Simple way to estimate the root of $x^5-x^ 4+2x^3+x^2+x+1=0$ How to give a mathematical proof that for all complex roots of $x^5-x^ 4+2x^3+x^2+x+1=0$, their real part is smaller than 1, and there is at least one root whose real part is larger than 0. If possible, not to solve any algebraic equation whose degree is larger than 3.
For the real roots, it would be easy to estimate them by observing the derivative and intermediate value theorem. What about the complex roots? I fail to find a way by trying Rouché's theorem.

Using the Routh-Hurwitz stability criterion you can tell how may roots of your system are in the open left-hand complex plane - i.e., the set $\{z\in\mathbb{C}: \operatorname{Re}(z) < 0\}$. 
In your case for the polynomial $p(x)=x^5 - x^4 + 2x^3 + x^2 + x + 1$ the Hurwitz matrix is:
    1.0000    2.0000    1.0000
   -1.0000    1.0000    1.0000
    3.0000    2.0000         0
    1.6667    1.0000         0
    0.2000         0         0
    1.0000         0         0

Indeed, there are two roots with nonnegative real part. We can verify that there are no imaginary roots simply by replacing $x=ic$ and try to determine $c\in\mathbb{R}$ so that $p(x)=0$.
Now, in order to determine whether all roots have a real part which is lower than $1$ we need to apply the Hurwitz criterion to the polynomial $q(x) = p(x+1)$. In fact, this is 
$$
q(x) = p(x+1) = x^5 + 4x^4 + 8x^3 + 11x^2 + 10x + 5
$$
for which the Hurwitz matrix is
1.0000    8.0000   10.0000
4.0000   11.0000    5.0000
5.2500    8.7500         0
4.3333    5.0000         0
2.6923         0         0
5.0000         0         0

therefore all roots of $q$ are in the open left-hand plane, thus all roots of $p$ have real parts which are smaller than $1$.
