a^2 + b^2 + 2ab
2006-10-01 01:42:10
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answer #1
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answered by prince to rule 1
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(a+b)^2
= (a+b)(a+b)
= a(a+b) + b(a+b)
= a^2 +a*b + a*b + b^2
= a^2 + 2*a*b + b^2
2006-10-01 03:54:06
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answer #2
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answered by K Sengupta 4
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a^2 + b^2
2006-10-01 01:39:40
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answer #3
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answered by Anonymous
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(a + b)^2 = a^2 + b^2 + 2ab
2006-10-01 01:41:46
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answer #4
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answered by Alan 2
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a^2 + 2ab + b^2
2006-10-01 02:39:58
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answer #5
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answered by Anonymous
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a^2+b^2+2ab
if the degree greater, use pascal triagle to find the coefficients.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
...............
2006-10-01 01:41:45
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answer #6
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answered by iyiogrenci 6
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(a+b)(a+b), which factors to a²+ab+ab+b², or a²+2ab+b²
I am so tired of people prescribing wikipedia for every single question. It isn't even a reliable source. Anyone can screw around with the entries and it usually takes days for it to be "corrected". From now on, I am immediately going to give a negative rating whenever I see someone cite that as their source or suggest that someone find their answers there.
2006-10-01 01:39:43
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answer #7
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answered by Anonymous
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a^2+2ab+b^2
2006-10-01 01:40:27
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answer #8
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answered by Noble 4
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(a+b)^2 = (a+b) (a+b)
Now you multiply such that each term multiplies both terms on the other expression.
i.e,
aa + ab + ab + bb = a^2 + 2ab + b^2
You can easily figure any (a+b)^n where n is an integer (or even fraction) with the binomial theorem without continuous multiplication.
The above link is worth visiting but for most purposes the simplest version of the theorem will suffice.
http://en.wikipedia.org/wiki/Binomial_theorem
2006-10-01 04:23:33
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answer #9
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answered by yasiru89 6
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FOIL Method
(a + b)²
(a + b)(a + b) = a² + ab + ab + b²= a² + 2ab + b²
The answer is a² + 2ab + b²
2006-10-01 08:08:58
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answer #10
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answered by SAMUEL D 7
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