Question

How do you find all the real and complex roots of `p(x) = x^5 + 4x^4 + 6x^3 + 4x^2 + 5x`?

Answer 1

The zeros are `x=0`, `x=+-i`, `x=-2+-i`

Explanation:

`p(x) = x^5+4x^4+6x^3+4x^2+5x`

First separate out the common factor `x` to find:

`p(x) = x(x^4+4x^3+6x^2+4x+5)`

Note that `x=0` is one of the zeros of `p(x)`.

Next, let `t = x+1`

This is called a Tschirnhaus transformation. This kind of simple Tschirnhaus transformation can be used normally to eliminate the `x^3` term, but in the case of this particular quartic it helps a lot more.

Then `t^4 = x^4+4x^3+6x^2+4x+1`

So `x^4+4x^3+6x^2+4x+5 = t^4+4`

This has zeros when `t^4 = -4`, that is when `t^2 = +-sqrt(-4) = +-2i`

That is when `t = +-1+-i`

If `t = 1+-i` then `x = t - 1 = +-i`

If `t = -1+-i` then `x = t - 1 = -2+-i`