# How do you find all the zeros of f(x)=x^3-4x^2+16x-64?

Aug 14, 2016

$f \left(x\right)$ has zeros $4$ and $\pm 4 i$

#### Explanation:

Note that the ratio of the first and second terms is the same as that of the third and fourth terms. So this cubic factors by grouping:

${x}^{3} - 4 {x}^{2} + 16 x - 64$

$= \left({x}^{3} - 4 {x}^{2}\right) + \left(16 x - 64\right)$

$= {x}^{2} \left(x - 4\right) + 16 \left(x - 4\right)$

$= \left({x}^{2} + 16\right) \left(x - 4\right)$

$= \left({x}^{2} - {\left(4 i\right)}^{2}\right) \left(x - 4\right)$

$= \left(x - 4 i\right) \left(x + 4 i\right) \left(x - 4\right)$

Hence zeros:

$\pm 4 i$ and $4$