The sum of the real roots of $x^4+x^3+x^2+x-1$ Consider a polynomial $f(x)=x^4+x^3+x^2+x-1$. The sum of the real roots of $f(x)$ lies in the interval ...


*

*$(0,1)$

*$(-1,0)$

*$(-2,-1)$

*$(1,2)$


Using Intermediate Value property, I know that one root exists between $0$ and $1$, but I am stuck here and can't do anything else.
Any hints on how should I proceed?

Let $$f(x)=x^4+x^3+x^2+x-1.$$
Thus, $$f(0.5)f(0.6)<0$$ and $$f(-1.3)f(-1.2)<0,$$
which says that there are two roots:
$$x_1\in(0.5,0.6)$$ and
$$x_2\in(-1.3,-1,2)$$ and
$$x_1+x_2\in(-0.8,-0.6).$$
But $$f''(x)=12x^2+6x+2>0,$$
which says that $f$ is a convex function.
Thus, our equation has at most two roots and the answer is $3)$.
By the Descartes' rule of signs (https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs) 
our equation has one positive root only and we can check a placing of the root by calculator or even by hand.
