# How do you find all the zeros of f(x) = x^2 + 6x + 18?

Mar 10, 2016

Complete the square and use the difference of squares identity to find zeros:

$x = - 3 \pm 3 i$

#### Explanation:

Complete the square then use the difference of squares identity:

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

with $a = \left(x + 3\right)$ and $b = 3 i$ as follows:

$f \left(x\right) = {x}^{2} + 6 x + 18$

$= {x}^{2} + 6 x + 9 + 9$

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

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

$= \left(\left(x + 3\right) - 3 i\right) \left(\left(x + 3\right) + 3 i\right)$

$= \left(x + 3 - 3 i\right) \left(x + 3 + 3 i\right)$

Hence $f \left(x\right) = 0$ when $x = - 3 \pm 3 i$