How do you FOIL #(5 - 2√2)^2#?

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Answer 1 · 2015-05-26T05:34:40

Problem: FOIL #(5-2sqrt2)^2# .

#(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#