<s>[INST] Interscholastic Mathematics League Senior B #12 Compute the product of the nonreal roots of the equation $x^4+4x^3+6x^2+1004x+1001=0$.
So here is what I have done so far. I got two of the roots to be zero and 4 since $x^3(x+4)+(6x^2+1004x+1001)=0$. But for the other two roots by using the quadratic formula I get $$\frac{-1004 \pm \sqrt{1004^2-24024}}{12}.$$ This contest was already completed but I couldn't solve that  quadratic equation during the competition. Can someone please post a easier way to solve this problem?
 [/INST] $4$ is not a root by the rational root theorem and $0$ is not a root since the polynomial is $1001$ at $x=0$.  We have $(x+1)^4+1000(x+1)=0$, so one real root is $-1$.  Taking that out we have a sum of cubes, $(x+1)^3+10^3=0$, which factors as $(x+11)((x+1)^2-10(x+1)+100)=0$.  Eliminating the real root $-11$ we are left with $(x+1)^2-10(x+1)+100=x^2-8x+91=0$.  The product of the roots of a monic quadratic is the constant term, so the product of the complex roots is $91$.  To see that the product of roots is the constant term, expand $(x-a)(x-b)$
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