## Algebra 1

$s^{2}+12st^{2}+36t^{4}$
$(s+6t^{2})^{2}$ Using the rule of $(a+b)^{2}$ = $a^{2}$+2ab+$b^{2}$ In this case, the a= s and b= $6t^{2}$ as substituting these gives us the original polynomial $a^{2}$+2ab+$b^{2}$ $s^{2}$+2(s)($6t^{2}$)+$(6t^{2})^{2}$ Thus, $s^{2}+12st^{2}+36t^{4}$