Factor $k^{4}+4k^{3}+8k^{2}+8k+4=0$ over $\mathbb C$ Any idea how to factor the polynomial $k^{4}+4k^{3}+8k^{2}+8k+4$ over $\mathbb C$?
Candidates for rational roots are $\pm1, \pm2, \pm4$ but none of them satisfies $k^{4}+4k^{3}+8k^{2}+8k+4=0$.

Note that
\begin{align*}
k^4 + 4k^3 + 8k^2 + 8k + 4
&= (k^4 + 4k^3 + 6k^2 + 4k + 1) + (2k^2 + 4k + 2) + 1 \\
&= (k + 1)^4 + 2(k+1)^2 + 1 \\
&\ge 1 \text{ for all } k,
\end{align*}
so the polynomial has no real roots.
However, it does factor into two quadratics.
From the above,
\begin{align*}
k^4 + 4k^3 + 8k^2 + 8k + 4
&= (k + 1)^4 + 2(k+1)^2 + 1 \\
&= \big[(k+1)^2 + 1\big]^2 \\
&= (k^2 + 2k + 2)^2.
\end{align*}
