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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)

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

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

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

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?

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. :)

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