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.

\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 :-)
