anonymous
  • anonymous
factor 3x^3+x^2-15x-5
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
use cardons method
lgbasallote
  • lgbasallote
use factorer's method
anonymous
  • anonymous
x^2 ( 3x + 1) -5 ( 3x + 1)

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jim_thompson5910
  • 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)
anonymous
  • anonymous
(3x +1) (x^2 - 5)
anonymous
  • anonymous
(x^2 -5) (3x +1)
anonymous
  • anonymous
So how would I then completely factor it out in order to get a positive and negative number?
jim_thompson5910
  • 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 \[\Large \left(x-\sqrt{5} \right )\left(x+\sqrt{5} \right )(3x+1)\]
anonymous
  • anonymous
Well, my book has the answer and it saw +- 5/2. But I dont know how to get to that
jim_thompson5910
  • jim_thompson5910
Is 3x^3+x^2-15x-5 = 0 the original problem?
anonymous
  • anonymous
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.
anonymous
  • anonymous
@jim_thompson5910 Yes
anonymous
  • anonymous
@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
jim_thompson5910
  • jim_thompson5910
There should be 3 solutions and they are x = -1/3, x = sqrt(5), or x = -sqrt(5)
jim_thompson5910
  • jim_thompson5910
So there's either a typo or something is missing.
anonymous
  • anonymous
alright then. Thank you all. :)
jim_thompson5910
  • 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.
anonymous
  • anonymous
yesh it just says, factor completely: 3x^3+x^2-15x-5
jim_thompson5910
  • jim_thompson5910
Well if you're just factoring over the rationals, then you would stop at (x^2-5)(3x+1)
anonymous
  • anonymous
Alright, that makes sense. Thank you.
jim_thompson5910
  • jim_thompson5910
you're welcome

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