Question

How do you find all the zeros of `f(x) = 4x^3 – 5x^2 + 9x – 6`?

Answer 1

Multiply by `432`, substitute `t = 12x-5`, then use Cardano's method,

Explanation:

`f(x) = 4x^3-5x^2+9x-6`

First multiply through by `3^3*4^2 = 432` to cut down on the number of fractions we need to work with.

`0 = 432 f(x)`

`= 1728x^3-2160x^2+3888x-2592`

`=(12x-5)^3+249(12x-5)-1222`

Substitute `t = 12x-5`

`t^3+249t-1222 = 0`

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

`u^3+v^3+(3uv+249)(u+v)-1222 = 0`

Add the constraint `v = -249/(3u)` to eliminate the term in `(u+v)`

`u^3-249^3/(27u^3)-1222 = 0`

Multiply through by `27u^3` and rearrange to get:

`27(u^3)^2-32994(u^3)-15438249 = 0`

Use the quadratic formula to find:

`u^3 = (32994+-sqrt(2755934928))/54`

`=(32994+-972 sqrt(2917))/54`

`=611+-18sqrt(2917)`

Due to the symmetry of the derivation in `u` and `v`, we can take one of these roots as the value of `u^3` and the other as `v^3` to get the Real root:

`t_1 = root(3)(611+18sqrt(2917)) + root(3)(611-18sqrt(2917))`

and Complex roots:

`t_2 = omega root(3)(611+18sqrt(2917)) + omega^2 root(3)(611-18sqrt(2917))`

`t_3 = omega^2 root(3)(611+18sqrt(2917)) + omega root(3)(611-18sqrt(2917))`

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

Then you can derive the roots of the original equation:

`x_1 = (t_1+5)/12=1/12(5+root(3)(611+18sqrt(2917)) + root(3)(611-18sqrt(2917)))`

`x_2 = 1/12(5+omega root(3)(611+18sqrt(2917)) + omega^2 root(3)(611-18sqrt(2917)))`

`x_3 = 1/12(5+omega^2 root(3)(611+18sqrt(2917)) + omega root(3)(611-18sqrt(2917)))`