# How do you use the rational roots theorem to find all possible zeros of # p(x)=-5x^4+x^3+2x^2-1#?

##### 1 Answer

Find that this quartic has no rational solutions.

Show how to solve it algebraically regardless...

#### Explanation:

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

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

#+-1/5# ,#+-1#

None of these work, so *rational* zeros.

In fact, this is a fairly typical quartic, which is very messy to solve algebraically.

Here's a sketch of how to start...

**Tschirnhaus transformation**

First transform the quartic into a monic one with no cube term, using a linear substitution:

#-32000p(x) = 160000x^4-32000x^3-64000x^2+32000#

#=(20x-1)^4-166(20x-1)^2-328(20x-1)+31837#

#=t^4-166t^2-328t+31837#

where

**Factor into quadratics**

Since our simplified quartic is monic with no cube term, it will factor as a product of two monic quadratics with opposite middle coefficients:

#t^4-166t^2-328t+31837#

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

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

Equating coefficients and rearranging a little, we get:

#{ (b+c = a^2-166), (b - c = -328/a), (bc = 31837) :}#

Hence we find:

#(a^2-166)^2 = (b+c)^2 = (b-c)^2+4bc = (-328/a)^2 + 4(31837)#

Expand the two ends to get:

#(a^2)^2-332(a^2)+27556 = 107584/((a^2))+127348#

Subtracting the right hand side from the left and multiplying through by

#(a^2)^3-332(a^2)^2-99792(a^2)-107584 = 0#

Note that if we can solve this cubic in

#b = 1/2(a^2-166-328/a)#

#c = 1/2(a^2-166+328/a)#

and hence have two quadratics to solve.

**Descriminant**

The discriminant

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

In our example,

#Delta = 1097659450331136+3975091880804352-15747873947648-312506560512-64158470627328 = 4992532480000000#

Since

**Tschirnhaus transformation**

To make the task of solving the cubic simpler, we make the cubic simpler with another Tschirnhaus transformation.

#0=27((a^2)^3-332(a^2)^2-99792(a^2)-107584)#

#=27(a^2)^3-8964(a^2)^2-2694384(a^2)-2904768#

#=(3(a^2)-332)^3-1228800(3(a^2)-332)-374272000#

#=s^3-1228800s-374272000#

where

**Next steps**

Since we know that this cubic has only Real zeros, methods like Cardano's method will result in expressions involving irreducible cube roots of Complex numbers. My preference in these cases is to solve using a trigonometric substitution:

#s = k cos theta#

where

#4 cos^3 theta - 3 cos theta = cos 3 theta#

I will stop here for now, but I think you will agree that such quartics are messy to solve.