Question 366509
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Hi, 
c^5-4c^3-c^2+4
Testing: 2,-2, 1 are roots
by repeated division (or by using a polynomial calculator) One finds:
(c-2)(c + 2)(c-1)[c^2 + c + 1] 

c^2 + c + 1 has two imaginary roots
{{{c = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{c = (-1 +- sqrt( -3 ))/(2) }}}
{{{c = (-1 +- i*sqrt(3))/(2) }}}