Show that the equation, $x^3+10x^2-100x+1729=0$ has at least one complex root $z$ such that $|z|>12.$ Show that the equation,

$x^3+10x^2-100x+1729=0$ has at least one complex root $z$ such that $|z|>12.$

 A: As $\displaystyle1728=12^3$  
and if the $a_1,a_2,a_3$ are the three roots with modulus $\le 12$
we have  $$|a_1a_2a_3|\le 12^3$$ and using Vieta's formula, $$a_1a_2a_3=(-1)^31729\implies |a_1a_2a_3|=+1729 $$ 
A: Let $~α_1,~α_2,~α_3~$ be roots of the equation. (These are complex numbers and existence is guaranteed by the Fundamental Theorem of Algebra). What do we already know about these roots? We know the sum of roots, sum of product of roots taken two at a time, and product of roots. These are expressible by the coefficients of the equation.
Suppose $~~|α_1|≤12,~~|α_2|≤12,~~|α_3|≤12~$.
Then $~|α_1α_2α_3|=|α_1||α_2||α_3|≤123=1728~$
But, the product of roots is $~−1729~$. That is, $~|α_1α_2α_3|=1729~.$
Therefore things go wrong if we suppose $~|α|≤12~$ for all roots $~α~$. Therefore the negation of this statement must be true. What is the negation? Of course, $~≤~$ is replaced by $~>~$ and for all is replaced by for some.
Hence the statement The equation $~x^3+10x^2−100x+1729=0~$ has at least one complex root α such that $~|α|>12~$ is True.
