How do you find all the zeros of g(x)= - 2x^3+5x^2-6x-10g(x)=−2x3+5x2−6x−10?
1 Answer
Use Cardano's method to find Real zero:
x_1 = 1/6(5+root(3)(685+6sqrt(13071))+root(3)(685-6sqrt(13071)))x1=16(5+3√685+6√13071+3√685−6√13071)
and related Complex zeros.
Explanation:
g(x) = -2x^3+5x^2-6x-10g(x)=−2x3+5x2−6x−10
Descriminant
The discriminant
Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd
In our example,
Delta = 900-1728+5000-10800-10800 = -17428
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=-108g(x)=216x^3-540x^2+648x+1080
=(6x-5)^3+33(6x-5)+1370
=t^3+33t+1370
where
Cardano's method
We want to solve:
t^3+33t+1370=0
Let
Then:
u^3+v^3+3(uv+11)(u+v)+1370=0
Add the constraint
u^3-1331/u^3+1370=0
Multiply through by
(u^3)^2+1370(u^3)-1331=0
Use the quadratic formula to find:
u^3=(-1370+-sqrt((1370)^2-4(1)(-1331)))/(2*1)
=(1370+-sqrt(1876900+5324))/2
=(1370+-sqrt(1882224))/2
=(1370+-12sqrt(13071))/2
=685+-6sqrt(13071)
Since this is Real and the derivation is symmetric in
t_1=root(3)(685+6sqrt(13071))+root(3)(685-6sqrt(13071))
and related Complex roots:
t_2=omega root(3)(685+6sqrt(13071))+omega^2 root(3)(685-6sqrt(13071))
t_3=omega^2 root(3)(685+6sqrt(13071))+omega root(3)(685-6sqrt(13071))
where
Now
x_1 = 1/6(5+root(3)(685+6sqrt(13071))+root(3)(685-6sqrt(13071)))
x_2 = 1/6(5+omega root(3)(685+6sqrt(13071))+omega^2 root(3)(685-6sqrt(13071)))
x_3 = 1/6(5+omega^2 root(3)(685+6sqrt(13071))+omega root(3)(685-6sqrt(13071)))