algabra help

[socialcoma]

(2X+Y)^5
can someone tell me how to expand this?

[Tom Mattson]

Just do successive multiplications:

(2x+y)(2x+y)(2x+y)(2x+y)(2x+y)

Multiply the first two:

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

Then the next two, etc. It's messy, but straightforward.

[socialcoma]

i know how to do that, but thanks for you replay. i am trying to find a faster way. possibly using factorials

[Tyger]

the Binomial Expansion, also known as Newtons Expansion.

[socialcoma]

how do you do newtons expansion?

[KLscilevothma]

Do you know Pascal Triangle? Or have you learnt combinations, Cnr, before?

[meteor]

Newton's binomial (a.k.a. Newton's expansion) is this:

(a+b)^n=(a^n)+(n*((a^(n-1))*b))+((n*(n-1)*(a^(n-2))*b^2)/(2!))+((n*(n-1)*(n-2)*(a^(n-3))*b^3)/(3!))+...+(b^n)

n can be any rational number

[I_am_hamster]

Or shorter

(a+b)^n=SUM (from m=0 to n) C(m out of n)*a^m*b^(n-m)

Damn can I turn on the HTML code?

[Thoth]

Perhaps I can help you with that.

This is how it goes from a simple binomial theorem:

(a+b)0=1 since anything to the power of zero is 1

(a+b)1=a+b

(a+b)2=a2+2ab+b2

(a+b)3=a3+3a2b+3ab2+b3

As we go on and on we can clearly see that a pattern is emerging. Look at the next post just following this.

[Thoth]

First, notice that if we add the powers for a and b the result is always equal to the original power given to the term.

For example in (a+b) 2= a 2+2ab+b 2 notice that power in each term is always equal to 2. The first a2, the second a 1 and b 1 and again 1+1 is equal 2. The same goes for a3+3a2b+3ab3+b3 and on and on.
Go to the next coming post.

[Thoth]

Second, it is apparent that power decreases from a, and increases in b as we go forward.

For example in (a+b)5=a 5 b 0+ a 4 b 1 + a 3 b 2 + a 2 b 3 + a 1 b 4 + a 0 b 5.
As it must become obvious from the above, a starts with power 5 and goes to power 0 and b starts with power 0 and goes to power 5. Of course in the above we are missing the coefficient for each term. Now I show you how to find them.

[Thoth]

To make writing the coefficients clear I rewrite the above powering using only coefficients.

i.e. how many of each kind of term:
(a + b)
1 1 0+
0 1 1
-------
1 2 1 0
(a + b)2
1 2 1 0+
0 1 2 1
----------
(a + b) 3
1 3 3 1 0 +
0 1 3 3 1
-------------
(a + b) 4
1 4 6 4 1 +

0 1 4 6 4 1
---------------------------
a + b) 5
1 5 10 10 5 1

This is what is known as Pascal's Triangle. The last thing that you have to do is substitute 2x for a and y for b in the above. Good luck
[;)]