Question

How do you find all the zeros of `x^3-x^2+2x+1`?

Answer 1

Use Cardano's method to find Real zero:

`x_1 = 1/3(1+root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2))`

and Complex zeros.

Explanation:

Premultiply by `3^3` to cut down on arithmetic involving fractions:

`0 = 3^3 (x^3-x^2+2x+1)`

`=27x^3-27x^2+54x+27`

`=(3x-1)^3+15(3x-1)+43`

Let `t = 3x-1` and solve:

`t^3+15t+43 = 0`

Using Cardano's method, let `t = u + v`

`u^3+v^3+3(uv+5)(u+v)+43 = 0`

Let `v = -5/u` to eliminate the term in `(u+v)`

`u^3-5^3/u^3+43 = 0`

Multiply through by `u^3` to get a quadratic in `u^3`:

`(u^3)^2+43(u^3)-125 = 0`

Use the quadratic formula to find:

`u^3 = (-43+-sqrt(43^2+(4*125)))/2`

`=(-43+-sqrt(1849+500))/2`

`=(-43+-sqrt(2349))/2`

`=(-43+-9sqrt(29))/2`

The derivation was symmetric in `u` and `v`, so we can use one of these roots for `u^3` and the other for `v^3` to find the Real root of our cubic equation in `t` is:

`t_1 = root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2)`

and Complex roots:

`t_2 = omega root(3)((-43+9sqrt(29))/2)+omega^2 root(3)((-43-9sqrt(29))/2)`

`t_3 = omega^2 root(3)((-43+9sqrt(29))/2)+omega root(3)((-43-9sqrt(29))/2)`

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

Then `x= 1/3(t+1)`, hence zeros of the original cubic in `x`:

`x_1 = 1/3(1+root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2))`

`x_2 = 1/3(1+omega root(3)((-43+9sqrt(29))/2)+omega^2 root(3)((-43-9sqrt(29))/2))`

`x_3 = 1/3(1+omega^2 root(3)((-43+9sqrt(29))/2)+omega root(3)((-43-9sqrt(29))/2))`