What are all the zeroes of #f(x)=5x^3+8x^2-4x+3#?

1 Answer
Nov 22, 2015

Answer:

The Real root is:

#x_1 = -8/15+root(3)((-4489+45sqrt(6185))/6750)+root(3)((-4489-45sqrt(6185))/6750)#

#~~ -2.11299545715260#

Complex roots as shown below.

Explanation:

#f(x)=5x^3+8x^2-4x+3#

This is of the form #ax^3+bx^2+cx+d# with #a=5#, #b=8#, #c=-4# and #d=3#

Its discriminant #Delta# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

#=1024+1280-6144-6075-8640=-18555#

Since #Delta < 0# this cubic has one Real zero and two Complex conjugate ones.

So we can use a Tschirnhaus transformation to eliminate the #x^2# term, then use Cardano's method.

Let #t = x+8/15#

Then:

#5t^3 = 5(x+8/15)^3 = 5x^3+8x^2+64/15x+512/675#

#-124/15t = -124/15(x+8/15) = -124/15x-992/225#

Hence:

#5t^3-124/15t+4489/675 = 5x^3+8x^2-4x+3#

Multiplying through by #675#, we want to solve:

#3375t^3-5580t+4489 = 0#

Let #t=u+v#

#3375u^3+3375v^3+(10125uv-5580)(u+v)+4489=0#

To eliminate the #(u+v)# term add the constraint:

#v = 5580/(10125u) = 124/(225u)#

Then we have:

#0 = 3375u^3+3375(124/(225u))^3+4489#

#=3375u^3+3375(1906624/(11390625u^3))+4489#

#=3375u^3+1906624/(3375u^3)+4489#

Multiply through by #3375u^3# to get:

#11390625(u^3)^2+15150375(u^3)+1906624 = 0#

Then by the quadratic formula:

#u^3 = (-15150375+-sqrt(15150375^2-(4*11390625*1906624)))/(2*11390625)#

#= (-15150375+-sqrt(142663306640625))/22781250#

#= (-15150375+-151875sqrt(6185))/22781250#

#= (-4489+-45sqrt(6185))/6750#

Since the derivation is symmetric in #u# and #v#, we can deduce:

#t = root(3)((-4489+45sqrt(6185))/6750)+root(3)((-4489-45sqrt(6185))/6750)#

#x_1 = -8/15+root(3)((-4489+45sqrt(6185))/6750)+root(3)((-4489-45sqrt(6185))/6750)#

The Complex roots are:

#x_2 = -8/15+omega root(3)((-4489+45sqrt(6185))/6750)+omega^2 root(3)((-4489-45sqrt(6185))/6750)#

#x_3 = -8/15+omega^2 root(3)((-4489+45sqrt(6185))/6750)+omega root(3)((-4489-45sqrt(6185))/6750)#

where #omega = -1/2+sqrt(3)/2i#