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

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    $\begingroup$ Plotting the function shows that it has no roots in $\mathbb R$. It's global minimum is at $k=-1$. $\endgroup$
    – AlexR
    Aug 6 '15 at 18:38
  • $\begingroup$ 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 :) $\endgroup$
    – etf
    Aug 6 '15 at 18:42
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    $\begingroup$ $$k^4+4k^3+8k^2+8k+4 =(k^2+2k+2)^2=((k+1)^2+1)^2$$ and pass for complex numbers if you want to $\endgroup$
    – Elaqqad
    Aug 6 '15 at 18:42
  • $\begingroup$ @Elaqqad Nice catch. You should turn that into an answer. OP just said that he wanted to find the complex roots. $\endgroup$
    – AlexR
    Aug 6 '15 at 18:44
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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*}

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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.

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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 $k^2+1$.

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    $\begingroup$ 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. $\endgroup$ Aug 6 '15 at 18:51

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