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

##### 1 Answer

Here is a solution for

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#### 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

That means that the only possible rational zeros are:

#+-1, +-23#

Note that

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 = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example,

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

Since

**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

**Cardano's method**

We want to solve:

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

Let

Then:

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

Add the constraint

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

Multiply through by

#(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

#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

Now

#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)#