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

Oct 17, 2016

$f \left(x\right)$ has $2$ Real zeros and $4$ non-Real Complex zeros.

#### Explanation:

$f \left(x\right) = 27 {x}^{6} + 208 {x}^{3} - 64$

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

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Descartes' Rule of Signs

The signs of the coefficients of $f \left(x\right)$ follow the pattern $+ + -$

With one change of sign, that means that $f \left(x\right)$ has exactly one positive Real zero.

The signs of the coefficients of $f \left(- x\right)$ follow the pattern $+ - -$

With one change of sign, that means that $f \left(x\right)$ has exactly one negative Real zero.

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

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Bonus - Direct solution

$27 f \left(x\right) = 27 \left(27 {x}^{6} + 208 {x}^{3} - 64\right)$

$\textcolor{w h i t e}{27 f \left(x\right)} = 729 {x}^{6} + 5616 {x}^{3} - 1728$

$\textcolor{w h i t e}{27 f \left(x\right)} = {\left(27 {x}^{3}\right)}^{2} + 2 \left(104\right) \left(27 {x}^{3}\right) + {\left(104\right)}^{2} - {112}^{2}$

$\textcolor{w h i t e}{27 f \left(x\right)} = {\left(27 {x}^{3} + 104\right)}^{2} - {112}^{2}$

$\textcolor{w h i t e}{27 f \left(x\right)} = \left(\left(27 {x}^{3} + 104\right) - 112\right) \left(\left(27 {x}^{3} + 104\right) + 112\right)$

$\textcolor{w h i t e}{27 f \left(x\right)} = \left(27 {x}^{3} - 8\right) \left(27 {x}^{3} + 216\right)$

$\textcolor{w h i t e}{27 f \left(x\right)} = 27 \left({\left(3 x\right)}^{3} - {2}^{3}\right) \left({x}^{3} + {2}^{3}\right)$

$\textcolor{w h i t e}{27 f \left(x\right)} = 27 \left(3 x - 2\right) \left(9 {x}^{2} + 6 x + 4\right) \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

$\textcolor{w h i t e}{27 f \left(x\right)} = 27 \left(3 x - 2\right) \left(3 x - 2 \omega\right) \left(3 x - 2 {\omega}^{2}\right) \left(x + 2\right) \left(x + 2 \omega\right) \left(2 + 2 {\omega}^{2}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$

So the Real zeros are:

$x = \frac{2}{3}$

$x = - 2$

and the non-Real Complex zeros are:

$x = \frac{2}{3} \omega$

$x = \frac{2}{3} {\omega}^{2}$

$x = - 2 \omega$

$x = - 2 {\omega}^{2}$