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

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Answer 1 · 2016-08-14T13:15:57

$f(x)$ has zeros $4$ and $+-4i$

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:

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

$=(x^3-4x^2)+(16x-64)$

$=x^2(x-4)+16(x-4)$

$=(x^2+16)(x-4)$

$=(x^2-(4i)^2)(x-4)$

$=(x-4i)(x+4i)(x-4)$

Hence zeros:

$+-4i$ and $4$

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