# How do you find all the real and complex roots of # P(x) = 3x^4 + x^3 + 2x^2 - 2#?

##### 1 Answer

It gets rather complicated, but it's doable...

#### Explanation:

You might try looking for rational zeros first, but you would not find any. This is a full blown quartic with

Normally with a quartic of the form

In our example, there's an easier method by substituting

#0 = 4t^4-4t^2-2t-6#

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

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

Equating coefficients and rearranging a little, we get:

#{ (b+c = (a^2-4)/2), (b-c=-2/a), (bc=-6) :}#

Then:

#((a^2-4)/2)^2 = (b+c)^2 = (b-c)^2+4bc = (-2/a)^2-24#

Hence:

#(a^2)^3-8(a^2)^2+112(a^2)-16 = 0#

Multiply through by

#0 = 27(a^2)^3-216(a^2)^2+3024(a^2)-432#

#=(3(a^2)-8)^3 + 816(3(a^2)-8)+6608#

Substitute

#w^3+816w+6608= 0#

Let

#u^3+v^3+3(uv+272)(u+v)+6608= 0#

Add the constraint

#(u^3)^2+6608(u^3)-20123648 = 0#

Since this derivation is symmetric in

#u^3 = (-6608+-sqrt(6608^2+4*20123648))/2#

#=(-6608+-sqrt(124160256))/2#

#=(-6608+-48sqrt(53889))/2#

#=(-6608+-48sqrt(53889))/2#

#=8(-413+-3sqrt(53889))#

Hence:

#3a^2-8 = w = u+v =2(root(3)(-413+3sqrt(53889))+root(3)(-413-3sqrt(53889)))#

Hence, choosing the positive root:

#a = sqrt(2/3(4+root(3)(-413+3sqrt(53889))+root(3)(-413-3sqrt(53889))))#

Then

Hence quadratics in

Then the zeros of the original polynomial are the reciprocals of the roots of these quadratics.

The resulting expressions are rather complicated, but tractable.