Question

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

Answer 1

`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`