# How do you find all the zeros of f(x)= x^3 - 2x^2 + 5x -10 with its multiplicities?

Aug 20, 2016

$f \left(x\right)$ has zeros $2$ and $\pm \sqrt{5} i$

#### Explanation:

This cubic factors by grouping then using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = x$ and $b = \sqrt{5} i$ as follows:

${x}^{3} - 2 {x}^{2} + 5 x - 10$

$= \left({x}^{3} - 2 {x}^{2}\right) + \left(5 x - 10\right)$

$= {x}^{2} \left(x - 2\right) + 5 \left(x - 2\right)$

$= \left({x}^{2} + 5\right) \left(x - 2\right)$

$= \left({x}^{2} - {\left(\sqrt{5} i\right)}^{2}\right) \left(x - 2\right)$

$= \left(x - \sqrt{5} i\right) \left(x + \sqrt{5} i\right) \left(x - 2\right)$

Hence zeros:

$\pm \sqrt{5} i$ and $2$