2007-12-13 01:44:45 · 9 answers · asked by (ƸӜƷ) 1 in Science & Mathematics ➔ Mathematics
(x+y)^2 = 2C0 x^2 + 2C1 x^1 y^1 + 2C2 y^2
= x^2 + 2 x^1 y^1 + y^2( as 2C0 = 2C2 =1 and 2C1 = 2 )
= x^2 + 2 x y + y^2
2007-12-13 01:53:00 · answer #1 · answered by bharat m 3 · 0⤊ 2⤋
x^2 + 2*x*y + y^2
2007-12-13 01:48:49 · answer #2 · answered by Anonymous · 0⤊ 1⤋
Use FOIL on the two binomials and then simplify by subtracting 4. First: (2)(1) = 2 Outer: (2)(3x) = 6x Inner: (x)(1) = x Last: (x)(3x) = 3x^2 2 + 6x + x + 3x^2 -4 -2 + 7x + 3x^2
2016-05-23 09:08:03 · answer #3 · answered by ? 3 · 0⤊ 0⤋
(x+y)^n = (n 0) x^n + (n 1) x^(n-1) y + ... + (n n-1) x y^(n-1) + (n n) y^n
Putting n=2, (x+y)^2 = (2 0) x^2 + (2 1) x^(2-1) y + (2 2) y^2
= x^2 + 2xy + y^2
2007-12-13 01:57:11 · answer #4 · answered by Deep B 2 · 0⤊ 1⤋
coefficients of power 2 are 1 21
(x+y)^2=1(x)^2(y)^0+2(x)^1(y)^1+1(x)^0(y)^2
=x^2+2xy+y^2
2007-12-13 02:03:43 · answer #5 · answered by kamjinga 2 · 0⤊ 2⤋
(x + y)^2
= (x + y)(x + y)
= x(x + y) + y(x + y)
= x^2 + 2xy + y^2
2007-12-13 01:52:06 · answer #6 · answered by Mathematica 7 · 1⤊ 1⤋
x^4 + 2x^2y+y^2
2016-03-16 07:48:58 · answer #7 · answered by bree 1 · 1⤊ 0⤋
(X+Y)^n=Summation k=0 to n[nCk X^(n-k)Y^k]
(X+Y)^2=2C0X^2Y^0+2C1X^1Y^1+2C2X^0Y^2
(X+Y)^2=X^2+2XY+Y^2
2007-12-13 02:06:27 · answer #8 · answered by Avi 2 · 0⤊ 2⤋
(x + y)(x + y)
x² + xy + yx + y²
x² + xy + xy + y²
x² + 2xy + y²
2007-12-13 01:57:19 · answer #9 · answered by Como 7 · 2⤊ 1⤋