# How do you find all the zeros of #f(x)=x^4-x^3-20x^2+10x+27#?

##### 1 Answer

#### Answer:

Long description of how to solve this (typical) quartic algebraically...

#### Explanation:

#f(x) = x^4-x^3-20x^2+10x+27#

This is a typical quartic, which is algebraically solvable, but gets very messy. It is possible to find approximations to the zeros numerically, but what do you learn from that?

Let's see at least the direction of how to solve this quartic algebraically...

**Tschirnhaus transformation**

First we simplify the quartic using a linear substitution called a Tschirnhaus transformation.

#256f(x) = 256x^4-256x^3-5120x^2+2560x+6912#

#=(4x-1)^4-326(4x-1)^2-8(4x-1)+7229#

#=t^4-326t^2-8t+7229#

where

Since this quartic in

#t^4-326t^2-8t+7229#

#=(t^2-at+b)(t^2+at+c)#

#=t^4+(b+c-a^2)t^2+(b-c)at+bc#

Equating coefficients and rearranging slightly we get:

#{ (b+c=a^2-326), (b-c=-8/a), (bc=7229) :}#

Then:

#(a^2-326)^2 = (b+c)^2 = (b-c)^2+4bc = 64/a^2+28916#

Expanding, this becomes:

#(a^2)^2-652(a^2)+106276 = 64/a^2+28916#

Multiply through by

#(a^2)^3-652(a^2)^2+77360(a^2)-64 = 0#

This is a cubic in

**Descriminant**

The discriminant

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

In our example,

#Delta = 726994944+126959616+2059485184-110592-2024136704 = 889192448#

Since

Since all three zeros are Real, methods like Cardano's method which express zeros in terms of

My preferred method for such cubics is trigonometric substitution, but first we want another Tschirnhaus transformation...

**Tschirnhaus transformation**

#27((a^2)^3-652(a^2)^2+77360(a^2)-64)#

#=27(a^2)^3-17604(a^2)^2+2088720(a^2)-1728#

#=(3a^2-652)^3-579072(3a^2-652)-100388864#

#=s^3-579072s-100388864#

where

**Trigonometric solution**

Let

Let

Then:

#0 = s^3-579072s-100388864#

#=(32sqrt(754) cos theta)^3-579072(32 sqrt(754) cos theta)-100388864#

#=6176768sqrt(754)(4cos^3 theta - 3cos theta)-100388864#

#=6176768sqrt(754)(cos 3 theta)-100388864#

#=4096(1508sqrt(754)(cos 3 theta)-24509)#

Hence:

#s_k = 32sqrt(754) cos(1/3 cos^(-1)(24509/(1508 sqrt(754))) + (2kpi)/3) " "# for#k = 0, 1, 2#

To cut a very long story a little short, we can choose any one of these roots to derive

#b = 1/2(a^2-326-8/a)#

#c = 1/2(a^2-326+8/a)#

leaving us with two quadratics to solve.