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 = 4–16 8 =– =– 3 2 4 x 2 –4x–15 =4( x– 5 2 ) ( x+ 3 2 ) 4 x 2 –4x–15 = 4 ( 2x–5 2 ) ( 2x+3 2 ) 4 x 2 –4x–15 =( 2x–5 ) ( 2x+3 )
  16. 4–13x–12 x 2 12 x 2 +13x–4 =0 12 x 2 –3x+16x–4 =0 3x( 4x–1 ) +4( 4x–1 ) =0 ( 4x–1 ) ( 3x+4 ) =0 4x–1 =0 4x =1 x 1 = 1 4 3x+4 =0 3x =–4 x 2 =– 4 3 12 x 2 +13x–4 =12( x– 1 4 ) ( x+ 4 3 ) 12 x 2 +13x–4 = 12 ( 4x–1 4 ) ( 3x+4 3 ) 12 x 2 +13x–4 =( 4x–1 ) ( 3x+4 )
  17. 72 x 2 –55x–7 72 x 2 –55x–7 =0 72 x 2 +8x–63x–7 =0 8x( 9x+1 ) –7( 9x+1 ) =0 ( 9x+1 ) ( 8x–7 ) =0 9x+1 =0 9x =–1 x 1 =– 1 9 8x–7 =0 8x =7 x = 7 8 72 x 2 –55x–7 =72( x+ 1 9 ) ( x– 7 8 ) 72 x 2 –55x–7 = 72 ( 9x+1 9 ) ( 8x–7 8 ) 72 x 2 –55x–7 =( 9x+1 ) ( 8x–7 )
  18. 6+31x–30 x 2 30 x 2 –31x–6 =0 30 x 2 +5x–36x–6 =0 5x( 6x+1 ) –6( 6x+1 ) =0 ( 6x+1 ) ( 5x–6 ) =0 6x+1 =0 6x =–1 x 1 =– 1 6 5x–6 =0 5x =6 x 2 = 6 5 30 x 2 –31x–6 =30( x+ 1 6 ) ( x– 6 5 ) 30 x 2 –31x–6 = 30 ( 6x+1 6 ) ( 5x–6 5 ) 30 x 2 –31x–6 =( 6x+1 ) ( 5x–6 )
  19. 10 x 2 +207x–63 x = –207 Β± 20 7 2 –4( 10 ) ( –63 ) 2( 10 ) x = –207 Β± 42849+2520 20 x = –207 Β± 45369 20 x = –207 Β± 213 20 x 1 = –207+213 20 = = 3 10 x 2 = –207–213 20 =– 20 =–24 10 x 2 +207x–63 =10( x– 3 10 ) ( x+24 ) 10 x 2 +207x–63 = 10 ( 10x–3 10 ) ( x–24 ) 10 x 2 +207x–63 =( 10x–3 ) ( x–24 )
  20. 100–15x– x 2 x 2 +15x–100 =0 ( x+20 ) ( x–5 ) =0 x+20 =0 x 1 =–20 x–5 =0 x 2 =5 x 2 +15x–100 =( x+20 ) ( x–5 )
  21. 18 x 2 +31x–49 18 x 2 +31x–49 =0 18 x 2 –18x+49x–49 =0 18x( x–1 ) +49( x–1 ) =0 ( x–1 ) ( 18x+49 ) =0 x–1 =0 x 1 =1 18x+49 =0 18x =–49 x 2 =– 49 18 18 x 2 +31x–49 =18( x–1 ) ( x+ 49 18 ) 18 x 2 +31x–49 = 18 ( x–1 ) ( 18x+49 18 ) 18 x 2 +31x–49 =( x–1 ) ( 18x+49 )
  22. 6 x 2 –ax–2 a 2 6 x 2 –ax–2 a 2 =0 6 x 2 +3ax–4ax–2 a 2 =0 3x( 2x+a ) –2a( 2x+a ) =0 ( 2x+a ) ( 3x–2a ) =0 2x+a =0 2x =–a x 1 =– a 2 3x–2a =0 3x =2a x 2 = 2a 3 6 x 2 –ax–2 a 2 =6( x+ a 2 ) ( x– 2a 3 ) 6 x 2 –ax–2 a 2 = 6 ( 2x+a 2 ) ( 3x–2a 3 ) 6 x 2 –ax–2 a 2 =( 2x+a ) ( 3x–2a )
  23. 5 x 2 +22xy–15 y 2 5 x 2 +22xy–15 y 2 =0 5 x 2 +25xy–3xy–15 y 2 =0 5x( x+5y ) –3y( x+5y ) =0 ( x+5y ) ( 5x–3y ) =0 x+5y =0 x 1 =–5y 5x–3y =0 5x =3y x 2 = 3y 5 5 x 2 +22xy–15 y 2 =5( x+5y ) ( x– 3y 5 ) 5 x 2 +22xy–15 y 2 = 5 ( x+5y ) ( 5x–3y 5 ) 5 x 2 +22xy–15 y 2 =( x+5y ) ( 5x–3y )
  24. 15 x 2 –32mx–7 m 2 15 x 2 –32mx–7 m 2 =0 15 x 2 +3mx–35mx–7 m 2 =0 3x( 5x+m ) –7m( 5x+m ) =0 ( 5x+m ) ( 3x–7m ) =0 5x+m =0 5x =–m x 1 =– m 5 3x–7m =0 3x =7m x 2 = 7m 3 15 x 2 –32mx–7 m 2 =15( x+ m 5 ) ( x– 7m 3 ) 15 x 2 –32mx–7 m 2 = 15 ( 5x+m 5 ) ( 3x–7m 3 ) 15 x 2 –32mx–7 m 2 =( 5x+m ) ( 3x–7m )
Salir de la versiΓ³n mΓ³vil