text: ## AudrianaS 3 years ago factor 3x^3+x^2-15x-5
  
  1. manishsatywali
  
  use cardons method
  
  2. lgbasallote
  
  use factorer's method
  
  3. Yahoo!
  
  x^2 ( 3x + 1) -5 ( 3x + 1)
  
  4. jim_thompson5910
  
  3x^3+x^2-15x-5 (3x^3+x^2)+(-15x-5) x^2(3x+1)-5(3x+1) (x^2-5)(3x+1)
  
  5. Yahoo!
  
  (3x +1) (x^2 - 5)
  
  6. manishsatywali
  
  (x^2 -5) (3x +1)
  
  7. AudrianaS
  
  So how would I then completely factor it out in order to get a positive and negative number?
  
  8. jim_thompson5910
  
  Not sure what you mean, but you could factor $$\Large x^2-5$$ further $\Large x^2-5$ $\Large x^2-\left(\sqrt{5} \right )^2$ $\Large \left(x-\sqrt{5} \right )\left(x+\sqrt{5} \right )$ So $\Large (x^2-5)(3x+1)$ then becomes $\Large \left(x-\sqrt{5} \right )\left(x+\sqrt{5} \right )(3x+1)$
  
  9. AudrianaS
  
  Well, my book has the answer and it saw +- 5/2. But I dont know how to get to that
  
  10. jim_thompson5910
  
  Is 3x^3+x^2-15x-5 = 0 the original problem?
  
  11. manishsatywali
  
  see @AudrianaS dear ..............ans. is -5/2 ,+5/2 and -1/3.. you must know that alternating sign. of solution is the property of cubic eqn.
  
  12. AudrianaS
  
  @jim_thompson5910 Yes
  
  13. AudrianaS
  
  @manishsatywali Yeah I know it is -5/2 and +5/2 but I dont know about -1/3. I can see how you got it, but its just not in the book
  
  14. jim_thompson5910
  
  There should be 3 solutions and they are x = -1/3, x = sqrt(5), or x = -sqrt(5)
  
  15. jim_thompson5910
  
  So there's either a typo or something is missing.
  
  16. AudrianaS
  
  alright then. Thank you all. :)
  
  17. jim_thompson5910
  
  I would double check the problem. Notice how you have a cubic equation. So if you have 2 real roots, then you must also have a 3rd real root.
  
  18. AudrianaS
  
  yesh it just says, factor completely: 3x^3+x^2-15x-5
  
  19. jim_thompson5910
  
  Well if you're just factoring over the rationals, then you would stop at (x^2-5)(3x+1)
  
  20. AudrianaS
  
  Alright, that makes sense. Thank you.
  
  21. jim_thompson5910
  
  you're welcome
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