How do you factor completely #x^3 – 2x^2 – 4x – 8#?

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

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Discriminant

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=-2#, #c=-4# and #d=-8#, so we find:

#Delta = 64+256-256-1728-1152 = -2816#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=27f(x)=27x^3-54x^2-108x-216#

#=(3x-2)^3-48(3x-2)-304#

#=t^3-48t-304#

where #t=(3x-2)#

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Cardano's method

We want to solve:

#t^3-48t-304=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-16)(u+v)-304=0#

Add the constraint #v=16/u# to eliminate the #(u+v)# term and get:

#u^3+4096/u^3-304=0#

Multiply through by #u^3# and rearrange slightly to get:

#(u^3)^2-304(u^3)+4096=0#

Use the quadratic formula to find:

#u^3=(304+-sqrt((-304)^2-4(1)(4096)))/(2*1)#

#=(304+-sqrt(92416-16384))/2#

#=(304+-sqrt(76032))/2#

#=(304+-48sqrt(33))/2#

#=152+-24sqrt(33)#

#=2^3*(19+-3(33))#

Since this is Real and the derivation is symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find Real root:

#t_1=2root(3)(19+3sqrt(33))+2root(3)(19-3sqrt(33))#

and related Complex roots:

#t_2=2omega root(3)(19+3sqrt(33))+2omega^2 root(3)(19-3sqrt(33))#

#t_3=2omega^2 root(3)(19+3sqrt(33))+2omega root(3)(19-3sqrt(33))#

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

Now #x=1/3(2+t)=2/3(1+t/2)#. So the zeros of our original cubic are:

#x_1 = 2/3(1+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))#

#x_2 = 2/3(1+omega root(3)(19+3sqrt(33))+omega^2 root(3)(19-3sqrt(33)))#

#x_3 = 2/3(1+omega^2 root(3)(19+3sqrt(33))+omega root(3)(19-3sqrt(33)))#