The value of $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor$ There was a multiple choices saying:

Find the value of  $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor$ knowing that $x^2+x<0$

The right answer is $-2$.
For solving this Maths test, I solved $x^2+x<0$ which is $(-1,0)$ and then find the value of each term of the sum.
My question is: Is there any approach more formal than I did? Thank you for your time.


*

*$\lfloor x\rfloor$ is  floor$(x)$

 A: $$x^2+x<0\Longrightarrow x(x+1)<0\Longleftrightarrow -1<x<0\Longrightarrow \left\{\begin{array}{}\;\;\; 0 <x^n<1&\,n\,\,\text{ is even}\\-1<x^n<0&\,n\,\,\text{ is odd}\end{array}\right. $$
Thus, passing to the floor function under the above condition:
$$\sum_{n=1}^4\lfloor x^n\rfloor=-1+0-1+0=-2$$
Of course, the above is just what you did slightly more fleshed out.
A: As $x^2+x<0, (x-0)(x-(-1))<0$ 
Now the product two terms is negative, so
either ($x-0>0$ and $x-(-1)<0$)  or ($x-0<0$ and $x-(-1)>0$).
If $x>0$ and $x<-1\implies -1>x>0\implies -1>0$, which is clearly impossible. 
If $x<0$ and $x>-1$, $-1<x<0$
$\implies -1<x^{2m+1}<0\implies  \bigl\lfloor x^{2m+1}\bigr\rfloor=-1$
$\implies 0<x^{2m}<1\implies  \bigl\lfloor x^{2m}\bigr\rfloor=0$.
So,  $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor=-1+0-1+0=-2$ 
In general, if $(x-a)(x-b)<0$  where $a<b$,
either $x<a$ and $x>b\implies a>x>b\implies a>b$, but $a<b$(given)
or $x>a$  and $x<b\implies a<x<b$  Here $a=-1,b=0$
