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

Jan 8, 2017

$f \left(x\right)$ has 9 Complex zeros, of which $3$ are Real and $6$ non-Real.

#### Explanation:

By the Fundamental Theorem of Algebra $f \left(x\right)$ 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:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Difference of cubes:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Sum of cubes:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

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

So we find:

$27 {x}^{9} + 8 {x}^{6} - 27 {x}^{3} - 8$

$= \left(27 {x}^{9} + 8 {x}^{6}\right) - \left(27 {x}^{3} - 8\right)$

$= {x}^{6} \left(27 {x}^{3} + 8\right) - 1 \left(27 {x}^{3} - 8\right)$

$= \left({x}^{6} - 1\right) \left(27 {x}^{3} + 8\right)$

$= \left({\left({x}^{3}\right)}^{2} - {1}^{2}\right) \left({\left(3 x\right)}^{3} + {2}^{3}\right)$

$= \left({x}^{3} - 1\right) \left({x}^{3} + 1\right) \left(3 x + 2\right) \left(9 {x}^{2} - 6 x + 4\right)$

$= \left(x - 1\right) \left({x}^{2} + x + 1\right) \left(x + 1\right) \left({x}^{2} - x + 1\right) \left(3 x + 2\right) \left(9 {x}^{2} - 6 x + 4\right)$

So there are three Real zeros:

$\left\{\begin{matrix}x = 1 \\ x = - 1 \\ x = - \frac{2}{3}\end{matrix}\right.$

And six non-Real Complex zeros:

$\left\{\begin{matrix}x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i \\ x = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i \\ x = \frac{1}{3} \pm \frac{\sqrt{3}}{3} i\end{matrix}\right.$