Roots of $z^5 + {\sqrt2}z^3 + 1 = 0$ What would be a good strategy to tackle this problem?
$$z^5 + {\sqrt2}z^3 + 1 = 0$$ where $z \in \Bbb C$.
This question comes out of my complex numbers text on finding roots of polynomials. I have tried to brute force this problem by factorising, however the $1$ hanging on the end leaves me confused.
Another approach I tried and couldn't work through any further was to express each complex variable in terms of the exponential function, $z=re^{i\theta}$ where $r$ is the modulus and $e^{i\theta}$ is our argument. 
Thus the above problem could be re-written as:
$$r^5e^{5i\theta} + \sqrt{2}r^3e^{3i\theta} + 1 = 0$$
I hope that helps give some context.
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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We can easily see that
  $$
\bbox[0.5em,#efe,border:0.1em groove navy]{\
r_{1} \equiv -0.78891677373356634396694435041559932834090342558\
}
$$
  is a 'good approximation' to the real root of
  $\ds{z^{5} + \root{2}z^{3} + 1 = 0}$.


Then, a 'fine approximation' is given by:
\begin{align}
z^{5} + \root{2}z^{3} + 1 & =
\pars{z - r_{1}}\pars{z^{4} + bz^{3} + cz^{2} + dz - {1 \over r_{1}}}
\\[5mm] & =
z^{5} + \pars{b - r_{1}}z^{4}+ \pars{c - br_{1}}z^{3}+ \pars{d - cr_{1}}z^{2}+ \pars{dr_{1} - {1 \over r_{1}}}z + 1
\end{align}
Now, we have an equation which can be solved analytically. Namely,
$$
z^{4} + r_{1}z^{3} + \pars{r_{1}^{2} + \root{2}}z^{2} + r_{1}\pars{r_{1}^{2} + \root{2}}z - {1 \over r_{1}} = 0
$$
