Factoring

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Factoring

by tonebeeze » Tue Apr 12, 2011 12:10 pm
Can someone walk me through the factoring process in statement 1. Please see attachment for Question.

OA =A
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Screen shot 2011-04-12 at 1.07.19 PM.png

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by clock60 » Tue Apr 12, 2011 12:48 pm
my try
4^m+4^m+4^m+4^m=4*4^m=2^2*2^(2*m)=2^(2*m+2)
2^(m+2)-2^(m+1)=2^m*2^2-2^m*2=2^m*(2^2-2)=2^m*2=2^(m+1)
and now divide
2^(2*m+2)/2^(m+1)=2^(2*m+2-m-1)=2^(m+1)
and 2^(m+1)=1=2^0
m+1=0, m=-1
i hope that i did not miss any traps that are commom in exponents problems

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by tonebeeze » Tue Apr 12, 2011 1:12 pm
No...you executed the problem well. Thanks!