# How do you use the rational roots theorem to find all possible zeros of # p(x)=x^5-4x^4+3x^3+2x-6#?

##### 1 Answer

#### Answer:

The only rational zero is

The other zeros are non-Real Complex roots.

#### Explanation:

The rational root theorem tells us that any rational zeros of

That is, the only possible rational zeros are:

#+-1# ,#+-2# ,#+-3# ,#+-6#

Trying each in turn, we find:

#p(3) = 243-324+81+6-6 = 0#

So

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

By the rational root theorem, the only remaining possible rational zeros are:

#+-1# ,#+-2#

None of these is a zero of

That's as far as we can go with the rational root theorem.

**Epilogue**

It is *possible* to express the solutions of

If you are interested, here's the beginning of the solution...

To solve

#0 = 4t^4-2t+2#

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

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

Equating coefficients and rearranging a little:

#b+c = a^2/2#

#b-c = 2/a#

#bc = 2#

Then:

#(a^2)^2/4 = (b+c)^2 = (b-c)^2+4bc = 4/(a^2)+8#

Hence:

#(a^2)^3-32(a^2)-16 = 0#

This cubic in

Substitute

Then:

#16 = (a^2)^3-32(a^2)#

#= (8/3 sqrt(6) cos theta)^3-32(4/3 sqrt(6) cos theta)^2#

#= 256/9sqrt(6)(4 cos^3 theta - 3 cos theta)#

#= 256/9sqrt(6) cos (3theta)#

So:

#cos(3theta) = (3sqrt(6))/32#

So:

#theta = +-1/3 (arccos((3sqrt(6))/32) + 2kpi)# for#k = 0, 1, 2#

So:

#a^2 = 8/3 sqrt(6) cos(1/3 (arccos((3sqrt(6))/32) + 2kpi))#

for

This gives a positive value (

So we can let

#a = sqrt(8/3 sqrt(6) cos(1/3 arccos((3sqrt(6))/32))#

Then

Hence we get quadratics to solve for