What are the zeros of #f(x) = x^3-140x^2+7984x-107584#?

2 Answers
George C.
Nov 2, 2016

Use Cardano's method to find Real zero:

#x_1 = 1/3(140+8 root(3)(1628+12sqrt(20589))+8 root(3)(1628-12sqrt(20589)))#

and two related Complex zeros.

Explanation:

Given:

#f(x) = x^3-140x^2+7984x-107584#

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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=-140#, #c=7984# and #d=-107584#, so we find:

#Delta = 1249387417600-2035736559616-1180841984000-312506560512+2164555653120 = -115142033408#

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^3-3780x^2+215568x-2904768#

#=(3x-140)^3+13056(3x-140)+1667072#

#=t^3+13056t+1667072#

where #t=(3x-140)#

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Cardano's method

We want to solve:

#t^3+13056t+1667072=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv+4352)(u+v)+1667072=0#

Add the constraint #v=-4352/u# to eliminate the #(u+v)# term and get:

#u^3-82426462208/u^3+1667072=0#

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

#(u^3)^2+1667072(u^3)-82426462208=0#

Use the quadratic formula to find:

#u^3=(-1667072+-sqrt((1667072)^2-4(1)(-82426462208)))/(2*1)#

#=(1667072+-sqrt(2779129053184+329705848832))/2#

#=(1667072+-sqrt(3108834902016))/2#

#=(1667072+-12288sqrt(20589))/2#

#=833536+-6144sqrt(20589)#

#=8^3(1628+-12sqrt(20589))#

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=8 root(3)(1628+12sqrt(20589))+8 root(3)(1628-12sqrt(20589))#

and related Complex roots:

#t_2=8 omega root(3)(1628+12sqrt(20589))+8 omega^2 root(3)(1628-12sqrt(20589))#

#t_3=8 omega^2 root(3)(1628+12sqrt(20589))+8 omega root(3)(1628-12sqrt(20589))#

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

Now #x=1/3(140+t)#. So the roots of our original cubic are:

#x_1 = 1/3(140+8 root(3)(1628+12sqrt(20589))+8 root(3)(1628-12sqrt(20589)))#

#x_2 = 1/3(140+8 omega root(3)(1628+12sqrt(20589))+8 omega^2 root(3)(1628-12sqrt(20589)))#

#x_3 = 1/3(140+8 omega^2 root(3)(1628+12sqrt(20589))+8 omega root(3)(1628-12sqrt(20589)))#

Douglas K.
Nov 2, 2016

#x ~~ 18.9, x ~~ 60 + 45i, and x ~~ 60 - 45i#

Explanation:

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