Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin. 
Prove that the zeros of the polynomial $P(z)=z^7+7z^4+4z+1$ lie inside the disk of radius $2$ centered at the origin.

Assuming the contrary that there exists an $|z|\geqslant2$ such that $P(z)=0$ I have that $$z^7+7z^4+4z+1 = 0 \implies 1+\frac{7}{z^3}+\frac{4}{z^6} +\frac{1}{z^7}=0 \implies 1+\frac{7}{8}+\frac{4}{64}+\frac{1}{128} = \frac{249}{128}\ne0$$
which is a contradiction so the zeros would lie inside the disk of radius $2$?

If there is a solution, $|z^7|=|7z^4+4Z+1|$ Suppose that $|z|\geq 2$,
$2^7=128>7 \times 2^4 +4 \times  2+1=127$
