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2006-10-01 01:38:05 · 11 answers · asked by pradeep_kumar19 1 in Science & Mathematics Mathematics

11 answers

a^2 + b^2 + 2ab

2006-10-01 01:42:10 · answer #1 · answered by prince to rule 1 · 0 0

(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 · answer #2 · answered by K Sengupta 4 · 0 0

a^2 + b^2

2006-10-01 01:39:40 · answer #3 · answered by Anonymous · 0 2

(a + b)^2 = a^2 + b^2 + 2ab

2006-10-01 01:41:46 · answer #4 · answered by Alan 2 · 0 0

a^2 + 2ab + b^2

2006-10-01 02:39:58 · answer #5 · answered by Anonymous · 0 0

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 · answer #6 · answered by iyiogrenci 6 · 1 0

(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 · answer #7 · answered by Anonymous · 0 0

a^2+2ab+b^2

2006-10-01 01:40:27 · answer #8 · answered by Noble 4 · 0 0

(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 · answer #9 · answered by yasiru89 6 · 0 1

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 · answer #10 · answered by SAMUEL D 7 · 0 0

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