Let $\displaystyle p(x)=x^3+ax^2+bx+c$, where a, b, and c are complex numbers. Suppose that $\displaystyle p(2009+9002{\pi}i)=p(2009)=p(9002)=0$
What is the number of nonreal zeros of $\displaystyle x^12+ax^8+bx^4+c$?
That last part is x^12 btw
Let $\displaystyle p(x)=x^3+ax^2+bx+c$, where a, b, and c are complex numbers. Suppose that $\displaystyle p(2009+9002{\pi}i)=p(2009)=p(9002)=0$
What is the number of nonreal zeros of $\displaystyle x^12+ax^8+bx^4+c$?
That last part is x^12 btw