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

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Answer 1 · 2016-08-14T13:18:02

$f(x)$ has zeros $1$ and $+-2i$

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

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

$=x^2(x-1)+4(x-1)$

$=(x^2+4)(x-1)$

$=(x^2-(2i)^2)(x-1)$

$=(x-2i)(x+2i)(x-1)$

Hence zeros: $+-2i$ and $1$

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