What are all the zeroes of #f(x) = 2x^3 - 5x^2 + 3x - 1#?

Given:

#f(x) = 2x^3-5x^2+3x-1#

First substitute #t = x-5/6#

Then:

#2t^3 = 2(x-5/6)^3 = 2(x^3-5/2x^2+25/12x-125/216)#

#=2x^3-5x^2+25/6x-125/108#

#color(white)()#

#-7/6t = -7/6x+35/36 = -7/6x+105/108#

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#2t^3-7/6t = 2x^3-5x^2+3x-5/27#

So:

#2t^3-7/6t-22/27 = 2x^3-5x^2+3x-1 = f(x)#

Multiply through by #54# to get integer coefficients:

#108t^3-63t-44 = 54 f(x)#

So we want to solve:

#108t^3-63t-44 = 0#

Use Cardano's method, letting #t = u + v#

#0 = 108(u+v)^3-63(u+v)-44#

#= 108u^3+108v^3+(324uv-63)(u+v)-44#

#= 108u^3+108v^3+9(36uv-7)(u+v)-44#

Next make the coefficient of the #(u+v)# term zero by adding the constraint: #36uv-7 = 0#, that is #v = 7/(36u)#

#= 108u^3+108(7/(36u))^3-44#

#= 108u^3+343/(432u^3)-44#

Multiply through by #432 u^3# to get a quadratic in #u^3#:

#46656(u^3)^2-19008(u^3)+343 = 0#

Then using the quadratic formula:

#u^3 = (19008+-sqrt(19008^2-4*46656*343))/(2*46656)#

#= (19008+-sqrt(361304064-64012032))/93312#

#= (19008+-sqrt(297292032))/93312#

#= (19008+-1296 sqrt(177))/93312#

#= (44+-3sqrt(177))/216#

Since the derivation was symmetric in #u# and #v#, one of these roots is suitable for #u^3# and the other for #v^3#.

Hence the Real root is:

#t = root(3)((44+3sqrt(177))/216)+root(3)((44-3sqrt(177))/216)#

#= 1/6 (root(3)(44+3sqrt(177))+root(3)(44-3sqrt(177)))#

and hence:

#x_1 = 5/6 + t = 1/6 (5 + root(3)(44+3sqrt(177))+root(3)(44-3sqrt(177)))#

The Complex roots are:

#x_2 = 1/6 (5 + omega root(3)(44+3sqrt(177))+omega^2 root(3)(44-3sqrt(177)))#

#x_3 = 1/6 (5 + omega^2 root(3)(44+3sqrt(177))+omega root(3)(44-3sqrt(177)))#

where #omega = -1/2 +sqrt(3)/2 i# is the primitive Complex cube root of #1#.