Question

What are all the zeroes of `f(x) = 2x^3 + 3x^2 + 4x - 10`?

Answer 1

`1.060419 and -1.280210+-i*1.753940`

Explanation:

Rewriting the function:
`f(x)=2*(x^3+1.5x^2+2x-5)`

The easy way (trying as root `+-1, +-5 and +-1/5`) doesn't help us to find the zeros.

Then we can go the hard way as described in
https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots

So for the general cubic equation

`a x^3 + b x^2 + c x + d = 0`

`x_k = -(1/(3a)) (b+u_k C+Delta_0/(u_k*C))`, `k=1,2, 3`

where
`u_1 = 1`, `u_2 = (-1 + i*sqrt(3))/2`, `u_3 =(-1 - i*sqrt(3))/2`

`C = root(3)((Delta_1 + sqrt(Delta_1^2 - 4*Delta_0^3))/2)` with
`Delta_0 = b^2-3 a c`
`Delta_1= 2 b^3-9 a b c+27 a^2 d`

In case:
`Delta_0=1.5^2-3*1*2=2.25-6=-3.75`
`Delta_1=2*3.375-9*1*1.5*2+27*1*(-5)=6.75-27-135=-155.25`

`C=root(3) ((-155.25+sqrt((-155.25)^2-4*(-3.75)^3))/2)`
`C=root(3) ((-155.25+sqrt(24313.5))/2)=root(3)(.338934)=.697223`

For `k=1 and u_1=1`
`x_1=(-1/3)(1.5+.697223+(-3.75)/.697223)=(-1/3)(2.197223-5.378480)` => `x_1=1.060419`

We could continue with this formula, but dividing `(x^3+1.5x^2+2x-5)` by `(x-1.060419)` we get `(x^2+2.560419x+4.715117)`, whose roots we can obtain in this way:
`(x^2+2.560419x+4.715117)=0`
`Delta=6.555745-18.860468=-12.304723`
`x=(-2.560.419+-i*3.507809)/2` => `x_2=-1.280210+i*1.753904`, `x_3=-1.280210-i*1.753904`

Answer 2

`x_1 = 1/6(-3 + root(3)(621+6sqrt(10806)) + root(3)(621-6sqrt(10806)))`

`x_2=1/6(-3 + omega root(3)(621+6sqrt(10806)) + omega^2 root(3)(621-6sqrt(10806)))`

`x_3=1/6(-3 + omega^2 root(3)(621+6sqrt(10806)) + omega root(3)(621-6sqrt(10806)))`

Explanation:

When a cubic has one Real and `2` non-Real Complex zeros, I like to use Cardano's method...

Given `f(x) = 2x^3+3x^2+4x-10`

Let `t_1 = 2x`

Let `t_2 = t_1+1`

Then:

`4f(x) = 8x^3+12x^2+16x-40`

`=(2x)^3+3(2x)^2+8(2x)-40`

`=t_1^3+3t_1^2+8t_1-40`

`=(t_1^3+3t_1^2+3t_1+1) + (5t_1+5)-46`

`=(t_1+1)^3+5(t_1+1)-46`

`=t_2^3+5t_2-46`

Next let `t_2 = u+v`

`0 = (u+v)^3+5(u+v)-46`

`= u^3+v^3+(3uv+5)(u+v)-46`

To eliminate the term in `(u+v)` add the constraint `v = -5/(3u)`

`= u^3 - (5/(3u))^3-46`

`= u^3 - 125/(27u^3)-46`

Multiply through by `27u^3` and rearrange to obtain this quadratic in `u^3`:

`27(u^3)^2-1242(u^3)-125 = 0`

Use the quadratic formula to find:

`u^3 = (1242+-sqrt(1242^2+(4*27*125)))/54`

`=(1242+-sqrt(1556064))/54`

`=(1242+-12sqrt(10806))/54`

`=(621+-6sqrt(10806))/27`

So the Real value of `u` is given by:

`u = root(3)((621+-6sqrt(10806))/27) =1/3 root(3)(621+-6sqrt(10806))`

Then since the derivation was symmetric in `u` and `v` we can deduce that one of these values is `u` and the other `v`.

`t_2 = u+v`

`= 1/3 root(3)(621+6sqrt(10806)) + 1/3 root(3)(621-6sqrt(10806))`

Hence the Real root of the original cubic is:

`x_1 = (t_2-1)/2`

`=1/6(-3 + root(3)(621+6sqrt(10806)) + root(3)(621-6sqrt(10806)))`

The Complex roots come from replacing the Real value of `u` with `omega u` or `omega^2 u`, where `omega = -1/2+sqrt(3)/2i` is the primitive cube root of `1` ...

`x_2=1/6(-3 + omega root(3)(621+6sqrt(10806)) + omega^2 root(3)(621-6sqrt(10806)))`

`x_3=1/6(-3 + omega^2 root(3)(621+6sqrt(10806)) + omega root(3)(621-6sqrt(10806)))`