How do you multiply #(x2y)^2#?
2 Answers
See a solution process below:
Explanation:
This is a special form of the quadratic:
Let:

#color(red)(a) = x# 
#color(blue)(b) = 2y#
Substituting gives:
Explanation:
We can rewrite
FOIL tells us that we multiply the first terms, outside terms, inside terms and last terms respectively. We get:
 Firsts
#(x*x)=x^2#  Outsides
#(x*2y)= 2xy#  Insides
#(2y*x)= 2xy#  Lasts
#(2y*2y)= 4y^2#
NOTE:Whenever we're multiplying binomials, we can use FOIL
Our new expression is as follows:
We can combine like terms to get:
Alternatively, we could have separated
We can distribute the
And we could combine like terms to get: