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 coefficients (just number).
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?
