www.mathcentre.ac.uk/.../Expanding_Removing%20Brackets.mp4

As well known investigation at
GCSE level called frogs.

Many of you might well have
attempted that particular

investigation and want to start
this particular section by

having a look at that
investigation, because it's

going to show us why we might
want to use brackets.

Now, how does the investigation
work? Well, what we have to do

is try and interchange these
coins at this side the pound

coins. With the 10 P coins at
this site and we have to do it

by using one of two kinds of
move, we can either slide into

an empty space, or we can hop or
jump over a coin of the opposite

kind. So here a pound coin
jumped over the 10 P coin. We

can never go back. We must keep

going. Forwards and so these 3
coins have with those rules to

interchange with these 3 coins
over here. So let's just see

what happens. Let's see if we
can keep count of the number of

moves, not only keep count of
the number of moves, but the

number of different kinds of
moves hops remember all slides.

That was a slide.

That's a hop, and the object is
to do it in the minimum number

of moves. Now when I do it, it
will be the minimum number of

moves. Don't worry about that,
but we need to keep count of how

many hops, how many slides and
how many moves altogether so.

Let's begin and will just keep a
tally here of hops and slides.

So we begin one slide.

Over there one hop.

One slide. And now with
that we want one.

Two hops

There one slide.

Now one.
2 three hops.

A slide gets us that one home.

And then two hops a slide
gets us that one home.

And then Hopkins that one home
and then a slide gets that one

home. So let's just talk
these up 123456 slides.

369 hops so altogether
a grand total of

15 moves. OK, when you were
doing this investigation at

school, or possibly you may not
even have done it. What you

would have done would be draw up
a table so we want the number of

coins. On

each side.
Hops?

Slides.

Moves Got our table
there. So number of coins on

each side. We could have one 2,
three which is the case. We did

four or five.

Hops? Well, let's fill in this
one for this table here. We had

nine hops, six slides, and that
gave us a total of 15 moves. Now

we could do it for one.

On each side, just do it

quickly. One slide.

One hop, one slide. So
altogether two slides and one

hop. And we can do it again.

Quickly for these.

A slide.

A hop.

A slide, so that's two slides
all together in one hop, and now

we have another two hops.

A slide. A hop.

And the slide. So altogether we

had there. Four hops and for
slides. I'm not going to go

ahead and do it for 4 coins.
It's quite easy to do and the

results that we get will be 16
and 8 and 25 and.

10 Giving us a total
number of moves of

three, 824 and 35.

Now the object with most of
these investigations is to try

and arrive at a prediction. What
would happen if we had any

number of coins? Can we say what
the result would be if we had 10

coins on each side? If we had 50
coins on each side, 'cause this

is the power that maths gives us
the power to model it?

In symbols and be able to use
those symbols to make our

predictions, so that's what
we're looking at. Can we make

that prediction? Well, let's
have a little look. Want to

concentrate here on this column
and here on this column.

I actually need a few more
columns because I want to look

at this set of numbers and I
want to look at how this set of

numbers is made up or created.
So what I'm going to do is now

we've got the data that we
need. I'm going to turn over

the sheet and I'm going to
rewrite this table on a

separate sheet and then we can
analyze it without having all

these bits and pieces around
the edge. So we pick up the

coins.

Put them to one side.

Turn over the sheet an begin
again. So here we've got the

coins and the number of coins
on each side. 12345 and we

have the number of hops that
was one 4 nine 1625. Then

we have the number of slides.

And that was 2468

ten. And then the total
number of moves which we

got from the total of the
hops and the slides. So

that's three 815-2435.

Now what we said was we want it
to be able to make a prediction

no matter how many coins we've
got on each side. We want it to

be able to say how many moves it
was going to take us to

interchange those two sets of

coins. Let's have a look at this

set of numbers. And let's try
and relate it to this set of

numbers. Well, if we look at
three an 15.

15 is 3 times by
5.

Just a lucky guess. Well, let's
have a look at the next one.

4 and 24 will 24
is 4 times by 6.

An again 5 and 35 that's five
times by 7, so eight is 8

the same relationship with two?
Yes, it is. That's two times by

4. And even for this top one
we've got one times by three.

So let's look at these numbers
here. These two columns. Here we

have this number. The number of

coins. The number of coins and
here we seem to have two more

than the number of coins, so if
this was an what we seem to be

saying is that here we've got N.

Multiplied by, I'm not going to
write anything in there at the

moment, and plus two now it's
the whole of this that we want

to multiply by N everything, not
just the N, not just the two,

but all of it, because when we
do one times by three, we

multiply all of the three by
one. When we do the two times by

4, we multiply all of the four
by the two, and so on. So this

has to be.

Help together in a bracket so
that we can show we are

multiplying everything inside
the bracket by what is outside

the bracket. So let's just leave
that there for the moment.

Let's have a look now at these
sets of numbers.

And what we do know is that
if we add these sets of

numbers together, we do get
the total number of moves.

Can we see any kind of

relationship between? These
numbers the number of coins and

these two was a fairly obvious
one. This is 1 squared. This is

2 squared, 3 squared, 4 squared
and five squared. In other

words, it's this number
multiplied by itself. So if we

end we can see we've got end

squared. If we look down here.

Each one of these relates in the
same way four is 2 * 2 six is

2 times by three, 8 is 2 times
by 410, two times by 5, so again

2 times by 1, two times by two 2
times by 3, two times by 4, two

times by 5. This is our N again,
so we must have two times by N

or two N.

Now the number of moves can't
change simply because we've

written it down differently.

We know the total number of
moves is the sum of these two

terms. That must be the same
as that the number of moves

simply can't change just because
it written down differently.

Doesn't mean that the value
changes, and So what we must

have is that the total number of

moves. Is equal to both of those
expressions, and so they must

both be equal. So that shows us
two things. First of all, it

shows us why we need brackets to
keep things together as entities

as a whole.

It also shows us how we can
remove brackets because we

multiply that with that to give
us the end squared and we can

multiply that with that to give
us the two N.

So bearing these things in mind,
what we're going to have a look

at is multiplying out brackets
and removing brackets from

expressions. Now we've seen why
we need it now. We need to be

able to work both ways across
this equal sign. We need to be

able to work from the brackets
to the expression and from the

expression to the brackets. Now
we're only going to go one way

in this particular clip. That's
from the brackets to the

multiplied out. Or expanded
xpression coming back this way

is factorizing and we look
at that in another section.

So let's take some examples
3 times by X +2.

Everything inside has to be
multiplied by what is outside,

so we need 3 times by X
and we need 3 times by two

which is 6. The plus sign,
because that's plus there.

X times X minus Y everything
inside must be multiplied by

what is outside, so we've X
times by XX squared X times by

Y is X why we are taking
the wire way so we've got X

times by minus Y is going to
give us minus XY.

Minus 3A squared.

And then in the bracket, 3 minus
B. Now we're multiplying

everything inside by what's
outside. What's outsiders got

this minus sign attached to it?
So we gotta take care. So we

have minus 3A squared times by 3
- 9 A squared.

Minus 3A squared times minus B,
so the minus minus together

gives us a plus and we have
3A squared B.

We can do this across a number
of terms, so if we minus two X

let's say X minus Y minus Z.

So we've minus two X times by X.
That's that times by that minus

two X squared we've.

Minus two X times Y minus
Y. That gives us +2 XY

and then minus two X times
Y minus said plus 2X Zedd.

OK, those are reasonably
straightforward and you can find

plenty of those in textbooks and
other sources for you to

practice. Let's now have a look
at what happens if we want to

multiply out expressions where
there are two brackets

multiplying each other.

So let's take X
+5 times by X

plus 10. What now?

Well, what I want just to go
back to is a simple one like

let's say 3 times by X plus 10.

If you remember what
we did, we said, well,

we'll take the three.

And multiply it with the X and
will take the three an. We will

multiply it by the 10.

So that this was the entity we
were multiplied by. Let's think

of this X +5 as being the three,
so for this one we would do 3

times by X. So for this one will
do X +5 times by X.

So the X +5 is
behaving exactly like the three.

Now here we would do 3 times by
10 and rather than right it is

30. I will write it as three
times by 10.

So this is taking the place of
the three, so it's going to be

X +5 times by 10.

Here we would have a
plus sign, so here we

will have a plus sign.

Now, the fact that I've got the
X and the 10 at the end doesn't

make any difference. I'm still
multiplying them together so the

same rules apply. I still have
to multiply everything inside

this bracket by what is outside
this bracket. So now I need that

with that, which gives me an X
squared and I need that with

that, which gives me 5X. I need
that with that, which gives me

10X and I need that with that,
which gives me.

50 and so we tidy up the middle

bit. 5X plus 10X is

15 X. And there we have.

Our expression.

OK.

This. Here. Is an
extra step we can manage without

it provided we keep track of
what we've done.

Notice I've got an arrow from
here going to each of these.

Alternatively, if you've got an
arrow going from that extra that

X and an arrow going from that

extra that 10. So every term is
got sort of two links associated

with it. So let's have a look at
doing that without going into

this step. Now we've seen how we
can do it can, we shorten the

process a little bit.

So we've X
+5 times by

X plus 10.

So we've got to multiply
everything in this bracket by

everything in this bracket. So
remember we had to have X by X,

so that's their X squared. Then
we have to have X times by 10,

so that's there 10X. Now we need
to make the same links here, so

that's X by 5.

Giving us 5X and then 5 by 10,
giving us 50 and so we end up

with exactly the same expression
as we had before.

Gain, we've got two hours
going into this bracket. Two

arrows going into this bracket
and then coming back. We've

got the same from 10. There
are two arrows linking back.

We can use this process for
brackets like this that leaders

to quadratic expressions an for
multiplying any pair of brackets

together, and in fact the arrows
help us to see that we've

completed the multiplication
'cause they help us to see that

if we've got two terms here,
then there are two arrows

linking this first term, one
with each term there, and so on.

So let's have another look at
some examples of this working

our way. Down some quite
complicated ones, but beginning

with some simple ones, very much
like we've just had a look at.

So first of all, we got X
takeaway 7 times by X takeaway

10 or X minus Seven times by X

minus 10. Let's expand this as
we've done before, so X times my

X that gives us X squared.

And X times by now this is minus
10, so it gives us minus 10X.

Minus 7 times by X minus 7X
and minus Seven times by minus

10. Plus 70 and
we can simplify that X

squared minus 17X plus 70.

Take X plus six times
by X minus 6.

Everything in this bracket must
be multiplied by everything in

this bracket, so we have X times
by X. That's how X squared.

X. Times by minus six
gives us minus 6X.

Plus six times by X their loss

6X. And then six times by minus
6 - 36 and now again we need

to simplify this. Looking at
these ex is so with X squared.

Now we have to take away 6X and
add on 6X. The net effect is

nothing so there are no excess
minus 36, so we don't always get

this middle term sometimes and
this is an example notice it

structure. Certain symmetry
about it, a 6 and a six hour

plus under minus.

And the ex is will disappear.

Take another one, 2X minus 3X
plus one. This time it's 2X not

just X, but again, everything in
here must be multiplied by

everything in there. So we have
two X times by X. That gives

us two X squared.

2X times by one that
gives us 2X.

Minus three times by X that's

minus 3X. And minus three times
by one is minus 3.

Here we've got some excess. We
must simplify them. We've got to

gather them together so we have
two X squared.

Plus 2X minus three X. We're
adding on 2X taking away 3X.

We still have their foreign X
to take away, minus three.

Now let's have a look at
just two or three more of

these. If we take 3X
minus two and three X +2.

Have we seen something like this
before? A minus sign in a plus

sign a two and a 2A3X and 3X.
Not just an expert of 3X.

Are we going to get the same

sorts of things? Symmetry
suggests that we might, so let's

have a look.

3X times by three X. That will
give us nine X squared 3 * 3

X times by X.

3X times by +2 is
plus 6X minus two times

by three. X is minus

6 X. And minus two
times by +2 is minus four. Gotta

tidy up these middle terms.

Nynex squared. Plus 6X minus 6X
no access, and we left with

minus four on the end. So yes,
it was another example of that

kind of expression where the
middle term is going to

disappear. Let's Make it
a little bit harder and

let's have a look at
something with three terms

in this bracket and only
two in this, so we make

that One X +2, but let's
put in X squared.

Minus two X cubed
+8.

Got three terms here. We need to
keep very careful track and make

sure that we multiply everything
in here by everything in there.

So let's begin.

X squared times by X?
That's X cubed X squared

times by two. That's two

X squared. Minus 2X now
minus two X cubed. Now times by

X is minus 2 X till the

4th. And minus two X cubed
times by two is minus 4X cubed,

and now we've got to do the 8.

So we've got 8 times by X
Plus 8X and we've got 8.

Times by two. So that's plus 60.

What we can see is that from
each of these terms we've got

two arrows coming.

And that's what we must have.
'cause we got two terms here so

we can be pretty happy that in
fact, we've done what we set out

to do. And now we need to
simplify. And it's usually

better to write down expressions
with the highest power in X

first. This here is the highest
power in X. It's X to the power

4. So we write that one down
first, minus 2X to the power 4.

What else we got? We got any ex
cubes. Well yes we have. We've A

plus X cubed an A minus 4X
cubed. So that's going to give

us minus three X cubed.

Sometimes helps just to put a
little tick on top of them, just

to show that, yeah, we've
actually dealt with those X

squared's only one that's there
plus two X squared. So we've

dealt with that one plus 8X.
We've dealt with that 1 + 16

we've dealt with that one, so
there our answer.

Let's see if we can just take
this a little bit further by

multiplying together 2
quadratic expressions.

Normally come up with something
as difficult as this, or perhaps

as tedious, but it's a good
exercise to see if you've

actually got the hang of what
you're supposed to be doing. So

we've got to multiply everything
in this bracket by everything in

that bracket. There are three
terms here, so if we've done it

right, we ought to have three
arrows coming from each of

these, associating with each of
these. If you think about that

ought to give us 9 terms.

In the same way that this one.

Three terms here, 2 terms
that gave us six terms

altogether. 123456 before we
simplified it, so we're going

to have nine terms here.

Let's work our way through it.

X squared times by X squared X
to the 4th.

X squared times by X
Plus X cubed.

X squared times by minus
6 - 6 X squared.

Finished with that one on to
this One X times by X squared

plus X cubed.

X times by X Plus X squared X
times by minus 6 - 6 X. Now this

one here minus two times by X
squared getting a bit fraught

with the brackets here with the
arrows linking them, but

nevertheless we can still use it
to help us check. So minus two

times by X squared minus two X

squared. Minus 2 times by X
so that gives us minus 2X and

then finally minus two times by
minus six gives us plus 12.

This now needs tidying up, so we
start with the highest powers

1st and write down the highest
power, then the next, then the

next and so on down the line we
take them off as we did here so

we can show that we've coped
with them. We've taken them into

account, so our highest term is
this one. Here X to the power

four, and there are no others.
Now we want the ex cubes. Well,

there's one there, and there's
one there. So that's plus 2X

cubed, so we can.

Take off those two X squared's.

Minus six X squared.

Plus, X squared minus two X
squared, so we've got minus six

X squared minus two X squared.
That's minus eight X squared,

and then we're adding on One X
squared, so altogether that's

minus Seven X squared, and we
can take them off. We've

accounted for them.

Minus six X minus two X the X
terms next altogether that's

minus 8X. And then at the end
we just got this number 12 that

we have not accounted for, and
so that's plus 12.

So that's multiplying out

brackets. Every pair of brackets
that needs to be multiplied out,

or indeed if it's three sets of
brackets, you do 2 together, and

then the third would. The
resulting expression is done

with the third one one. I want
to have a look at now is just

simply removing brackets.
Sometimes we get collections of

expressions nested in various
sets of brackets.

So let's have a look at the
sorts of horrible, awful

expressions that people write
down in textbooks and say have a

go at these and remove the

brackets. So here we are, whole
set of letters jumbled together

with brackets in and what we're
being asked to do is just remove

these brackets and simplify. So
the first thing is remove the

brackets, get rid of them. So
let's have a look. OK, now we're

taking away B minus C, so we've
got to take away everything

that's in the brackets, and
we've got to take away the B and

then takeaway minus see which of
course is the same as adding see

on. Plus a no problem there plus
B because we've got to add on

everything that's in this
bracket. So plus be plus minus C

which is just minus C.

Plus B and then again. We've
gotta take away everything

that's in this bracket. So we
gotta take away, see an then

takeaway? Minus A and that's the
same as adding a on.

Now we have to simplify, so we
have to look for all the days,

all the bees and all the Seas.
So let's take the ace Pursuivant

A&A&A and they're all, plus. So
that's three a we can just take

them off to show that we've
dealt with them.

Bees minus B Plus B plus

B. Minus B Plus B know bees
and then another be to add

on plus be. So we've dealt
with all the bees.

Plus C minus C minus C Plus
C minus C, no seas and the

sea still to take away minus C.

So that was fairly
straightforward. No nesting of

brackets, so let's have a look
at one where we've actually got

some brackets nested so minus.

5X inside that bracket
minus 11 Y.

Minus 3X.
Minus.

5 Y minus three
X minus six Y.

Now that looks pretty
awful. What's a brackets?

And if we look, we've got a pair
of brackets here covering

everything. We've got an
expression in a pair of

brackets. There got another pair
of brackets here and inside.

We've got another pair of
brackets. Where do we start?

Looks a bit tricky, but what you
do is you start right inside the

brackets so you look inside each
set of brackets and you say what

do I need to do to get rid of
one set of brackets and usually

go right inside the most complex

bit? And this bit here is by far
the most complex bit. So let's

begin by going right inside and
looking at this.

Really nested embedded set of
brackets and get rid of them to

begin with. So if we're going
to do that, we write the rest

down just as it was.

We won't touch it at
all.

We're just going to look at
this bit. We're taking away

everything there, so we're
taking away 3X and we're

taking away minus six Y, so
altogether. That's takeaway

3X, and then the minus minus
is plus six Y. And then that

Square bracket and that one.

It's beginning to look a little
bit simpler. We've now got this

big set of square brackets
outside, but we've got that one

there and that one there, and we
can deal with both of these

together. Be'cause, they're no
different from what we've got up

there, really. So equals minus
5X minus 11 Y and then minus

minus is A plus 3X minus five Y

minus. Minus 3X is A plus, 3X
minus plus six Y, so we're going

to have minus six Y and restore
the Big Square bracket.

I think we want to tidy this up.
We've got lots of axes and wise

floating about so really we've
got to tighten this up, so let's

have a look. And we'll do it
simply inside the bracket.

So we don't need to think at the
moment about the effect of this

minus sign with just looking at
that bit there. That's inside

the bracket. So we 5X3X
and 3X, that's 11X.

Minus 11 Y minus five Y
minus six Y all together we

are taking away 22Y and close
the big bracket and now we've

got to remove this final bracket
so we have minus 11X plus

minus takeaway minus 22 Y.

So what? Looked quite an awful
lump of algebra has now come

down to be something that's
really quite small. An quite

manageable will just take one
more 'cause it's helpful to get

this focus on getting right
inside those brackets and what

you will notice, I think, is
that we tend to use different

sets of brackets, different
types of brackets.

To draw our attention tool, the
different lumps of algebra that

we've got to deal with.

So here in this expression we've
got three sets of brackets, the

curly brackets, the square
brackets, and the round

brackets, and notice I've worked
my way in, so I'm now right

inside, right in the middle of
this expression. So this is the

one I'm going to go for first.
Everything else for the moment

is going to stay the same, not
going to touch anything else,

just this round bracket here. So
I've got to multiply.

By two so 2 * 10 A is
20 A and two times by minus B

is minus 2B. Put my square
bracket in. Put my curly bracket

here. What now? I think
it would be a good idea to

try and tidy this up before we
went any further, so we've got

5A Minus Square bracket. I've
got 6A plus 28, that's 26 a

takeaway to be.

Close up the brackets at the

end. Equals 3B, Now I've gotta
go for the square bracket. Now

these are inside the curly
brackets so it's the square

brackets we tackle. I've made
that look like a six. Let's make

it more clearly. AB minus.

5A. Take away, take
away everything inside this

bracket. So takeaway 26 a
takeaway minus 2B so it's the

same as adding on to be and
keep the curly brackets. Now

we can simplify this.

3B
minus.

5A takeaway 26. So that's take
away altogether or minus 21 A

plus 2B. Close the bracket.

And now come to the final step.
Almost the final step. Anyway,

we can get rid of these curly
brackets 3B. Take away

everything in the curly
brackets. Minus minus gives us

plus 21 A minus +2 BS. That's
taking away to be and then. Now

we just need to simplify this
and what we've got is to be

taken 3 be sorry 3B takeaway to
beat is B.

Plus 21 a so we were able
to reduce that.

Huge lump of algebra with three
sets of brackets to a very

simple expression. And again,
notice where did we start right

in the middle with the most
nested set of brackets that we

could find right in the middle.
And we simplified that. And then

we gradually worked our way

outwards. Dealing with each
set of brackets intern and

simplifying as we went along
so as to make the job easier

for ourselves and be more
sure that we've got the right

answer at the end.