Compute the product of the nonreal roots of

ruililorNokmncf

ruililorNokmncf

Answered question

2022-03-28

Compute the product of the nonreal roots of the equation
x4+4x3+6x2+1004x+1001=0

Answer & Explanation

paganizaxpo3

paganizaxpo3

Beginner2022-03-29Added 8 answers

We are solving the equation
x4+4x3+6x2+1004x+1001=0
Note that
(x+1)4=x4+4x3+6x2+4x+1
So we are looking at
(x+1)4+1000x+1000=0
Let 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's essentially over.
cineworld93uowb

cineworld93uowb

Beginner2022-03-30Added 16 answers

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)210(x+1)+100)=0. Eliminating the real root -11 we are left with (x+1)210(x+1)+100=x28x+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 (xa)(xb)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?