Question

How do you find all the zeros of `4x^3-2x^2-x+6`?

Answer 1

Use Cardano's method to find Real zero:

`x_1 = 1/6(1+root(3)((-313+3sqrt(10857))/2) + root(3)((-313-3sqrt(10857))/2))`

and related Complex zeros.

Explanation:

`f(x) = 4x^3-2x^2-x+6`

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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=4`, `b=-2`, `c=-1` and `d=6`, so we find:

`Delta = 4+16+192-15552+864=-14476`

Since `Delta < 0` this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

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Tschirnhaus transformation

To simplify the problem, eliminate any square term using a linear substitution. This is a simple form of what is called a Tschirnhaus transformation. To reduce the amount of arithmetic involving fractions, first multiply by `2*3^3 = 54` ...

`54 f(x) = 216x^3-108x^2-54x+324`

`=(6x-1)^3-12(6x-1)+313`

`=t^3-12t+313`

where `t = 6x-1`

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Cardano's method

To solve `t^3-12t+313 = 0`, use the substitution `t=u+v` to get:

`u^3+v^3+3(uv-4)(u+v)+313 = 0`

To eliminate the term in `(u+v)` add the constraint `v = 4/u` to get:

`u^3+64/u^3+313 = 0`

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

`(u^3)^2+313(u^3)+64 = 0`

Using the quadratic formula, we find:

`u^3 = (-313+-sqrt(313^2-(4*1*64)))/(2*1)`

`=(-313+-sqrt(97969-256))/2`

`=(-313+-sqrt(97713))/2`

`=(-313+-3sqrt(10857))/2`

Since these roots are Real and the derivation was 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 = root(3)((-313+3sqrt(10857))/2) + root(3)((-313-3sqrt(10857))/2)`

and related Complex roots:

`t_2 = omega root(3)((-313+3sqrt(10857))/2) + omega^2 root(3)((-313-3sqrt(10857))/2)`

`t_3 = omega^2 root(3)((-313+3sqrt(10857))/2) + omega root(3)((-313-3sqrt(10857))/2)`

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

Then `x = 1/6(1+t)`, hence zeros of our original cubic:

`x_1 = 1/6(1+root(3)((-313+3sqrt(10857))/2) + root(3)((-313-3sqrt(10857))/2))`

`x_2 = 1/6(1+omega root(3)((-313+3sqrt(10857))/2) + omega^2 root(3)((-313-3sqrt(10857))/2))`

`x_3 = 1/6(1+omega^2 root(3)((-313+3sqrt(10857))/2) + omega root(3)((-313-3sqrt(10857))/2))`