How do you FOIL ${(5 - 2 \sqrt 2)}^{2}$ $(5 - 2√2)^2$ ?

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How do you FOIL ${(5 - 2 \sqrt 2)}^{2}$ $(5 - 2√2)^2$ ?

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Answer 1 · 2015-05-26T05:34:40

Problem: FOIL ${(5 - 2 \sqrt{2})}^{2}$ $(5-2sqrt2)^2$ .
![http://math.tutorvista.com/algebra/foil-method.html

$(5 - 2 \sqrt{2}) (5 - 2 \sqrt{2})$ $(5-2sqrt 2)(5-2sqrt 2)$

Firsts: $5 \cdot 5 = 25$ $5*5=25$

Outers: $5 \cdot - 2 \sqrt{2} = - 10 \sqrt{2}$ $5*-2sqrt 2=-10sqrt 2$

Inners: $- 2 \sqrt{2} \cdot 5 = - 10 \sqrt{2}$ $-2sqrt2*5=-10sqrt2$

Lasts: $2 \sqrt{2} \cdot 2 \sqrt{2} = 4 \sqrt{2} \sqrt{2} = 4 \cdot \sqrt{2 \times 2} = 4 \sqrt{4} = 4 \cdot 2 = 8$ $2sqrt 2*2sqrt 2=4sqrt 2sqrt 2=4*sqrt(2xx2)=4sqrt4=4*2=8$

Combine the terms.

$25 - 10 \sqrt{2} - 10 \sqrt{2} + 8$ $25-10sqrt2-10sqrt2+8$ =

$25 - 20 \sqrt{2} + 8$ $25-20sqrt 2+8$ =

$33 - 20 \sqrt{2}$ $33-20sqrt 2$

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How do you FOIL ${(5 - 2 \sqrt 2)}^{2}$ $(5 - 2√2)^2$ ?

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