Question

How do you find all the zeros of `x^3 + 2x^2 + 5x +1`?

Answer 1

Use Cardano's method...

Explanation:

`f(x) = x^3+2x^2+5x+1`

To cut down on the number of fractions we need to work with, multiply by `3^3 = 27` first...

`0 = 27f(x)`

`=27x^3+54x^2+135x+27`

`=(3x+2)^3+33(3x+2)-47`

Substitute `t = 3x + 2` ...

`=t^3+33t-47`

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

`=u^3+v^3+3(uv+11)(u+v)-47`

Add the constraint `v = -11/u` to eliminate the `(u+v)` term ...

`=u^3-(11^3/u^3)-47`

`=u^3-1331/u^3-47`

Multiply through by `u^3` to get this quadratic in `u^3` ...

`(u^3)^2-47(u^3)-1331 = 0`

Solve using the quadratic formula to get:

`u^3 = (47+-sqrt(47^2+4*1331))/2`

`= (47+-sqrt(7533))/2`

`= (47+-9sqrt(93))/2`

Since 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 derive the Real root:

`t_1 = root(3)((47+9sqrt(93))/2) + root(3)((47-9sqrt(93))/2)`

and the Complex roots:

`t_2 = omega root(3)((47+9sqrt(93))/2) + omega^2 root(3)((47-9sqrt(93))/2)`

`t_3 = omega^2 root(3)((47+9sqrt(93))/2) + omega root(3)((47-9sqrt(93))/2)`

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

Then `x = (t-2) / 3` hence zeros of the original function:

#x_1 = 1/3(-2+root(3)((47+9sqrt(93))/2) + root(3)((47-9sqrt(93))/2) )#

`x_2 = 1/3(-2 + omega root(3)((47+9sqrt(93))/2) + omega^2 root(3)((47-9sqrt(93))/2))`

`x_3 = 1/3(-2 + omega^2 root(3)((47+9sqrt(93))/2) + omega root(3)((47-9sqrt(93))/2))`