<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] We are solving the equation
$$x^4+4x^3+6x^2+1004x+1001=0.$$ 
Note that
$$(x+1)^4=x^4+4x^3+6x^2+4x+1.$$
Interesting! So we are looking at
$$(x+1)^4 +1000x+1000=0.$$
Let $y=x+1$. Our equation can be rewritten as $y^4+1000y=0$. The real roots are $y=0$ and $y=-10$, and the other two roots are non-real.  Now it's essentially over.
Added: For completeness, note that two of the roots are $x=-1$ and $x=-11$. The product of the roots of the original equation is $1001$. So the product of the two non-real roots is $1001/11$.
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