What are the zeros of #x^3-8x-4#?

#f(x) = x^3-8x-4#

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Descriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

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

In our example, #a=1#, #b=0#, #c=-8# and #d=-4#, so we find:

#Delta = 0+2048+0-432+0 = 1616#

Since #Delta > 0# this cubic has #3# Real zeros.

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Trigonometric method

We use a substitution #x = k cos theta#, where #k# is chosen to squeeze #f(x)# into a form containing #4cos^3 theta - 3 cos theta = cos 3theta#.

Let #k=4/3 sqrt(6)#

Then:

#0 = f(x) = x^3-8x-4#

#=(k cos theta)^3 - 8 (k cos theta) - 4#

#=k (k^2 cos^3 theta - 8 cos theta) - 4#

#=4/3 sqrt(6) (32/3 cos^3 theta - 8 cos theta) - 4#

#= 32/9 sqrt(6) (4 cos^3 theta - 3 cos theta) - 4#

#= 32/9 sqrt(6) cos 3 theta - 4#

Add #4# to both ends and transpose to get:

#32/9 sqrt(6) cos 3 theta = 4#

Multiply both sides by #sqrt(6)# to get:

#64/3 cos 3 theta = 4 sqrt(6)#

Multiply both sides by #3/64# to get:

#cos 3 theta = 3/16 sqrt(6)#

So:

#3 theta = +-cos^(-1)(3/16 sqrt(6))+2kpi#

So:

#theta = +-1/3 cos^(-1)(3/16 sqrt(6))+(2kpi)/3#

So:

#cos theta = cos(1/3 cos^(-1)(3/16 sqrt(6))+(2kpi)/3)#

This takes three distinct values, for which we can use #k=0, 1, 2#.

Hence zeros of our original cubic:

#x_k = 4/3 sqrt(6) cos(1/3 cos^(-1)(3/16 sqrt(6))+(2kpi)/3) " " k = 0,1,2#

#x_0 ~~ 3.051374241731#

#x_1 ~~ -2.534070196723#

#x_2 ~~ -0.517304045008#