How do you find the zeros of #f(z) = z^4+2z^35z^27z+20# ?
1 Answer
Use a numerical method to find approximations to the zeros:
#x_(1,2) ~~ 1.49358+0.773932i#
#x_(3,4) ~~ 2.49358+0.921833i#
Explanation:
#f(z) = z^4+2z^35z^27z+20#
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+1# ,#+2# ,#+4# ,#+5# ,#+10# ,#+20#
None of these works (in fact they all give positive values), so
This is a fairly typical quartic with two complex conjugate pairs of roots. It is possible to solve algebraically, but gets very messy.
The way I would approach it algebraically would be:

Use a Tschirnhaus transformation to derive a simplified quartic of the form:
#t^4+pt^2+qt+r = 0# 
Consider a factorisation of this simplified quartic of the form
#(t^2at+b)(t^2+at+c)# 
Equate coefficients and use
#(b+c)^2 = (bc)^2+4bc# to get a cubic in#a^2# 
If the resulting cubic has three Real irrational roots then use a
#t = k cos theta# substitution to find the solutions in trigonometric form. Otherwise use Cardano's method to find the one Real zero. 
Use the positive zero of the cubic as
#a^2# and the positive square root of that as#a# . 
Derive
#b# and#c# to get two quadratics to solve. 
Solve the quadratics using the quadratic formula.
As I say, this (usually) gets rather messy.
Alternatively we can use a numerical method such as Durand Kerner to find approximations to the zeros:
#x_(1,2) ~~ 1.49358+0.773932i#
#x_(3,4) ~~ 2.49358+0.921833i#
Here's the C++ program I used (implementing the Durand Kerner algorithm)...