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

Aug 14, 2016

$f \left(x\right)$ has zeros $1$ and $\pm 2 i$

Explanation:

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:

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

$= {x}^{2} \left(x - 1\right) + 4 \left(x - 1\right)$

$= \left({x}^{2} + 4\right) \left(x - 1\right)$

$= \left({x}^{2} - {\left(2 i\right)}^{2}\right) \left(x - 1\right)$

$= \left(x - 2 i\right) \left(x + 2 i\right) \left(x - 1\right)$

Hence zeros: $\pm 2 i$ and $1$