# Math Help - hard problem

1. ## hard problem

Let $p(x)=x^3+ax^2+bx+c$, where a, b, and c are complex numbers. Suppose that $p(2009+9002{\pi}i)=p(2009)=p(9002)=0$

What is the number of nonreal zeros of $x^12+ax^8+bx^4+c$?

That last part is x^12 btw

2. $x^{12}+ax^8+bx^4+c=p(x^4)$

The real zeros of the equation are $\pm\sqrt[4]{2009}, \ \pm\sqrt[4]{9002}$

So, there are 8 nonreal zeros.