How do you find the number of complex zeros for the function #f(x)=27x^9+8x^6-27x^3-8#?

1 Answer
Jan 8, 2017

Answer:

#f(x)# has 9 Complex zeros, of which #3# are Real and #6# non-Real.

Explanation:

By the Fundamental Theorem of Algebra #f(x)# has exactly #9# Complex (possibly Real) zeros, counting multiplicity, since it is of degree #9#.

We can find the zeros by factoring by grouping and using some identities:

Difference of squares:

#a^2-b^2 = (a-b)(a+b)#

Difference of cubes:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

Sum of cubes:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Note that in these latter two identities, the resulting quadratic factors have no linear factor with Real coefficients.

So we find:

#27x^9+8x^6-27x^3-8#

#= (27x^9+8x^6)-(27x^3-8)#

#= x^6(27x^3+8)-1(27x^3-8)#

#= (x^6-1)(27x^3+8)#

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

#= (x^3-1)(x^3+1)(3x+2)(9x^2-6x+4)#

#= (x-1)(x^2+x+1)(x+1)(x^2-x+1)(3x+2)(9x^2-6x+4)#

So there are three Real zeros:

#{ (x=1),(x=-1),(x=-2/3) :}#

And six non-Real Complex zeros:

#{ (x = -1/2+-sqrt(3)/2i), (x = 1/2+-sqrt(3)/2i), (x = 1/3+-sqrt(3)/3i) :}#