# Algabra help

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

Tom Mattson
Staff Emeritus
Gold Member
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

You use

the Binomial Expansion, also known as Newtons Expansion.

socialcoma
how do you do newtons expansion?

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

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

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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?

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Thoth

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 