Question

How do you list all possible roots and find all factors of `x^4+2x^3-8x^2+16x-23`?

Answer 1

Here is a solution for `x^4-2x^3+8x^2+16x-23=0`

(`x^4+2x^3-8x^2+16x-23=0` is much more complicated)

Explanation:

Assuming typographical errors in the problem, let us solve:

`x^4-2x^3+8x^2+16x-23=0`

By the rational roots theorem, any rational zeros of this quartic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-23` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-23`

Note that `x=1` is a solution, since the sum of the coefficients is `0`, i.e. `1-2+8+16-23 = 0`

Then:

`x^4-2x^3+8x^2+16x-23= (x-1)(x^3-x^2+7x+23)`

To find the zeros for the remaining cubic, proceed as follows:

`f(x) = x^3-x^2+7x+23`

Discriminant

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 example, `a=1`, `b=-1`, `c=7` and `d=23`, so we find:

`Delta = 49-1372+92-14283-2898 = -18412`

Since `Delta < 0` this cubic has `1` Real zero and `2` non-Real Complex zeros, which are Complex conjugates of one another.

Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

`0=27f(x)=27x^3-27x^2+189x+621`

`=(3x-1)^3+60(3x-1)+682`

`=t^3+60t+682`

where `t=(3x-1)`

Cardano's method

We want to solve:

`t^3+60t+682=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv+20)(u+v)+682=0`

Add the constraint `v=-20/u` to eliminate the `(u+v)` term and get:

`u^3-8000/u^3+682=0`

Multiply through by `u^3` and rearrange slightly to get:

`(u^3)^2+682(u^3)-8000=0`

Use the quadratic formula to find:

`u^3=(-682+-sqrt((682)^2-4(1)(-8000)))/(2*1)`

`=(682+-sqrt(465124+32000))/2`

`=(682+-sqrt(497124))/2`

`=341+-3sqrt(13809)`

Since this is Real and the derivation is symmetric in `u` and `v`, we can use one of these roots for `u^3` and the other for `v^3` to find Real root:

`t_1=root(3)(341+3sqrt(13809))+root(3)(341-3sqrt(13809))`

and related Complex roots:

`t_2=omega root(3)(341+3sqrt(13809))+omega^2 root(3)(341-3sqrt(13809))`

`t_3=omega^2 root(3)(341+3sqrt(13809))+omega root(3)(341-3sqrt(13809))`

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

Now `x=1/3(1+t)`. So the roots of our original cubic are:

`x_1 = 1/3(1+root(3)(341+3sqrt(13809))+root(3)(341-3sqrt(13809)))`

`x_2 = 1/3(1+omega root(3)(341+3sqrt(13809))+omega^2 root(3)(341-3sqrt(13809)))`

`x_3 = 1/3(1+omega^2 root(3)(341+3sqrt(13809))+omega root(3)(341-3sqrt(13809)))`

Then the complete factorisation of our original quartic takes the form:

`x^4-2x^3+8x^2+16x-23 = (x-1)(x-x_1)(x-x_2)(x-x_3)`