The sum of the real roots of $x^4+x^3+x^2+x-1$

Question

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 ...

1. $(0,1)$
2. $(-1,0)$
3. $(-2,-1)$
4. $(1,2)$

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?

Answer 1

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.