(X+Y)^n = ∑ C(n,r)*X^r * Y^(n-r)
(X+Y)^4 = X^4 + 4X^3Y + 6x^2Y^2 + 4Xy^3 + Y^4
If you know about Pascal's triangle you could also have just substituted the coefficients.
2007-06-03 14:51:08
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answer #1
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answered by Dr D 7
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use the 4th row of Pascal's triangle (1, 4, 6, 4, 1) and expand the binomial with the binomial theorem:
1x^4 + 4x^3y + 6x²y² + 4xy^3 + y^4
2007-06-03 14:52:09
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answer #2
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answered by Kathleen K 7
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First, you are going to get something that looks like
ax^4 + bx^3y + cx^2y^2 + dx^y3 + ey^4
Using Pascal's triangle, we can find the values of a through e
1
1 . 1
1 . 2 . 1
1 . 3 . 3 . 1
1 . 4 . 6 . 4 . 1
So
x^4 + 4x^3y + 6x^2y^2 + 4x^y3 + y^4
2007-06-03 14:55:44
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answer #3
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answered by TychaBrahe 7
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Either manually using FOIL, or using Binomial expansion:
(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
2007-06-03 14:51:35
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answer #4
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answered by Kemmy 6
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Use Pascal's triangle
1
11
121
1331
14641
The x exponents go down, the y go up
x^4+4x^3y+6x^2y^2+4xy^3+y^4
2007-06-03 14:51:23
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answer #5
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answered by llllarry1 5
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can be done effectively using binomial theorem, or as simple by
(X+Y)^4= ((X+Y)^2) *((X+Y)^2)
2007-06-03 14:56:03
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answer #6
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answered by logesh 2
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