i am trying to work out the expansion of (2x+y)^4?

(2x + y) (2x + y) (2x + y) (2x + y)
(4x² + 4xy + y² ) (4x² + 4xy + y² )

16x^4 + 16x³y + 4x²y²
----------16x³y + 16x²y² + 4xy³
---------------------4x²y² + 4xy³ + y^4

16x^4 + 32x³y + 24x²y² + 8xy³ + y^4

Generally there are ways to find the coefficients. The easiest is pascal's triangle.
Let (a + b)^n be the binomial we want to expand.
You begin with 1
then 121, and then for each step you add the two numbers above, putting number 1 in the beginning and the end.
In the 4th step you have:
.........1
......1...2...1
...1...3...3...1
1...4...6...4..1
......
So we found the coefficients.
Then with the first coefficient (1) we put a in the power n,
then the next number we but the 2nd coefficient (4), then a^n-1*b1, etc... reducing the exponent of a by 1, and increasing the exponent of be by 1.

so for n=4 we have:
(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
If we put a = 2x and b = y we take:
(2x + y)⁴ = (2x)⁴ + 4 (2x)³ y + 6 (2x)² y² + 4 (2x) y³ + y⁴ =
16x⁴ + 32x³ y + 24x² y² + 8xy³ + y⁴

For more info you can view wikipedia.
Hope this helps!

(2x + y)^4
= (2x + y)(2x + y)(2x + y)(2x + y)
= (4x^2 + 4xy + y^2)(4x^2 + 4xy + y^2)
= 16x^4 + 16x^3y + 4x^2y^2 + 16x^3y + 16x^2y^2 + 4xy^3 + 4x^2y^2 + 4xy^3 + y^4
= 16x^4 + 16x^3y + 16x^3y + 4x^2y^2 + 16x^2y^2 + 4x^2y^2 + 4xy^3 + 4xy^3 + y^4
= 16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4

hi,for solving your problem first of all you have to expand the term (2x+y)^2 by putting the square sign over the end of the bracket. eg. (4x^2 + 4xy + y^2 )^2.now you have to expand the bracket what you got as a rule of (a+b+c)^2 so finally you will get the answer like given below,
(16x^4+16x^2y^2 + y^4 +32x^3y + 8xy^3 +8x^y^2 ) by simplifing this you will get ,
(16x^4 + 24x^2y^2 +32x^3y + 8xy^3 )

The easiest way to do this is to write the definition of an exponent. You know that 3^4 = 3 * 3 * 3 * 3. Just do the same with your polynomial.

(2x + y)^4 = (2x+y) * (2x+y) * (2x+y) * (2x+y)

The you would just multiply the first two (2x+y) together with the last pair, then multiply those again:
(2x+y) * (2x+y) * (2x+y) * (2x+y)
(1) = [(2x+y) * (2x+y)] * [(2x+y) * (2x+y)] by associativity

(2) = [4x² + 2xy + 2xy + y²] * [ 4x² + 2xy + 2xy + y²] multiply them out

(3) = [4x² + 4xy + y²] * [4x² + 4xy + y²] add common terms

(4) = 16x^4 + 16yx^3 + 4x²y² + 16yx^3 + 16x²y² + 4xy^3 + 4x²y² + 4xy^3 + y^4 (by multiplying them out)

(5) = 16x^4 + 32yx^3 + 24x²y² + 8xy^3 + y^4 (by combining common terms)

Or, alternatively, to speed up time just say that:

(2x + y)^4 = (2x+y)² * (2x+y)²

= (4x² + 4xy + y²) + (4x² + 4xy + y²) since (a+b)² = (a² + 2ab + b²)

From there you can see that it's identical to step (3) and just continue the rest :)

(4x^2+y^2+4xy)x(4x^2+y^2+4xy)=16x^4+y^4+24x^2y^2+32x^3y+8xy^3

square it twice:
(4x^2+4xy+y^2)(4x^2+4xy+y^2)

then multiply by foiling
16x^4+16x^3y+4x^2y^2+16x^3y+16x^2y^2+4xy^3+4x^2y^2+4xy^3+y^4

then simplify:
16x^4 + 32x^3y + 22x^2y^2 + 8xy^3 + y^4

use Pascal's triangle

..............1
.........1.....1
.......1...2....1
....1...3....3....1
.1...4...6...4...1
so it is
1(2x)^4 + 4(2x)^3*y+6(2x)^2*y^2+4(2x)*y^3+y^4

ans...
16x^4 + 32x^3 y + ...

you can do it!

Lets do this the long way:

(2x+y)(2x+y) = 4x^2+4xy+y^2. Now lets multiply that by itself:

(4x^2+4xy+y^2)(4x^2+4xy+y^2) = 16x^4+16x^3y+4x^2y^2+16x^3y+16x^2y^2+4xy^3+4x^2y^2+4xy^3+y^4.

Yea, its messy but lets combine like terms and you get your answer