How to solve an equation with 6 degree polynomial? Does anyone have an idea to solve the following equation if it is possible? It's the best to get analytic solution, but if you can help me to show when equation has all real root with certain conditions of parameters, it is also alright. 
Variable is $\Omega$, and parameters are $M_1 >0$ and $r >0, r \neq 1$.
I am trying to find when it has all real roots.
$f(\Omega)=(r-1)\Omega^6 - (r+1) \Omega^5 - \frac{(r-1)}{4} \Omega^4 + \left(\frac{(r+1)}{2}+ \frac{4}{M_1^2}\right)\Omega^3 -\frac{(r-1)}{16} \Omega^2 + \left(\frac{1}{M_1^2}- \frac{r+1}{16}\right)\Omega + \frac{(r-1)}{64}=0.$
Thank you very much for any help!
 A: The following is a summary and expansion of some of my comments above, and by no means a complete answer to the question.

If $r<1$ then by Descartes' rule of signs $f$ has either $2$ or $0$ positive real roots, and either $4$, $2$ or $0$ negative real roots, and  clearly $\Omega=0$ is not a root. It is not true in general that $f$ has at least $4$ roots.
If $r\neq1$ then setting $t:=\frac{1}{16}+\tfrac{1}{8(r-1)}$ and $s=\tfrac{1}{M_1^2}$ we might as well find the roots of 
\begin{eqnarray*}
\tfrac{f(x)}{r-1}&=&x^6-\left(1+\tfrac{2}{r-1}\right)x^5-\tfrac14x^4
+\left(\tfrac12+\tfrac{1}{r-1}+\tfrac{4}{M_1^2}\right)x^3-\tfrac{1}{16}x^2 + \left(\tfrac{1}{M_1^2}-\tfrac{1}{16}-\tfrac{1}{8(r-1)}\right)x+\tfrac{1}{16}\\
&=&x^6-16tx^5-\tfrac14x^4+(8t+4s)x^3-\tfrac{1}{16}x^2+(s-t)x+\tfrac{1}{16},
\end{eqnarray*}
with the conditions that $s>0$ and $t\notin[-\tfrac{1}{16},\tfrac{1}{16}]$, or even the nicer looking
$$\frac{64f(\tfrac x2)}{r-1}=x^6-32tx^5-x^4+32(s+2t)x^3-x^2+32(s-t)x+4.$$
Then if $t<0$ and $s>-t$ then most coefficients are positive; setting the above equal to $0$ yields
$$x^6+32(-t)x^5+32(s+2t)x^3+32(s+(-t))x+4=x^4+x^2,$$
where the terms are sorted left and right so that all coefficients are positive. But it is not hard to check that for all real $x$ we have
$$x^6+4>x^4+x^2,$$
so $f$ has no positive real roots if $t<0$ and $s>-t$, or equivalently $r<1$ and
$$\frac{1}{M_1^2}>-\left(\frac{1}{16}+\frac{1}{8(r-1)}\right),$$
which is in turn equivalent to $M_1^2>16\frac{1-r}{1+r}$.
Similarly, if $t>0$ and $s<t$ then all coefficients in the equation
$$(-x)^6+32t(-x)^5+32(s-t)(-x)+4=(-x)^4+32(s+2t)(-x)^3+(-x)^2,$$
are positive. If $c<1.94316$ then for all real $x$ we have
$$x^6+4>x^4+cx^3+x^2,$$
so if $32(s+2t)<1.94316$ then the above has no real solutions, so $f$ has no negative real roots. In particular, by Descartes' rule of signs it has at most $2$ real roots. 
EDIT: The inequality $32(s+2t)<1.94316$ is not satisfied as $t>0$ implies $t>\tfrac{1}{16}$. This shows that $f$ has at least two negative real roots if $r>1$.
