Number of roots of $z^6-5z^2+8z+2$ in closed unit disk This problem, of course, screams Rouche's Theorem. Unfortunately, the coefficients were chosen in such a way that getting the strict inequality necessary for application of the theorem is not easy. 
Since we have a disk of radius less than or equal to 1, I know the general thing to do is to compare the polynomial with term with highest coefficient. So we'd look at $f(z) = z^6-5z^2+8z+2$ and $g(z) = 8z$. (Indeed, wolframalpha tells us there is 1 root in the unit disk so this seems like the right way to go about it). 
Then for all $|z|=1$, $$ |f(z)-g(z)| \le 1+5+2 = 8 = |g(z)|,$$
which is not enough. I've also considered looking at disks of radius $1-\epsilon$, but that doesn't seem to work either. Any suggestions? (Hopefully I'm not missing something super obvious)
 A: We have that
$$|f(z)-g(z)|\leq |z^6+1|+|-5z^2+1|$$
Now note that for $|z|=1$, $|-5z^2+1|\leq 6$ and equality holds iff $z^2=-1$ that is $z=\pm i$ which implies that $|z^6+1|=|(-1)^3+1|=0<2$.
Hence, for $|z|=1$,
$$|f(z)-g(z)|\leq |z^6+1|+|-5z^2+1|<2+6=8=|g(z)|.$$
A: Use the symmetric version of Rouché: $f$ and $g$ have the same number of roots inside the closed contour $\Gamma$ if
$$ |f(z) - g(z)| < |f(z)| + |g(z)|$$
on $\Gamma$.  In this case, on  the unit circle $\Gamma$,  $|f(z) - g(z)| \le 8 = |g(z)|$ so it's enough to show $f(z) \ne 0$ on $\Gamma$. 
For $f(z) = 0$ with $z \in \Gamma$ we'd need $z^6$,  $- 5 z^2$  and  $2$ to have the same argument, which is the opposite to the argument of $8z$.  For the arguments of $2$ and $8z$ to be opposite,  $8z$ must be negative real, but then $-5 z^2$ is also a negative real, so this can't happen.
A: Adding a factor with a known root at $z=\frac14$ to balance the root close to $-\frac14$ resulting from the binomial  $8z+4$ of the lowest order terms gives the modified polynomial
$$
(4z-1)f(z)=(4z-1)(z^6−5z^2+8z+2)=4z^7-z^6-20z^3+37z^2-2
$$
which now has a clearly dominant quadratic term.
A: The strict inequality is satisfied for $|z|=\frac{1}{2}$ which is well inside the given contour
