# 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?

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 If you only need the product, you won't have to take the square root... – AMPerrine Oct 27 '11 at 3:47

$4$ 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 $(x-a)(x-b)$
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 $1001/11$.