How do you find the number of complex zeros for the function #f(x)=27x^6+208x^3-64#?

1 Answer
George C.
Oct 17, 2016

#f(x)# has #2# Real zeros and #4# non-Real Complex zeros.

Explanation:

#f(x) = 27x^6+208x^3-64#

By the Fundamental Theorem of Algebra, #f(x)# has exactly #6# Complex (possibly Real) zeros counting multiplicity, since it is of degree #6#.

#color(white)()#
Descartes' Rule of Signs

The signs of the coefficients of #f(x)# follow the pattern #+ + -#

With one change of sign, that means that #f(x)# has exactly one positive Real zero.

The signs of the coefficients of #f(-x)# follow the pattern #+ - -#

With one change of sign, that means that #f(x)# has exactly one negative Real zero.

The other #4# zeros must be non-Real Complex.

#color(white)()#
Bonus - Direct solution

#27f(x) = 27(27x^6+208x^3-64)#

#color(white)(27f(x)) = 729x^6+5616x^3-1728#

#color(white)(27f(x)) = (27x^3)^2+2(104)(27x^3)+(104)^2-112^2#

#color(white)(27f(x)) = (27x^3+104)^2-112^2#

#color(white)(27f(x)) = ((27x^3+104)-112)((27x^3+104)+112)#

#color(white)(27f(x)) = (27x^3-8)(27x^3+216)#

#color(white)(27f(x)) = 27((3x)^3-2^3)(x^3+2^3)#

#color(white)(27f(x)) = 27(3x-2)(9x^2+6x+4)(x+2)(x^2-2x+4)#

#color(white)(27f(x)) = 27(3x-2)(3x-2omega)(3x-2omega^2)(x+2)(x+2omega)(2+2omega^2)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#

So the Real zeros are:

#x = 2/3#

#x = -2#

and the non-Real Complex zeros are:

#x = 2/3omega#

#x = 2/3omega^2#

#x = -2omega#

#x = -2omega^2#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
1704 views around the world
You can reuse this answer
Creative Commons License