How to make Brackets work

Let’s start with something simple, like

many (this + that) is many this + many that

similarly, if we have lots instead of many, then

lots (this + that) = lots this + lots that

so if we put both lines together using brackets, so many and lots becomes (many + lots), then

(many + lots) times (this + that)

= many times this + many times that + lots times that + lots times that

and if I use x for ‘times”, then if you see the first bracket as having two terms inside it and we expand that one first into ‘many’ of (this&that) followed by ‘lots’ of (this&that), we get

(many + lots) x (this + that) =many x (this + that) + lots x (this + that)

= many x this + many x that + lots x this + lots x that

or, using algebra instead of words and I pick the letters a,b,c,d, then

(a + b) x (c + d) = a x c + a x d + b x c + b x d = ac + ad + bc + bd

and if we have some x and y, while it looks more complicated, it’s just the same really: I’ve done this as slow as I can so you can see how the parts can be manipulated. I might expect every second line in a current homework and just the last one with practice.

Can we agree that (this&that) is exactly the same as (that&this)? so you don’t mind swapping them around occasionally? I remember that some students find this a problem — these are the same people who see three fours as somehow different from four threes. I recognise this as also someone who genuionely thinks that six of one and half a dozen of the other are actually different. I blame your primary education when this happens and I hope we can fix it. In algebra, and mix of letters without signs separating them are therefore multiplied together, so xy and yx are *the same thing*. This is not dyslexia; this is not a mistake; this is correct. In the same way, abc, bca, cab, acb, cba, bac are each identical to the others. The same applies to addition as applies to addition [a+b is the same as b+a; a+b+c is the same as b+c+a]. But subtraction and division are different and for those oparations the order *does* matter. This should have been clear to you back in Y5.

(ax+ by) x (cx + dy) = ax x cx + ax x dy + by x cx + by x dy

=axcx + axdy + bycx + bydy =acx² + ad xy + bc xy + bdy²

=acx² + (ad + bc) xy + bdy²

There is a pretty (effective, stupid, silly, all of these) way of writing this as a ‘face’, two eyebrows, a nose and a mouth, which I managed to reproduce on the paper version of this page. It should be in your exercise book by now.

Notice that every term in (ax+ by) x (cx + dy) = acx² + (ad + bc) xy + bdy² has four letters in it. If x and y had dimension like length, then ax+by has length too, and length (of some x plus some y) times length (of some other x plus some y) makes area. Similarly, each of x² , y² and xy must be area. So the whole is consistent — meaning that all the terms are area and it, the algebra, therefore makes sense. Proper vocabulary (and convention) says that x & y are * variables* (stand for something including a something with dimension), while each or a,b,c and d are

*(just number).*

__coefficients__This page is written for those who feel a need for a different way of looking at this topic. The practice with numbers is on other sheets...

You might notice I sometimes use a pair of brackets. Remember that a term is a thing separated by + or - signs, so while a term might include a bracketed part, it will include terms inside the brackets.

Can you see a need for some better term-inology?