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2007-06-03 14:48:04 · 6 answers · asked by Jenna 2 in Science & Mathematics Mathematics

6 answers

(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 · answer #1 · answered by Dr D 7 · 1 0

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 · answer #2 · answered by Kathleen K 7 · 0 0

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 · answer #3 · answered by TychaBrahe 7 · 0 0

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 · answer #4 · answered by Kemmy 6 · 0 0

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 · answer #5 · answered by llllarry1 5 · 0 0

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 · answer #6 · answered by logesh 2 · 0 0

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