Question 367535
{{{2x^4 + x^3 - x^2 -x}}}
 
{{{ x(2x^3 + x^2 - x - 1)}}}
 
 
the polynomial inside is factorizable over R : 
 
2x^3 +
{{{2x (x^3 + x^2/2 - x/2 - 1/2)}}}
 
{{{2x ( (x + 1/6)^3 - x/12 - x/2 - 1/2 - 1/216)}}}
 
{{{2x ( (x + 1/6)^3 - 7x/12 - 109/216 ) }}}


let : {{{x + 1/6 = y}}}, we get in the parenthesis : 
 
{{{y^3 - (7/12)*y - 109/216 + 7/12 = y^3 - (7/12)*y + 27/216}}}
   
 

writing {{{y = p + q}}}, 
 
{{{p^3 + q^3 + 3pq (p+q) - (7/12)*(p+q) + 27/216 = 0}}}
 
we get : 
 
{{{p^3 + q^3 = -27/216}}}
{{{3pq = 7/12}}}

the latter gives : {{{q = 7/36*1/p}}}, if substituted in the equation before : 
 
{{{p^3 + (7/36)^3/p^3 = -27/216}}} times {{{p^3}}}
 
{{{p^6 + (27/216)*p^3 + (7/36)^3 = 0}}}

with 2nd degree we find : 

{{{p^3 = (-27/216 +- sqrt((27/216)^2 - 4*(7/36)^3))/2}}}
 
{{{q = 7/36*1/p}}} and {{{x  + 1/6 = p + 7/36*1/p}}}
 
with {{{p = ((-27/216 +- sqrt((27/216)^2 - 4*(7/36)^3))/2)^(1/3)}}}
 
 
hence the solution is {{{x = ((-27/216 +- sqrt((27/216)^2 - 4*(7/36)^3))/2)^(1/3) + (7/36)(2/(-27/216 +- sqrt((27/216)^2 - 4*(7/36)^3)))^(1/3) -1/6}}}
 
where 2 roots are identical.
 
we can write the polynomial as : 2x(x-x[++])(x-x[+-])(x-x[--])