How do you find all the zeros of # x^3-4x^2+3x-1#?
1 Answer
Use Cardano's method to find Real zero:
#1/3(4+root(3)((-47+3sqrt(93))/2)+root(3)((-47-3sqrt(93))/2))#
and Complex conjugate pair of related zeros.
Explanation:
#f(x) = x^3-4x^2+3x-1#
Descriminant
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 144-108-256-27+216 = -31#
Since
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^3-108x^2+81x-27#
#=(3x-4)^3-21(3x-4)-47#
#=t^3-21t-47#
where
Cardano's method
We want to solve:
#t^3-21t-47=0#
Let
Then:
#u^3+v^3+3(uv-7)(u+v)-47=0#
Add the constraint
#u^3+343/u^3-47=0#
Multiply through by
#(u^3)^2-47(u^3)+343=0#
Use the quadratic formula to find:
#u^3=(47+-sqrt((-47)^2-4(1)(343)))/(2*1)#
#=(-47+-sqrt(2209-1372))/2#
#=(-47+-sqrt(837))/2#
#=(-47+-3sqrt(93))/2#
Since this is Real and the derivation is symmetric in
#t_1=root(3)((-47+3sqrt(93))/2)+root(3)((-47-3sqrt(93))/2)#
and related Complex roots:
#t_2=omega root(3)((-47+3sqrt(93))/2)+omega^2 root(3)((-47-3sqrt(93))/2)#
#t_3=omega^2 root(3)((-47+3sqrt(93))/2)+omega root(3)((-47-3sqrt(93))/2)#
where
Now
#x_1 = 1/3(4+root(3)((-47+3sqrt(93))/2)+root(3)((-47-3sqrt(93))/2))#
#x_2 = 1/3(4+omega root(3)((-47+3sqrt(93))/2)+omega^2 root(3)((-47-3sqrt(93))/2))#
#x_3 = 1/3(4+omega^2 root(3)((-47+3sqrt(93))/2)+omega root(3)((-47-3sqrt(93))/2))#