Question 330385
{{{12x^3+8x^2-x-1=0}}}

Let's try breaking it up, grouping and factoring. One could use the Rational Root Theorem as well.

Rewrite:

{{{12x^3+12x^2-4x^2-4x+3x-1=0}}}

Group:

{{{12x^3+12x^2+3x-4x^2-4x-1}}}

Factor out 3x from the first 3 terms and -1 from the last 3 terms: 

{{{3x(4x^2+4x+1)-(4x^2+4x+1)}}}

{{{(3x-1)(4x^2+4x+1)}}}

{{{(3x-1)(2x+1)^2=0}}}

What makes 3x-1=0?. x=1/3

What makes 2x+1=0?. x=-1/2

Since x=-1/2 has multiplicity 2, we have three roots as required of the cubic.