How do you find all the zeros of x^3-4x^2+3x-1x3−4x2+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))13⎛⎝4+3√−47+3√932+3√−47−3√932⎞⎠
and Complex conjugate pair of related zeros.
Explanation:
f(x) = x^3-4x^2+3x-1f(x)=x3−4x2+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))