How do you FOIL #(5 - 2√2)^2#?

1 Answer
May 26, 2015

Problem: FOIL #(5-2sqrt2)^2# .
http://math.tutorvista.com/algebra/foil-method.html?view=simple

#(5-2sqrt 2)(5-2sqrt 2)#

Firsts: #5*5=25#

Outers: #5*-2sqrt 2=-10sqrt 2#

Inners: #-2sqrt2*5=-10sqrt2#

Lasts: #2sqrt 2*2sqrt 2=4sqrt 2sqrt 2=4*sqrt(2xx2)=4sqrt4=4*2=8#

Combine the terms.

#25-10sqrt2-10sqrt2+8# =

#25-20sqrt 2+8# =

#33-20sqrt 2#