Question

How do you find the zeros of `f(x) = 7x^3 + 4x^2 - 3x + 3`?

Answer 1

Use a substitution to get into the form `t^3+pt+q = 0` then Cardano's method...

Explanation:

When a cubic has one irrational Real zero and two non-Real Complex zeros, I like to solve using Cardano's method.

First, I would like to have a cubic with integer coefficients and no square term. To get there from our `f(x)`, multiply by `3^3*7^2 = 1323` and substitute `t = 21x+4` as follows:

`1323(7x^3+4x^2-3x+3)`

`=9261x^3+5292x^2-3969x+3969`

`=(9261x^3+5292x^2+1008x+64)-(4977x+948)+4853`

`=(21^3x^3+(3*21^2*4)x^2+(3*21*4^2)x+4^3)-((237*21)x+(237*4))+4853`

`=(21x+4)^3-237(21x+4)+4853`

`=t^3-237t+4853`

So we want to solve `t^3-237t+4853 = 0`

Next substitute `t = u + v` to get:

`0 = (u+v)^3-237(u+v)+4853`

`=u^3+v^3+(3uv-237)(u+v)+4853`

`=u^3+v^3+3(uv-79)(u+v)+4853`

Add the constraint `v = 79/u` and multiply through by `u^3` to get:

`(u^3)^2+4853(u^3)+493039 = 0`

Use the quadratic formula to find:

`u^3 = (-4853+-sqrt(4853^2-4*1*493039))/2`

`=(-4853+-sqrt(21579453))/2`

`=(-4853+-63 sqrt(5437))/2`

Since the derivation was symmetric in `u` and `v`, one of these roots can be taken as `u^3` and the other `v^3` to give us the Real root:

`t_1 = root(3)((-4853+63 sqrt(5437))/2)+root(3)((-4853-63 sqrt(5437))/2)`

and the Complex roots:

`t_2 = omega root(3)((-4853+63 sqrt(5437))/2)+ omega^2 root(3)((-4853-63 sqrt(5437))/2)`

`t_3 = omega^2 root(3)((-4853+63 sqrt(5437))/2)+ omega root(3)((-4853-63 sqrt(5437))/2)`

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

Then `x = (t-4)/21`, so the zeros of `f(x)` are:

`x_1 = 1/21(-4+ root(3)((-4853+63 sqrt(5437))/2)+root(3)((-4853-63 sqrt(5437))/2))`

`x_2 = 1/21(-4+ omega root(3)((-4853+63 sqrt(5437))/2)+ omega^2 root(3)((-4853-63 sqrt(5437))/2))`

`x_3 = 1/21(-4+ omega^2 root(3)((-4853+63 sqrt(5437))/2)+ omega root(3)((-4853-63 sqrt(5437))/2))`