# How do you find all the zeros of x^2 - 8x - 48 = 0 with its multiplicities?

Apr 16, 2016

$x = 12$ or $x = - 4$, each with multiplicity $1$

#### Explanation:

Complete the square then use the difference of squares identity:

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

with $a = \left(x - 4\right)$ and $b = 8$ as follows:

$0 = {x}^{2} - 8 x - 48$

$= {\left(x - 4\right)}^{2} - 16 - 48$

$= {\left(x - 4\right)}^{2} - {8}^{2}$

$= \left(\left(x - 4\right) - 8\right) \left(\left(x - 4\right) + 8\right)$

$= \left(x - 12\right) \left(x + 4\right)$

So $x = 12$ or $x = - 4$, each with multiplicity $1$