Answer
$${\left( {x + 2y} \right)^4} = {x^4} + 8{x^3}y + 24{x^2}{y^2} + 32x{y^3} + 16{y^4}$$
Work Step by Step
$$\eqalign{
& {\left( {x + 2y} \right)^4} \cr
& {\rm{Apply \,the\, binomial\, theorem}} \cr
& {\left( {x + 2y} \right)^4} = {\left( x \right)^4} + \left( \matrix{
4 \hfill \cr
1 \hfill \cr} \right){\left( x \right)^3}\left( {2y} \right) + \left( \matrix{
4 \hfill \cr
2 \hfill \cr} \right){\left( x \right)^2}{\left( {2y} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( \matrix{
4 \hfill \cr
3 \hfill \cr} \right)\left( x \right){\left( {2y} \right)^3}\,\, + {\left( {2y} \right)^4} \cr
& {\rm{Evaluate \,each\, binomial\,coefficient \,use }}\left( \matrix{
c \hfill \cr
r \hfill \cr} \right) = {{n!} \over {\left( {n - r} \right)!r!}} \cr
& {\left( {x + 2y} \right)^4} = {\left( x \right)^4} + {{4!} \over {3!1!}}{\left( x \right)^3}\left( {2y} \right) + {{4!} \over {2!2!}}{\left( x \right)^2}{\left( {2y} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{4!} \over {1!3!}}\left( x \right){\left( {2y} \right)^3}\, + {\left( {2y} \right)^4} \cr
& {\rm{Simplify}} \cr
& {\left( {x + 2y} \right)^4} = {\left( x \right)^4} + 4{\left( x \right)^3}\left( {2y} \right) + 6{\left( x \right)^2}{\left( {2y} \right)^2} + 4\left( x \right){\left( {2y} \right)^3}\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {2y} \right)^4} \cr
& {\left( {x + 2y} \right)^4} = {x^4} + 8{x^3}y + 24{x^2}{y^2} + 32x{y^3} + 16{y^4} \cr} $$
