Show that the equation `x^(3)+2x^(2)+x+5=0` has only one real root, such that `[alpha]=-3`, where `[x]` denotes the integral point of `x`

Explanation

We have `x^(3)+2x^(2)+x+5=0`
`impliesx^(3)+2x^(2)+x=-5`
Let `f(x)=x^(3)+2x^(2)+x` and `g(x)=-5`
`:'f'(x)=0implies3x^(2)+4x+1=0`
`impliesx=-1,-1/3` and `f''(x)=6x+4`
`:.f'(-1)=-2lt0` and `f''(-1/3)=-2+4=2gt0`
`:.f(x)` local maximum at `x=-1` and local minimum at `x=-1/3`
and `f(-1)=0,f(-1/3)=-4/27`