# How do you factor 4t^3-64t^2-128t?

$4 t \left(t - 8 + 4 \sqrt{6}\right) \left(t - 8 - 4 \sqrt{6}\right)$
$4 {t}^{3} - 64 {t}^{2} - 128 t = 4 t \left({t}^{2} - 16 t - 32\right) = 4 t \left({\left(t - 8\right)}^{2} - 64 - 32\right) = 4 t \left({\left(t - 8\right)}^{2} - 96\right) = 4 t \left({\left(t - 8\right)}^{2} - {\left(4 \sqrt{6}\right)}^{2}\right) = 4 t \left(\left(t - 8 + 4 \sqrt{6}\right) \left(t - 8 - 4 \sqrt{6}\right)\right)$[Ans]