How do you multiply #(2r+9s)^2#?

1 Answer
Aug 29, 2016

Answer:

#4r^2+36rs+81s^2#

Explanation:

Consider #(2r+9s)^2=(2r+9s)(2r+9s)#

We must ensure that each term in the 2nd bracket is multiplied by each term in the first bracket.This can be achieved as follows.

#(color(red)(2r+9s))(2r+9s)#

#=color(red)(2r)(2r+9s)+color(red)(9s)(2r+9s)#

distribute the brackets : #4r^2+18rs+18rs+81s^2#

and simplifying to obtain : #4r^2+36rs+81s^2#

#rArr(2r+9s)^2=4r^2+36rs+81s^2#
#"----------------------------------------------------------"#
Alternatively there is the FOIL method.

#Fto" First terms"#
#Oto" Outer terms"#
#Ito" Inner terms"#
#Lto" Last terms"#

F-multiply the first terms in each bracket together.
O-multiply the outer terms together.
I-multiply the inner terms together.
L-multiply the last terms together.

#rArr(2r+9s)(2r+9s)#

#=(2rxx2r)+(2rxx9s)+(2rxx9s)+(9sxx9s)#

#=4r^2+18rs+18rs+81s^2=4r^2+36rs+81s^2#