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

1 Answer
Apr 16, 2016

Answer:

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)))))#