## 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