How do you find all the zeros of #(3x^6+x^2-4)(x^3+6x+7)#?

1 Answer
Jul 16, 2016

Answer:

This product has zeros:

  • #1#

  • #-1# (with multiplicity #2#)

  • #+-(sqrt((4sqrt(3)-3)/12))+-(sqrt((4sqrt(3)+3)/12))i#

  • #1/2+-(3sqrt(3))/2 i#

Explanation:

The zeros of #(3x^6+x^2-4)(x^3+6x+7)# are all the values of #x# which are zeros of #(3x^6+x^2-4)# or #(x^3+6x+7)#.

#color(white)()#
Zeros of #bb(3x^6+x^2-4)#

Note that all of the degrees are even and the sum of the coefficients is #0#.

Hence this sextic has zeros #+-1# and quadratic factor:

#(x-1)(x+1) = x^2-1#

We find:

#3x^6+x^2-4 = (x^2-1)(3x^4+3x^2+4)#

Treating the remaining quartic factor as a quadratic in #x^2# and using the quadratic formula, we find:

#x^2 = (-3+-sqrt(3^2-4(3)(4)))/(2*3)#

#=(-3+-sqrt(9-48))/6#

#=(-3+-sqrt(-39))/6#

#=-1/2+-sqrt(39)/6i#

The square roots of #a+-bi# are:

#+-(sqrt((sqrt(a^2+b^2)+a)/2)) +- (sqrt((sqrt(a^2+b^2)-a)/2))i#

(see https://socratic.org/s/aw38evei)

So, putting #a=-1/2# and #b=sqrt(39)/6# we find:

#a^2+b^2 = (-1/2)^2+(sqrt(39)/6)^2 = 1/4+39/36 = 48/36 = 4/3#

So:

#sqrt(a^2+b^2) = sqrt(4/3) = sqrt(4/9*3) = (2sqrt(3))/3#

Hence:

#x = +-(sqrt(((2sqrt(3))/3-1/2)/2))+-(sqrt(((2sqrt(3))/3+1/2)/2))i#

#=+-(sqrt((4sqrt(3)-3)/12))+-(sqrt((4sqrt(3)+3)/12))i#

#color(white)()#
Zeros of #bb(x^3+6x+7)#

Notice that #-1# is a zero, so #(x+1)# is a factor and we find:

#x^3+6x+7 = (x+1)(x^2-x+7)#

The remaining quadratic factor has zeros given by the quadratic formula:

#x = (1+-sqrt((-1)^2-4(1)(7)))/(2*1)#

#=(1+-sqrt(-27))/2#

#=1/2+-(3sqrt(3))/2 i#