Question

How do you find all the zeros of `f(x)= -4x^8-9x^7+9x-4` with its multiplicities?

Answer 1

See explanation...

Explanation:

This seems a strangely difficult problem to be given, but we can at least reduce it from an octic to a quartic, using the semi-symmetry of the coefficients...

`f(x) = -4x^8-9x^7+9x-4`

Then:

`f(iz) = -4(iz)^8-9(iz)^7+9(iz)-4`

`=-4z^8+9iz^7+9iz-4`

`=z^4(-4z^4+9iz^3+9iz^(-3)-4z^(-4))`

Let `t = z+1/z`

`f(iz)/z^4 = -4t^4+9it^3+16t^2-27it-8`

This quartic is not very simple to solve, but it is at least possible to solve algebraically. Each zero `t_1, t_2, t_3, t_4` of this quartic gives you a quadratic in `z` to solve and hence find the zeros of the original octic polynomial.

For example, one of the zeros of this quartic is:

`t = (9 i)/16-1/16 sqrt(1/3 (269-1424/(3 sqrt(29534277)-16282)^(1/3)+16 (3 sqrt(29534277)-16282)^(1/3)))-1/2 sqrt(269/96+89/(12 (3 sqrt(29534277)-16282)^(1/3))-1/12 (3 sqrt(29534277)-16282)^(1/3)+(1881 i)/(32 sqrt(1/3 (269-1424/(3 sqrt(29534277)-16282)^(1/3)+16 (3 sqrt(29534277)-16282)^(1/3)))))`