# How do you find the solutions to the equation #x^4-6x^3+5x^2+30x-50=0#?

##### 1 Answer

#### Answer:

This quartic has roots:

#### Explanation:

#f(x) = x^4-6x^3+5x^2+30x-50#

By the rational root theorem, any *rational* zeros of

That means that the only possible *rational* zeros are:

#+-1, +-2, +-5, +-10, +-25, +-50#

None of these work. So *rational* zeros.

**Tschirnhaus transformation**

First make a linear substitution called a Tschirnhaus transformation in order to simplify the quartic:

#16f(x) = 16x^4-96x^3+80x^2+480x-800#

#=(2x-3)^4-34(2x-3)^2+144(2x-3)-143#

#=t^4-34t^2+144t-143#

where

Since this quartic in

#t^4-34t^2+144t-143#

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

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

Equating coefficients and rearranging a little, we get:

#{ (b+c = a^2-34), (b-c = 144/a), (bc = -143) :}#

So we find:

#(a^2-34)^2 = (b+c)^2 = (b-c)^2+4bc = 20736/a^2-572#

Expanding, this becomes:

#(a^2)^2-68(a^2)+1156 = 20736/((a^2))-572#

Multiply through by

#(a^2)^3-68(a^2)^2+1728(a^2)-20736=0#

By the rational roots theorem, the possible rational roots of this cubic are any factors of

That's rather a lot to try, but there are various methods to speed up the search and eventually find

Let

Then:

#b = 1/2(a^2-34+144/a) = 1/2(36-34+24) = 13#

#c = 1/2(a^2-34-144/a) = 1/2(36-34-24) = -11#

Hmmm. We could have spotted that one fairly easily from the simultaneous equations by looking at possible factorings of

Well however we get there we now have two quadratics to solve:

#0 = t^2-6t+13 = (t-3)^2 + 2^2 = (t-3-2i)(t-3+2i)#

#0 = t^2+6t-11 = (t+3)^2-(2sqrt(5))^2 = (t+3-2sqrt(5))(t+3+2sqrt(5))#

Then

So we get solutions to our original quartic:

#x=1/2((3+-2i)+3)=3+-i#

#x=1/2((-3+-2sqrt(5))+3) = +-sqrt(5)#