Question

How do you find all the zeros of `f(x)=x^3+3x^2-5x+8`?

Answer 1

Use Cardano's method to find Real zero:

`x_1 = 1/3(-3+root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))`

and related Complex zeros.

Explanation:

`f(x) = x^3+3x^2-5x+8`

From the rational root theorem we can deduce that the only possible rational zeros are:

`+-1, +-2, +-4, +-8`

None of these work, so there are no rational zeros.

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=3`, `c=-5` and `d=8`, so we find:

`Delta = 225+500-864-1728-2160=-4027`

Since `Delta < 0`, this cubic has one Real zero and two non-Real Complex zeros. As a result, Cardano's method will work well.

We can simplify the problem by making the substitution `t=x+1`:

`x^3+3x^2-5x+8`

`= (x^3+3x^2+3x+1)-8(x+1)+15`

`=(x+1)^3-8(x+1)+15`

`= t^3-8t+15`

Using Cardano's method, let `t=u+v`, then we want to solve:

`u^3+v^3+(3uv-8)(u+v)+15 = 0`

Add the constraint `v = 8/(3u)` to eliminate the term in `(u+v)` and get:

`u^3+512/(27u^3)+15 = 0`

Multiply through by `27u^3` and rearrange slightly to get a quadratic in `u^3`:

`27(u^3)^2+405(u^3)+512 = 0`

Use the quadratic formula to find:

`u^3=(-405+-sqrt(405^2-4(27)(512)))/(2*27)`

`=(-405+-sqrt(164025-55296))/54`

`=(-405+-sqrt(108729))/54`

`=(-405+-3sqrt(12081))/54`

Hence Real zero:

`t_1 = root(3)((-405+3sqrt(12081))/54) + root(3)((-405-3sqrt(12081))/54)`

`= 1/3(root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))`

and related Complex zeros:

`t_2 = 1/3(omega root(3)((-405+3sqrt(12081))/2) + omega^2 root(3)((-405-3sqrt(12081))/2))`

`t_3 = 1/3(omega^2 root(3)((-405+3sqrt(12081))/2) + omega root(3)((-405-3sqrt(12081))/2))`

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

Then `x = t-1`, hence zeros of `f(x)`:

`x_1 = 1/3(-3+root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))`

`x_2 = 1/3(-3+omega root(3)((-405+3sqrt(12081))/2) + omega^2 root(3)((-405-3sqrt(12081))/2))`

`x_3 = 1/3(-3+omega^2 root(3)((-405+3sqrt(12081))/2) + omega root(3)((-405-3sqrt(12081))/2))`