# How do you find all the zeros of f(x) = x^2 - 4x - 60 with its multiplicities?

Apr 6, 2016

$f \left(x\right)$ has zeros at $x = 6$ and $x = - 4$ (there are only $2$ zeros including multiplicity)

#### Explanation:

Factoring
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = {x}^{2} - 4 x - 60$

$\textcolor{w h i t e}{\text{XXXXX}} = \left(x + 4\right) \left(x - 6\right)$

When $f \left(x\right) = 0$
either
$\textcolor{w h i t e}{\text{XXX}} \left(x + 4\right) = 0 \rightarrow x = - 4$
or
$\textcolor{w h i t e}{\text{XXX}} \left(x - 6\right) = 0 \rightarrow x = + 6$

Since $f \left(x\right)$ is a quadratic (i.e. of degree $2$)
it has (including multiplicity and possible complex values) exactly $2$ zeroes.