How can I guarantee the unique positive root of this polynomial? How can I guarantee the unique positive root of this polynomial?
I have two polynomial, 
$$
x^{n+1} + x^n - 1 =0 
$$
and 
$$
x^{n+1} - x^n - 1 =0 
$$
respectively, where $n\in\mathbb{N}$. I have tried for the cases from $n=1$ to $100$. For every calculation, I have found unique positive root for each polynomial with the help of MATHEMATICA, but I couldn't prove it mathematically. How can I prove these polynımials have unique positive root ?
 A: The derivative of this polynomial is
$$(n + 1) x^n + n x^{n - 1} = x^{n - 1} \big((n + 1) x + n\big)$$
This has zeros at $x = 0$ and $x = -\frac{n}{n + 1} < 0$; in particular, it is positive for $x > 0$. Thus the original polynomial is strictly increasing from the point $(0, -1)$, and has a unique positive root.
A: From $x^n = \frac{1}{( x \pm 1 )}$
explore the interception of two functions:
     $x^n$ (always positive growing for $x>0$ ) whatever $n>0$
and  $\frac{1}{x+1}$ (always decreasing from $1$ to $0$ for $x>0$ )
  or $\frac{1}{x-1}$ (always decreasing from $\infty$ to $0$ for $x>1$  ) 
the interception belongs always and necessarily to the first quadrant 
( explore the graphs in wolframalpha.com , varying the $n$ )
solve $x^n = \frac{1}{( x + 1 )}$  and $n=2$
solve $x^n = \frac{1}{( x - 1 )}$  and $n=2$ 
A: Descartes's sign rule ensures that both polynomials have exactly $1$ positive root for every $n$.
A: Consider $f(x)=x^{n+1}+ x^n -1.$ Note that $f(0)=-1.$ Now, since $\lim_{x\to \infty} f(x)=+\infty$ there exists $a>0$ such that $f(a)>0.$ Since $f$ is continuous in $[0,a]$ it follows from Bolzano's theorem that there is a root in $(0,a).$ A similar argument works with $f(x)=x^{n+1}- x^n -1.$
Now, if $f(x)=x^{n+1}+ x^n -1$ then $f'(x)=(n+1)x^n+nx^{n-1}>0$ on $(0,\infty).$ Thus, $f$ is strictly increasing, from where it has at most one root. Since it has one, it has exactly one. 
Now, if $f(x)=x^{n+1}- x^n -1$ then $f(x)<0$ in $(0,1).$ And $f'(x)=(n+1)x^n-nx^{n-1}>0$ on $(1,\infty).$ Thus, $f$ is strictly increasing in $(1,\infty),$ from where it has at most one root. Since it has one, it has exactly one. 
