Compute the product of the nonreal roots of the equationx4+4x3+6x2+1004x+1001=0

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paganizaxpo3 · Accepted Answer

We are solving the equationx4+4x3+6x2+1004x+1001=0Note that(x+1)4=x4+4x3+6x2+4x+1So we are looking at(x+1)4+1000x+1000=0Let y=x+1. Our equation can be rewritten as y4+1000y=0. The real roots are y=0 and y=−10, and the other two roots are non-real. Now it&#039;s essentially over.

cineworld93uowb · Accepted Answer

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+103=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=x2−8x+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)

Compute the product of the nonreal roots of

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