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AudrianaS Group Title

factor 3x^3+x^2-15x-5

  • 2 years ago
  • 2 years ago

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  1. manishsatywali Group Title
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    use cardons method

    • 2 years ago
  2. lgbasallote Group Title
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    use factorer's method

    • 2 years ago
  3. Yahoo! Group Title
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    x^2 ( 3x + 1) -5 ( 3x + 1)

    • 2 years ago
  4. jim_thompson5910 Group Title
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    3x^3+x^2-15x-5 (3x^3+x^2)+(-15x-5) x^2(3x+1)-5(3x+1) (x^2-5)(3x+1)

    • 2 years ago
  5. Yahoo! Group Title
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    (3x +1) (x^2 - 5)

    • 2 years ago
  6. manishsatywali Group Title
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    (x^2 -5) (3x +1)

    • 2 years ago
  7. AudrianaS Group Title
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    So how would I then completely factor it out in order to get a positive and negative number?

    • 2 years ago
  8. jim_thompson5910 Group Title
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    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 \[\Large \left(x-\sqrt{5} \right )\left(x+\sqrt{5} \right )(3x+1)\]

    • 2 years ago
  9. AudrianaS Group Title
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    Well, my book has the answer and it saw +- 5/2. But I dont know how to get to that

    • 2 years ago
  10. jim_thompson5910 Group Title
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    Is 3x^3+x^2-15x-5 = 0 the original problem?

    • 2 years ago
  11. manishsatywali Group Title
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    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.

    • 2 years ago
  12. AudrianaS Group Title
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    @jim_thompson5910 Yes

    • 2 years ago
  13. AudrianaS Group Title
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    @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

    • 2 years ago
  14. jim_thompson5910 Group Title
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    There should be 3 solutions and they are x = -1/3, x = sqrt(5), or x = -sqrt(5)

    • 2 years ago
  15. jim_thompson5910 Group Title
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    So there's either a typo or something is missing.

    • 2 years ago
  16. AudrianaS Group Title
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    alright then. Thank you all. :)

    • 2 years ago
  17. jim_thompson5910 Group Title
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    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.

    • 2 years ago
  18. AudrianaS Group Title
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    yesh it just says, factor completely: 3x^3+x^2-15x-5

    • 2 years ago
  19. jim_thompson5910 Group Title
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    Well if you're just factoring over the rationals, then you would stop at (x^2-5)(3x+1)

    • 2 years ago
  20. AudrianaS Group Title
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    Alright, that makes sense. Thank you.

    • 2 years ago
  21. jim_thompson5910 Group Title
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    you're welcome

    • 2 years ago
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