# How do you factor 18b^2+24bt+8t^2?

Given $18 {b}^{2} + 24 b t + 8 {t}^{2}$
If you extract the obvious common factor of $2$
$\textcolor{w h i t e}{\text{XXXX}}$$2 \left(9 {b}^{2} + 12 b t + 4 {t}^{2}\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$= 2 \left({\left(3 b\right)}^{2} + 2 \left(3 b\right) \left(2 t\right) + {\left(2 t\right)}^{2}\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$2 {\left(3 b + 2 t\right)}^{2}$