MAMEWorld Forums - The Loony Bin

> Multiply [3 + (2a + b)]²
>
> Exactly what is says in the book. Even shows how to do it... but I don't understand
> it!

It's easy once you understand where the (a+b)^2=a^2+2ab+b^2 property is coming from.

First you have a layer of right/left distributivity:
(a+b)^2 = (a+b)(a+b) = a.(a+b) + b.(a+b) = aa + ab + ba + bb
(a+b+c)^2 = (a+b+c)(a+b+c) = a(a+b+c) + b(a+b+c) + c(a+b+c) = aa+ab+ac+ba+bb+bc+ca+cb+cc

What you get there is one term for every possible pair of letters, getting 4 of them for 2 initial terms, 9 for three, 16 for four, etc.

Now with the commutativity of multiplication and addition you can fold them together:
- for one letter, you get the pair with two times that letter once
- for two letters you get the pair twice, one in one order and the second the other way around (e.g. ab and ba)

So it generalizes as (a+b+...) = (sum of every term squared) + (sum of 2*every pair of terms in alphabetical order)

In the 3-term case that ends up (a+b+c)^2 = a^2+b^2+c^2+2ab+2ac+2bc

So (2a+b+3)^2 = 4a^2+b^2+9+4ab+12a+6b

Of course, that's not the way they want you to do it. They want you to take the (a+b)^2=a^2+2ab+b^2 as an article of faith, and use a extremely useful method called variable change to reuse it.

What is variable change, you ask? It just means that letters in an equation are just placeholders for whatever you want to put in them (hece the name variable, what they represent can vary). For clarity let's rewrite the equation as (X+Y)^2=X^2+2XY+Y^2.

You want to compute (3+(2a+b))^2, so you can say (and they didn't put the inner parenthesis there per chance):
X = 3
Y = 2a+b

The (3+(2a+b))^2 = (X+Y)^2, which you immediatly turn into X^2 + 2XY + X^2.
You can now substitute X and Y with the values you decided for them, giving:
(3+(2a+b))^2 = 3^2 + 2.3.(2a+b) + (2a+b)^2

Of course you're not going to stop it, so you can simplify and distribute once:
(3+(2a+b))^2 = 9 + 12a + 6b + (2a+b)^2

and the last term is yet another case of applying the property with X=2a and Y=b, giving
(3+(2a+b))^2 = 9 + 12a + 6b + (2a)^2 + 2.2a.b + b^2
(3+(2a+b))^2 = 9 + 12a + 6b + 4a^2 + 4ab + b^2

Which, oh wonder, is the same than the previous result.

OG.