How do you find all the zeros of #f(x)=(6x^7+4x^2+6)(x^4+5x-6)# with its multiplicities?

1 Answer
Jun 24, 2016

Answer:

Find zeros of #(x^4+5x-6)# algebraically and approximate zeros of #(6x^7+4x+6)# numerically.

Explanation:

#f(x) = (6x^7+4x^2+6)(x^4+5x-6)#

The zeros of #f(x)# are the zeros of #(6x^7+4x^2+6) = 2(3x^7+2x^2+3)# and the zeros of #(x^4+5x-6)#

#color(white)()#
Zeros of #bb (x^4+5x-6)#

Notice that the sum of the coefficients is #0#. That is: #1+5-6 = 0#.

Hence #x=color(blue)(1)# is a zero and #(x-1)# a factor.

#x^4+5x-6 = (x-1)(x^3+x^2+x+6)#

By the rational root theorem, the possible rational zeros of #x^3+x^2+x+6# are: #+-1, +-2, +-3#. Note that #color(blue)(-2)# is a zero and #(x+2)# a factor:

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

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

#x = (1+-sqrt(-11))/2 = color(blue)(1/2+-sqrt(11)/2i)#

#color(white)()#
Zeros of #bb (6x^7+4x^2+6 = 2(3x^7+2x^2+3))#

This septic has no simple factorisation and no rational zeros.

About the best you can do is use a numerical algorithm to find approximations for the #1# Real and #6# non-Real Complex zeros.

The Durand-Kerner method is straightforward to code and gives all of the approximate zeros at once.

#x ~~ -1.08646#

#x ~~ -0.539746+-0.832444i#

#x ~~ 0.15543+-0.896241i#

#x ~~ 0.927548+-0.519443i#