Bet you can't bash this out!

The problem is a simple application of Descartes' rule of signs

Lets take this poly to be $f(x)=2014x^{2014}+3021x^{1006}-4028x$ .

We find that it can be factored to $1007x(2x^{2013}+3x^{1005}-4).$ Thus one root is $x=0$ .

Let $g(x)=2x^{2013}+3x^{1005}-4$

$\bullet$ $g(x)$ has $1$ sign change, in the coefficient changing from $3$ to $-4$ . Thus $g(x)$ has 1 positive root and including 0, there are 2 $\implies p=2$

$\bullet$ $g(-x)=-2x^{2013}-3x^{1005}-4$ , this has $0$ sign changes, to give us $0$ negative roots.

$\bullet$ Thus it has $2012$ complex roots. (2014 total roots-2 positive real roots) $\implies z=2012$

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Now all we gotta do is to simply simplify the above expression $12072^{2014} \pmod {10000}$ which is a piece of cake using Euler's totient theorem.

Answer is $\boxed{2944}$

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