Answer
$${x^{12}} + 24{x^{11}} + 264{x^{10}} + 1760{x^9}$$
Work Step by Step
$$\eqalign{
& {\rm{first\, four\, terms of }}{\left( {x + 2} \right)^{12}} \cr
& {\rm{Apply\, the\, binomial\, theorem\, to\, the\, first\, four \,terms}} \cr
& {\left( {x + 2} \right)^{12}} = {\left( x \right)^{12}} + \left( \matrix{
12 \hfill \cr
1 \hfill \cr} \right){\left( x \right)^{11}}\left( 2 \right) + \left( \matrix{
12 \hfill \cr
2 \hfill \cr} \right){\left( x \right)^{10}}{\left( 2 \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( \matrix{
12 \hfill \cr
3 \hfill \cr} \right){\left( x \right)^9}{\left( 2 \right)^3} \cr
& {\left( {x + 2} \right)^{12}} = {\left( x \right)^{12}} + {{12!} \over {11!}}{\left( x \right)^{11}}\left( 2 \right) + {{12!} \over {10!2!}}{\left( x \right)^{10}}{\left( 2 \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{12!} \over {9!3!}}{\left( x \right)^9}{\left( 2 \right)^3} \cr
& {\rm{Simplify}} \cr
& {\left( {x + 2} \right)^{12}} = {x^{12}} + 12{\left( x \right)^{11}}\left( 2 \right) + 66{\left( x \right)^{10}}{\left( 2 \right)^2} + 220{\left( x \right)^9}{\left( 2 \right)^3} \cr
& {\left( {x + 2} \right)^{12}} = {x^{12}} + 24{x^{11}} + 264{x^{10}} + 1760{x^9} \cr} $$
