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