Question 406075
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Hi,
 x = 1 is a real root. Dividing by (x-1) to find the other two roots:
      3x^2 + 5x +6
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x -1 | 3x^3+2x^2+x-6

(x-1)(3x^2 + 5x +6)= 0
     (3x^2 + 5x +6)= 0
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-5 +- sqrt(-47 ))/(6) }}}
{{{x = (-5 +- i*sqrt(47 ))/(6) }}}
the remaining two roots of 3x^3+2x^2+x-6 are complex roots.