<s>[INST] Regarding complex roots of a polynomial I am having difficulty in finding the roots of the following polynomial when it is given that all the roots are complex:-
$$f(x)= x^4+4x^3+8x^2+8x+4$$
How can I factorize the polynomial to get its roots?
 [/INST] Hint:
$$
\begin{align}
f(x)= x^4+4x^3+8x^2+8x+4 & = x^2\left(x^2+\frac{4}{x^2} + 4\left(x + \frac{2}{x}\right) +8\right) \\
 & = x^2\left(\left(x+\frac{2}{x}\right)^2 + 4\left(x + \frac{2}{x}\right) +4\right) \\ 
 & = x^2\left(x+\frac{2}{x} + 2\right)^2 = \;\cdots
\end{align}
$$

[ EDIT ]  The above is based on the observation that the ratio between "symmetric" coefficients of the polynomial $\;\color{blue}1\cdot x^4+\color{red}4\cdot x^3+8\cdot x^2+\color{red}8\cdot x+\color{blue}4 \cdot x^0\;$ follows the powers of $\frac{1}{2}$ from center outwards: $\;\cfrac{8}{8} = \cfrac{1}{2^0}\,, \;\color{red}{\cfrac{4}{8}}=\cfrac{1}{2^1}\,, \;\color{blue}{\cfrac{1}{4}}=\cfrac{1}{2^2}\,$.

This indicates that the polynomial is amenable to a reciprocal polynomial with the substitution $x=\sqrt{2}\,y\,$, and indeed $\color{blue}4\,y^4+\color{red}{8 \sqrt{2}}\,y^3 + 16\,y^2+\color{red}{8\sqrt{2}}\,y+\color{blue}4\,$ is reciprocal.
Lastly, a reciprocal polynomial of even degree $2n$ can be factored as $y^n$ times a polynomial of degree $n$ in $y+\cfrac{1}{y}=\cfrac{1}{\sqrt{2}}\left(x+\cfrac{2}{x}\right)\,$.
This explains why the posted answer started by factoring out $x^2$ then focused on the $x+\cfrac{2}{x}$ terms.
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