Question

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

Answer 1

use cardons method

Answer 2

use factorer's method

Answer 3

x^2 ( 3x + 1) -5 ( 3x + 1)

Answer 4

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


(3x^3+x^2)+(-15x-5)


x^2(3x+1)-5(3x+1)


(x^2-5)(3x+1)

Answer 5

(3x +1) (x^2 - 5)

Answer 6

(x^2 -5) (3x +1)

Answer 7

So how would I then completely factor it out in order to get a positive and negative number?

Answer 8

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

\[\Large \left(x-\sqrt{5} \right )\left(x+\sqrt{5} \right )(3x+1)\]

Answer 9

Well, my book has the answer and it saw +- 5/2. But I dont know how to get to that

Answer 10

Is 3x^3+x^2-15x-5 = 0 the original problem?

Answer 11

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.

Answer 12

@jim_thompson5910 Yes

Answer 13

@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

Answer 14

There should be 3 solutions and they are x = -1/3, x = sqrt(5), or x = -sqrt(5)

Answer 15

So there's either a typo or something is missing.

Answer 16

alright then. Thank you all. :)

Answer 17

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.

Answer 18

yesh it just says, factor completely: 3x^3+x^2-15x-5

Answer 19

Well if you're just factoring over the rationals, then you would stop at (x^2-5)(3x+1)

Answer 20

Alright, that makes sense. Thank you.

Answer 21

you're welcome