Algebra question from practice GRE exam The following is a question from the GRE exam GR9367:
Let $n > 1$ be an integer. Which of the following conditions guarantee that the equation $x^n = \sum_{i=0}^{n-1} a_ix^i$ has at least one root in the interval $(0,1)$?
I. $a_0 > 0$ and $\sum_{i=0}^{n-1}a_i < 1$
II. $a_0>0$ and $\sum_{i=0}^{n-1}a_i > 1$
III. $a_0<0$ and $\sum_{i=0}^{n-1}a_i > 1$
Through some guess work and luck I was able to get the right answer but I would prefer to understand the reasoning behind the problem. What is the underlying idea here? It feels like the right thing to use is Decartes' rule of signs but I  couldn't find a way of using the condition on $\sum_{i=0}^{n-1}a_i$.
 A: The underlying idea here is the intermediate value theorem. The function $f(x) = \sum_{i=0}^{n-1} a_ix^i - x^n$ is continuous on $[0,1]$ with $f(0) = a_0$ and $f(1) = \sum_{i = 0}^{n-1} a_i - 1$. If $f(0)$ has opposite sign to $f(1)$, then $f$ has a zero in $(0,1)$ by IVT. So if either $a_0 > 0$ and $\sum_{i = 0}^{n-1} a_i < 1$ or $a_0 < 0$ and $\sum_{i = 0}^{n-1} a_i > 1$, then the equation $x^n = \sum_{i = 0}^{n-1} a_i x^i$ has a root in $(0,1)$.
A: The answer hy kobe shows that I.and III. are sufficient. To see that II.is not,consider $f(x)=1+x-x^2$.
A: The key to this question lies in three known facts. Namely:

*

*A continuous function $f$ has a root in an open interval $]a,b[$ if $f(a)<0<f(b)$ or $f(b)<0<f(a)$

*All power series are continuous.

and


*$1>0$
So, to prove that a power series $\sum_{i=0}^\infty a_i x^i$ has a root between 0 and 1, wee need to show that either

*

*$a_0<0$ and $\sum_{i=0}^\infty a_i >0$
or


*$a_0>0$ and $\sum_{i=0}^\infty a_i <0$
If we look at condition III, we see that it satisfies $a_0<0$ and $\sum_{i=0}^\infty a_i >1>0$. So Condition III ensures a root between 0 and 1. The other two, however, don't. The "power series" $\frac{1}{2}$ and $2$ satisfy conditions I and II respectively, but neither have roots.
