Prove that the sum of the moduli of the roots of $x^4-5x^3+6x^2-5x+1=0$ is $4$ Prove that the sum of the moduli of the roots of $x^4-5x^3+6x^2-5x+1=0$ is $4$.

I tried various methods to find its roots,but no luck.Is there any method to solve it without actually finding the roots of the equation.Please help me.
Thanks.
 A: I'll give you an answer which is the application of a simple method for finding the solutions of symmetric polynomials:
Because $0$ is not a root, divide by $x^2$. Then use the substitution 
$$
t=x+\frac{1}{x}.
$$
This yields
$$
t^2-5t+4=0,
$$
which you can solve in a simple matter and then put the solutions back to the substitution.
EDIT: This is the method for when such equations are of even degree. When it's of odd degree, then $-1$ is always one solution. Then use polynomial division (Horner's algorithm or something you prefer) and you always get a symmetric polynomial of even degree as a result of division, so then you can use the method from my answer again.
In general, when your equation is of even degree, you divide it by the "middle" power of $x$, i.e. with $x$ to the power of the highest degree divided by $2$.
A: \begin{align}
x^4-5x^3+6x^2-5x+1&=(\color{red}1\times x^4-\color{red}4\times x^3+\color{red}1\times x^2)\\
&-(\color{red}1\times x^3-\color{red}4\times x^2+\color{red}1\times x)\\
&+(\color{red}1\times x^2-\color{red}4\times x+\color{red}1)\\
&=x^2(x^2-4x+1)-x(x^2-4x+1)+(x^2-4x+1)\\
&=(x^2-4x+1)(x^2-x+1)
\end{align}
In general we have 
$$1 \pm n x + (1 + n) x^2 \pm n x^3 + x^4=\pm\left(x^2\pm x+1\right) \left(n x\pm x^2-x\pm 1\right)$$
for example we have $$x^4+\sqrt{2} x^3+\left(1+\sqrt{2}\right) x^2+\sqrt{2} x+1=\left(x^2+x+1\right) \left(x^2+\sqrt{2} x-x+1\right)$$
Something similar you may find for $1  \pm n x + (1 - n) x^2 \pm n x^3 + x^4$. You could find similar rules for higher order polynomials, but the rules quickly get complicated :-)
