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

1 Answer
May 5, 2016

Answer:

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

Explanation:

#P(x) = 3x^4+x^3+2x^2-2#

You might try looking for rational zeros first, but you would not find any. This is a full blown quartic with #2# Real roots and #2# Complex roots, all irrational.

Normally with a quartic of the form #ax^4+bx^3+cx^2+dx+e#, the first thing you would do would be to substitute something like #t=x+b/(4a)# in order to get a quartic in #t# with no #t^3# term.

In our example, there's an easier method by substituting #t = 1/x# and multiplying by #-2t^4# to get:

#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 #3^3=27# to get:

#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=3a^2-8# to get:

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

Let #w=u+v# to get:

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

Add the constraint #v = -272/u# to eliminate the term in #(u+v)# and multiply through by #u^3# to get:

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

Since this derivation is symmetric in #u# and #v#, we can use the quadratic formula and hence the two Real roots of this quadratic as #u^3# and #v^3#, to provide a Real value for #w# and hence #a^2#...

#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 #b = (a^3-4a-4)/(4a)# and #c = (a^3-4a+4)/(4a)#.

Hence quadratics in #t# to solve.

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

The resulting expressions are rather complicated, but tractable.