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

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

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Answer 1 · 2016-08-14T13:15:57

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

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:

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

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

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

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

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

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

Hence zeros:

$\pm 4 i$ $+-4i$ and $4$ $4$

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

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Explanation:

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How do you find all the zeros of f(x)=x3−4x2+16x−64f(x)=x^3-4x^2+16x-64?

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Explanation:

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