Question

What are the zeros of `x^3-8x-4`?

Answer 1

`x_k = 4/3 sqrt(6) cos(1/3 cos^(-1)(3/16 sqrt(6))+(2kpi)/3) " " k = 0,1,2`

Explanation:

`f(x) = x^3-8x-4`

`color(white)()`
Descriminant

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=0`, `c=-8` and `d=-4`, so we find:

`Delta = 0+2048+0-432+0 = 1616`

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

`color(white)()`
Trigonometric method

We use a substitution `x = k cos theta`, where `k` is chosen to squeeze `f(x)` into a form containing `4cos^3 theta - 3 cos theta = cos 3theta`.

Let `k=4/3 sqrt(6)`

Then:

`0 = f(x) = x^3-8x-4`

`=(k cos theta)^3 - 8 (k cos theta) - 4`

`=k (k^2 cos^3 theta - 8 cos theta) - 4`

`=4/3 sqrt(6) (32/3 cos^3 theta - 8 cos theta) - 4`

`= 32/9 sqrt(6) (4 cos^3 theta - 3 cos theta) - 4`

`= 32/9 sqrt(6) cos 3 theta - 4`

Add `4` to both ends and transpose to get:

`32/9 sqrt(6) cos 3 theta = 4`

Multiply both sides by `sqrt(6)` to get:

`64/3 cos 3 theta = 4 sqrt(6)`

Multiply both sides by `3/64` to get:

`cos 3 theta = 3/16 sqrt(6)`

So:

`3 theta = +-cos^(-1)(3/16 sqrt(6))+2kpi`

So:

`theta = +-1/3 cos^(-1)(3/16 sqrt(6))+(2kpi)/3`

So:

`cos theta = cos(1/3 cos^(-1)(3/16 sqrt(6))+(2kpi)/3)`

This takes three distinct values, for which we can use `k=0, 1, 2`.

Hence zeros of our original cubic:

`x_k = 4/3 sqrt(6) cos(1/3 cos^(-1)(3/16 sqrt(6))+(2kpi)/3) " " k = 0,1,2`

`x_0 ~~ 3.051374241731`

`x_1 ~~ -2.534070196723`

`x_2 ~~ -0.517304045008`