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 $k^{4}+4k^{3}+8k^{2}+8k+4=0$.

• Plotting the function shows that it has no roots in $\mathbb R$. It's global minimum is at $k=-1$. Aug 6 '15 at 18:38
• It has roots in C, k1/2=k3/4=-1+-i, but question is how to get them by hand. This is solution using software :)
– etf
Aug 6 '15 at 18:42
• $$k^4+4k^3+8k^2+8k+4 =(k^2+2k+2)^2=((k+1)^2+1)^2$$ and pass for complex numbers if you want to Aug 6 '15 at 18:42
• @Elaqqad Nice catch. You should turn that into an answer. OP just said that he wanted to find the complex roots. Aug 6 '15 at 18:44

The equation does not have real roots because: $$k^4+4k^3+8k^2+8k+4 =(k^2+2k+2)^2=((k+1)^2+1)^2\geq 1> 0$$ But this equation shows that we can find its complex roots : $$k^4+4k^3+8k^2+8k+4 =((k+1)^2+1)^2 = (k+1+i)^2(k+1-i)^2$$ which gives the complete factorization of the equation and its complex roots.
A nice trick is to notice that with the substitution $k\to k-1$ we get the polynomial $k^4+2k^2+1$ that is just the square of $k^2+1$.
• And this is a natural substitution to try because it automatically removes the $k^3$ term and one would then look for an expression as the difference of two squares - except that's not necessary here. Aug 6 '15 at 18:51