Question

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

Answer 1

`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) :}`