# How do you use the rational roots theorem to find all possible zeros of #P(x) = 2x^4 + 10x^3 -3x^2 -8x +4#?

##### 1 Answer

We find there are no rational zeros, but it is possible to solve algebraically.

#### Explanation:

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

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

#+-1/2, +-1, +-2, +-4#

Evaluating *rational* zeros.

In fact, it has two Real zeros and two non-Real Complex zeros. It is a little messy to solve algebraically, but it can be done...

**Tschirnhaus transformation**

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

#128P(x) = 256x^4+1280x^3-384x^2-1024x+512#

#=(4x+5)^4-174(4x+5)^2+984(4x+5)-683#

#=t^4-174t^2+984t-683#

where

**Factorisation into quadratics**

Since this quartic in

#t^4-174t^2+984t-683#

#=(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^2-174), (b-c=984/a), (bc=-683) :}#

Hence:

#(a^2-174)^2 = (b+c)^2 = (b-c)^2+4bc = (984/a)^2+4(-683)#

Expanded:

#(a^2)^2-348(a^2)+30276 = 968256/((a^2))-2732#

Multiply through by

#(a^2)^3-348(a^2)^2+33008(a^2)-968256 = 0#

We can solve this cubic in

#a^2 = 4/3(87+2root(3)(-6885+60sqrt(1761))+2root(3)(-6885-60sqrt(1761))) ~~ 16.1274#

So we can let

#a = sqrt(4/3(87+2root(3)(-6885+60sqrt(1761))+2root(3)(-6885-60sqrt(1761))))#

Then:

#b = 1/2(a^2-174+984/a)#

#c = 1/2(a^2-174-984/a)#

Hence two quadratics to solve to find

Messy, but not difficult.