Show that $z^6 + 5z^4 - z^3 + 3z$ has at least two real roots given that all roots are distinct. 
Show that $z^6 + 5z^4 - z^3 + 3z$ has at least two real roots given
  that all roots are distinct. Also, show that $|3z - z^3 + 5z^4| &lt |z^6|$ when $|z| > 3$.

I can see that 0 is a real root; however, I am having trouble starting this one. I couldn't seem to see a way to factorize this. Other than testing points in the derivative and maybe using the intermediate value theorem, I don't know what direction to take on this one.
 A: Let $P(z) = z^6+5z^4-z^3+3z$, then $P(-1/2) = -67/64$, but the highest power of $P$ is even with positive coefficient, so $\lim_{x\to+\infty}P(x) = +\infty$ and $\lim_{x\to-\infty}P(x) = +\infty$, so by the intermediate value property of continuous functions $P$ possess at least two real roots.
Edit: And for the second part, looks like triangle inequality suffices.
A: You have the root $z=0$, so any other root would also be a root of $z^5 + 5z^3 - z^2 + 3$.  Call that expression $g(z)$.
$g(0)=+3$ while $g(-3)= -384$ so by continuity and the intermediate value theorem there is a real root of $g(z)$ in $(-3,0)$, and this will be a real root of your original expression.
Meanwhile for $|z| \gt 3$ you have $$|3z - z^3 + 5z^4| \le  3|z| +|z^3| +5|z^4| \lt \left(\frac{3}{3^5}+\frac{1}{3^3}+\frac{5}{3^2}\right)|z^6| \lt |z^6|.$$
A: Since the coefficients are real, all roots have conjugate roots as well. This means that excluding the $z=0$ root, you have five other distinct roots, so at least one of them is real. So we have at least two real roots.
A: $P(-\frac{1}{2})=- \frac{67}{64}$ and $P(-1)=4$.  Therefore the other real root is in the interval $(-1,-\frac{1}{2})$. In any case, when you factor out zero you get a polynomial of fifth degree. A polynomial of odd degree has at least one real root.  That root in our case is not zero therefore it has at least two distinct real roots.
A: The derivative at zero is non-zero, so it must be strictly negative at some point $a$ close to $0$. Also, the polynomial tends to $+\infty$ in both directions. So it must have another root, with the same sign as $a$.
A: For the first part of the question:
We do not need to know that the polynomial has distinct roots. $0$ is an obvious simple root, dividing by $z$ gives a polynomial of degree 5, which must have a (clearly different) real root by the intermediate value theorem -it tends to both $\pm\infty$.
nbubis' proof is very nice but uses algebraic closure of $\mathbb C$ -the fundamental theorem of algebra. The intermediate value theorem is simpler to prove.
EDIT: Well I should have read Henry's answer, mine overlaps alot with his. The main thing I add perhaps is that we do not need the distinct roots assumption.
