Question

How do I find all real and complex zeros of `x^5+x+1`?

Answer 1

Factor into the product of a quadratic and a cubic, then find the roots of those using the quadratic formula and Cardano's method.

Explanation:

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

`x^2+x+1 = 0` has solutions `x = color(blue)((-1+-isqrt(3))/2)`

To solve `x^3-x^2+1 = 0` first substitute `t = x-1/3`

Then

`t^3 = (x-1/3)^3 = x^3-x^2+1/3x-1/27`

`-1/3t = -1/3(x-1/3) = -1/3x+1/9`

So

`t^3-1/3t+25/27 = x^3-x^2+1/3x-1/27-1/3x+1/9+25/27`

`=x^3-x^2+1`

So our cubic becomes:

`t^3-1/3t+25/27 = 0`

Multiply through by `27` to get:

`27t^3-9t+25 = 0`

Next let `t = u + v` to get

`0 = 27(u^3+3u^2v+3uv^2+v^3)-9(u+v)+25`

`=27u^3+27v^3+(81uv-9)(u+v)+25`

Add the constraint that `v = 1/(9u)`

Then `81uv-9 = 9(9uv-1) = 9(1-1) = 0`, so our cubic becomes:

`0 = 27u^3+27(1/9u)^3+25 = 27u^3 + 1/(27u^3)+25`

Multiply through by `27u^3` to get a quadratic in `u^3`:

`0 = 729(u^3)^2+675(u^3)+1`

So:

`u^3 = (-675+-sqrt(675^2-4*729))/(2*729)`

`=(-675+-27sqrt(625-4))/(27*54)`

`=(-25+-sqrt(621))/54`

`=(-25+-sqrt(9*69))/54`

`=(-25+-3sqrt(69))/54`

Since the derivation was symmetric in `u` and `v`, the Real root is given by:

`t = root(3)((-25+3sqrt(69))/54) + root(3)((-25-3sqrt(69))/54)`

`= 1/3root(3)((-25+3sqrt(69))/2) + 1/3root(3)((-25-3sqrt(69))/2)`

and

`x = t+1/3 = color(blue)(1/3+1/3root(3)((-25+3sqrt(69))/2) + 1/3root(3)((-25-3sqrt(69))/2))`

`~~ -0.75487766625`

The Complex roots are found using `omega = -1/2+isqrt(3)/2` as:

`x = color(blue)(1/3+omega/3 root(3)((-25+3sqrt(69))/2) + omega^2/3 root(3)((-25-3sqrt(69))/2))`

`x = color(blue)(1/3+omega^2/3 root(3)((-25+3sqrt(69))/2) + omega/3 root(3)((-25-3sqrt(69))/2))`

That is

`x~~0.87744+0.74486 i`

`x~~0.87744-0.74486 i`

Postscript

Note that in general, quintic equations of the form `x^5+x+a = 0` are not so easily (!) solved. They are normally not expressible in terms of ordinary square and cube roots or any kind of ordinary radical.

There is a special kind of radical called a Bring Radical, which represents the principal root of this quintic. That is:

`BR(a)` is the principal root of `x^5+x+a = 0`

If `a in RR` then `BR(a) in RR` is the Real root.