How many integral values of $k$ such that $2x^3+3x^2+6x+k=0$ has exactly three real roots. I am unable to see how I'd start this question. A small hint, or the entire solution, both will be highly appreciated!
The discriminant of $2x^3+3x^2+6x+k$ is $-108 (k^2 - 5 k + 13)=-27 (2 k - 5)^2 - 729 $, which is always negative. Therefore, there is one real root and two complex conjugate roots (see Wikipedia).