Question

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

Answer 1

Real zeros: `1`, `1/3(1+root(3)((83+9sqrt(85))/2)+root(3)((83-9sqrt(85))/2))`

and related Complex zeros.

Explanation:

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

I suspect there may be a typo here. Was `x^4` supposed to be `x^3`? I will proceed assuming that it is correct as given...

First note that since the sum of the coefficients of `f(x)` is zero, `f(1) = 0` and `(x-1)` is a factor:

`x^4-x^2-3x+3=(x-1)(x^3+x^2-3)`

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 case `a=1`, `b=1`, `c=0` and `d=-3`, so we find:

`Delta = 0+0+12-243+0 = -231 < 0`

Since `Delta < 0` this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

We could use the substitution `t=1/x` to get a simplified cubic, but on this occasion, let's multiply `(x^3+x^2-3)` by `27` and use `t=3x-1`. (That will make it slightly easier to find the roots in `a+bi` form.)

`27(x^3-x^2-3)`

`=27x^3-27x^2-81`

`=(3x)^3-3(3x)^2-81`

`=(3x-1)^3-3(3x-1)-83`

`=t^3-3t-83`

So we want to solve `t^3-3t-83 = 0`

Use Cardano's method:

Let `t=u+v`

`u^3+v^3+3(uv-1)(u+v)-83 = 0`

Eliminate the `(u+v)` term by letting `v = 1/u` to find:

`u^3+1/u^3-83 = 0`

Multiply through by `u^3` to get:

`(u^3)^2-83(u^3)+1 = 0`

By the quadratic formula:

`u^3 = (83+-sqrt(83^2-4(1)(1)))/(2*1)`

`=(83+-sqrt(6889-4))/2`

`=(83+-sqrt(6885))/2`

`=(83+-9sqrt(85))/2`

Since this is Real valued and our derivation was symmetric in `u` and `v`, we can use one of these roots for `u^3` and the other for `v^3` to deduce the roots of our cubic in `t`.

Then we can use `x = 1/3(t+1)` to find the Real zero of our original cubic:

`x_1 = 1/3(1+root(3)((83+9sqrt(85))/2)+root(3)((83-9sqrt(85))/2))`

and related Complex zeros:

`x_2 = 1/3(1+omega root(3)((83+9sqrt(85))/2)+omega^2 root(3)((83-9sqrt(85))/2))`

`x_3 = 1/3(1+omega^2 root(3)((83+9sqrt(85))/2)+omega root(3)((83-9sqrt(85))/2))`

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