Solve (x+y)^5?

Question

that is (x+y) to the power of 5

. show working

binomial expansion

Answer 1

I assume you mean to multiply this out...

a( x^5) + b (x^4)y + c( x^3)(y^2) + d( x^2)(y^3) + e(x)( y^4) + f y^5

where a = 5 ! / ( 5! 0!) = 1 ,

b = 5 ! / ( 4!1!) = 5 c = 5! / ( 3!2!) = 5*4 / 2 = 10
d = 5 ! / ( 2!3!) = 10 e = 5! / ( 1! 4! ) = 5
f = 5! / ( 0! 5! ) = 1
............
factorials... 5 ! = 5* 4 * 3 *2 *1
so 5! / ( 3! 2! ) = [ 5* 4* 3* 2* 1 ] / [ ( 3 * 2* 1) * ( 2 * 1 ) ] = 5 * 4 / ( 2 * 1 ) = 20 / 2 = 10
etc........
the coeff a, b, c, etc... come from the exponent, 5, and the two factorials in the denom come from the exponents on the x and y terms... thus ( x^4) ( y ) ----> 4 ! * 1 !

x^5 implies x^5 * y^0 , thus the denom is 5 ! * 0 !

finally, 1 ! = 1, and 0 ! = 1
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don't forget to choose one of these posted answers as the Best Answer... thanks !!

Answer 2

x^5+5x^4y+10 x^3 y^2+ 10 x^2 y^3+ 5 x y^4 +y^5.

In these binomial expansions, if you have (x+y)^m, any term x^j y^k where j+k=m has the coefficient of m!/(j!k!)

Answer 3

Have you heard of Pascal's triangle? It goes like this

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

The next row is computed by adding each pair of numbers from the row above (with 1's on the ends). Then

(x + y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5

Answer 4

x^5 + 5*x^4*y + 10*x^3*y^2 + 10*x^2*y^3 +5*x*y^4 +y^5

Answer 5

35 and y squared

Answer 6

(x + y)^5
= (x+y)(x+y)(x+y)(x+y)(x+y)

=(x^2+2xy+y^2)
(x^2+2xy+y^2)(x+y)

=(x^2+2xy+y^2)
(x^3+2x^2y+xy^2
+x^2y+2xy^2+y^3)

=(x^2+2xy+y^2)
(x^3+3x^2y+3xy^2+y^3)

=(x^5+3x^4y+3x^3y^2+x^2y^3
+2x^4y+6x^3y^2+6x^2y^3+2xy^4
+x^3y^2+3x^2y^3+3xy^4+y^5)

combine like terms
=
x^5 + 5x^4y + 10x^3y^2+ 10x^2y^3 + 5xy^4 + y^5

Answer 7

There is no solving this....no equation given.

If you just want to expand it...just use pascals triangle

1 to the 0
1,1 to the 1st
1,2,1 to the 2nd
1,3,3,1 to the 3rd
1,4,6,4,1 4th
1,5,10,10,5,1 5th
1,6,15,20,15,6,1 6th


got the pattern?

Answer 8

You can't solve that on its own, x an y could be any number at all.

Answer 9

(x+y)^5=5C0*x^5+5C1*x^(5-1)*y^1+5C2*x^(5-2)*y^2+5C3*x^(5-3)*y^3+5C4*x^(5-4)*y^4+5C5y^5

= x^5+5x^4*y+10x^3*y^2+10x^2*y^3+5xy^4+y^5