# How do you factor (x^2-9)×(x^3-8)?

$\left({x}^{2} - 9\right) = \left(x - 2\right) \left(x + 2\right)$
$\left({x}^{3} - 8\right) = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$
(note that $\left({x}^{2} + 2 x + 4\right)$ has no Real factors; this may have been an intentional trick)
$\left({x}^{2} - 9\right) \times \left({x}^{3} - 8\right)$
${\left(x - 2\right)}^{2} \left(x + 2\right) \left({x}^{2} + 2 x + 4\right)$