Question

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

Answer 1

Use a trigonometric solution to find:

`x_k = 1/3(-2+2sqrt(13) cos(1/3(arccos((11sqrt(13))/338) + 2kpi)))`

`k = 0, 1, 2`

Explanation:

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

Using the rational root theorem we find that the only possible rational zeros are:

`+-1`, `+-3`

None of these work, so none of the zeros are rational.

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=1`, `b=2`, `c=-3` and `d=-3`, so we find:

`Delta = 36+108+96-243+324 = 321`

Since `Delta > 0` this cubic has `3` Real zeros.

As a result, pure algebraic methods like Cardano's method will yield results containing irreducible cube roots of Complex numbers. So choose to use a trigonometric solution instead.

First, use a Tschirnhaus transformation to eliminate the square term...

`27f(x) = 27x^3+54x^2-81x-81 = (3x+2)^3-39(3x+2)-11`

Let `t = (3x+2)`

We want to solve:

`t^3-39t-11 = 0`

Next, use a substitution of the form:

`t = 2sqrt(13) cos theta`

The multiplier `2sqrt(13)` is chosen so that our cubic contains a multiple of `4cos^3theta - 3cos theta = cos 3 theta` as follows:

`0 = (2sqrt(13) cos theta)^3 - 39 (2sqrt(13) cos theta) - 11`

`=26sqrt(13)(4 cos^3 theta - 3 cos theta) - 11`

`=26sqrt(13)cos 3theta - 11`

So:

`cos 3 theta = 11/(26sqrt(13)) = (11sqrt(13))/338`

`theta = +-1/3(arccos((11sqrt(13))/338) + 2kpi)`

`cos theta = cos(1/3(arccos((11sqrt(13))/338) + 2kpi))`

`t = 2sqrt(13) cos(1/3(arccos((11sqrt(13))/338) + 2kpi))`

`x_k = 1/3(-2+2sqrt(13) cos(1/3(arccos((11sqrt(13))/338) + 2kpi)))`

which has distinct values for `k = 0, 1, 2`

`x_0 ~~ 1.4605048700187635`

`x_1 ~~ -2.699628148275318`

`x_2 ~~ -0.7608767217434455`