# How do you find all the zeros of f(x)=x^3-x^2+49x-49?

Aug 19, 2016

$f \left(x\right)$ has zeros $\pm 7 i$ and $1$

#### Explanation:

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

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

with $a = x$ and $b = 7 i$ as follows:

$f \left(x\right) = {x}^{3} - {x}^{2} + 49 x - 49$

$= \left({x}^{3} - {x}^{2}\right) + \left(49 x - 49\right)$

$= {x}^{2} \left(x - 1\right) + 49 \left(x - 1\right)$

$= \left({x}^{2} + 49\right) \left(x - 1\right)$

$= \left({x}^{2} - {\left(7 i\right)}^{2}\right) \left(x - 1\right)$

$= \left(x - 7 i\right) \left(x + 7 i\right) \left(x - 1\right)$

Hence $f \left(x\right)$ has zeros $\pm 7 i$ and $1$