www.mathcentre.ac.uk/.../Factorising_Quadratic_Equations.mp4

When we have two

brackets. X +2 times by X
+3 and we know how to multiply

these two brackets out.

We have X kinds by X that
gives us X squared. We have X

times by three, gives us 3X.

2 times by X gives us 2X
and then two times by three

gives us 6.

And we can simplify these two
terms. 3X plus 2X gives us 5X.

This is an example of a
quadratic expression or

quadratic function. It's gotta
termine ex squared, which it

must have to be a quadratic

expression. It's got a term in
X which it might or might not

have, and it's got a constant
term and there are no other

possibilities, so our most
general quadratic expression

would be AX squared plus BX plus

C. What we're going to have a
look at is how we factorise

expressions like this in others.
How we go back from this kind of

expression. To this now, why
might we want to do that? Well,

let's just take this X squared
plus 5X plus six. And let's say

it's not just an expression, but
it's an equation and it says

equals 0. What are the values of

X? That will make it equal to
0 that are answers to that

equation. One of the things we
can do is to rewrite this form.

By this so we can say
X +2 times by X +3

equals 0. When we have
two numbers that multiply

together to give zero and one of
the things that must be true is

that one of them zero or the
other one zero, or they're both

0. So in this case, X +2 equals
0 or X +3 equals 0, and so

X would be minus two, or X would
be minus three.

So being able to factorize
actually helps us to solve a new

kind of equation.

So we're going to be having a
look in this video that how you

factorise this kind of function.
This kind of expression a

quadratic expression. Now I'm
going to start by going back to

this little piece of work again.

So let's write it down

X. Plus 3
* 5 X +2

and again. Will
multiply out the

brackets X times by X
is X squared.

X times by two is 2 X
3 times by X is 3X.

3 times by two is 6.
This simplifies to X squared

plus 5X or 6.

So we've gone one way. What
happens if we want to go back

the other way?

Let's have a look where this six
came from. We know it came from

3 times by two.

Where did this five come from?
Where it came from 2 + 3?

So if we were to reverse this
process, we be looking for two

numbers that multiply together
to give us six an 2 numbers that

added together to give us 5.

The obvious ones that go in
there are three 2.

So if we began.

With this

We would.

Be looking to
break that 5X down

as X squared plus
3X plus 2X plus 6.

Then we could look at these two
and see if there was a common

factor and there is X leaving us
with X Plus three. Then we would

look at these two. Is there a
common factor and there is 2.

Leaving us again with

X +3. And then we've
got this common factor of X plus

three in each of these two lumps
of algebra, so we can take out

that X +3, and we've got the
other factor left X times by X

plus three and two times by X
+3, and so we've arrived at that

factorization. Those brackets
that we started off with.

Now. That's what we've done and
what we need to do is to be able

to repeat this process of
looking for numbers that

multiply together to give the
constant term and numbers that

will add together to give the
exterm. So let's have a little

bit of practice at that.

Let's look at X squared
minus 7X plus 12.

So we want two numbers that will
multiply together to give us.

12 Times together to give
us 12 and will add together

to give us minus 7.

Minus four times minus three is
12 and minus four plus minus

three gives us minus Seven, so
let's just write those in minus

four times, minus three gives us
plus 12 and minus four plus

minus three gives us minus
Seven, so that's given us a way

of breaking down this minus
Seven XX squared minus 4X Minus

3X. Plus 12 so now we look at
these two at the front.

Take out X as a common factor
that gives us X minus four, and

now we look at these two.

Well, I want to make sure I get
the same factor X minus 4.

Clearly, in these two terms,
that is a factor of three. But

here I've got minus three, so I
think I'm going to have to make

the factor, not three, but minus
three. So that's minus three

times X, so that insures when I
multiply these two together

minus three times by ex. I do
get minus 3X, but now I need

minus three times by something
that's got to give me plus 12.

So that will have to be minus
four and close the bracket. Now

again I've got two lumps of
algebra, and in each one that is

the same factor. This common
factor of X minus 4X minus four,

so I'll take that as my common
factor X minus four. Then I've

got X minus four times by X&X
minus four times by minus three.

That's my factorization of

that. Let's take another

one. X

squared Minus 5X
minus 14. So now

looking for two
numbers to multiply

together to give minus
14 and add together to

give minus 5.

Fairly obvious factors of 14 R.
Seven and two. So can we play

with Seven and two? Well, if we
made it minus Seven and plus

two, then minus Seven times my
plus two would give us minus 14

and minus Seven, plus the two
would give us minus five, so

minus Seven and two look like
the two numbers that we need.

So let's breakdown this minus
5X as minus 7X Plus 2X.

And let's not forget the
minus 14 that we had again.

Let's look at these two. The
front two terms. Common

factor. Yes, it's X. Take
that out X minus 7.

And here a common factor of +2.

Let's take that out and X

minus 7. Two lumps of algebra
that one and that one and each

one's got the same. Factoring
this X minus Seven, so we'll

take that as our common factor.

So X minus Seven is
multiplying X and it's auto

multiplying +2. So again, there
we've arrived at a factorization

of this X squared minus 5X
minus 14 factorizes as X minus

Seven X +2.

Type X
squared minus

9X plus
20.

Now, from what we've got
already, it might be that some

of you watching this might
think, well, do I need to do

that every time? The answer is
no. Sometimes you may be able to

do these by inspection, which
means looking at it.

And doing the working out in
your head rather than on the

paper. So to do it by
inspection. What we might do is

right down the pair of brackets
to begin with. Recognize X

squared means we're going to
have to have an X and then X.

Recognize 20 as being four times
by 5 and of course 4 + 5 would

give me 9, but I want minus
nine, so perhaps what I need is

minus four and minus five,
because minus four times by

minus five is going to give me
plus 20 and minus 4X.

Minus 5X is going to give me
minus 9X, so I've done that one

by inspection. But I could have
done it in exactly the same way

as I did the other two. Let's
take X squared minus nine X

minus 22. And again, let's try
this one out by inspection. So

pair of brackets.

X&X in front of each bracket.

Let's have a look at minus 22
two numbers to multiply together

to give minus 22 will likely
candidates are minus 11 and two.

Or minus 2 and 11, but at the
end of the day I need minus 9X

and that kind of suggests that
perhaps we've got to have the

bigger of 11 and two as being
negative and the smaller one as

being positive. Let's just check
minus 11 times +2 gives me minus

22, and then I have minus 11X
and 2X, which gives me minus 9X.

So again, we've done that one by

inspection. Again, you don't
have to do it by inspection. You

can use the previous method.

If we have quadratic expressions
which don't have a unit

coefficient, now this is one
that has a unit coefficient,

'cause this is One X squared.

It could be 2 X squared. It
could be 6 X squared, could be

11 X squared, could be anything
times by X squared. That would

be harder to do.

So let's have a look at how we
might tackle some of those. So

we take three X squared.

Plus 5X minus
2.

I'm going to use a method that's
very similar to the first method

that we saw. I'm going to look
for two numbers that multiply

together to give, well, let's
leave that unsaid for them in

it, but these two numbers are
going to add together to give I.

They're going to add together to
give this +5 the Exterm, so that

hasn't changed. We're looking
for two numbers that will add

together to give us the
coefficient of X.

What do the two numbers have to
multiply together to give us?

Well, they have to multiply
together to give us 3 times by

minus two, so we don't just take
the constant term, we multiply

it by the coefficient of the X
squared and three times by minus

two is minus 6, and I'll just
write that here at the side that

the minus six came from the
three times by the minus two.

And if you think about it,
that's actually consistent with

what we were doing before.

Because in the previous
examples, this number in front

of the X squared had been one.

And so one times by minus two

would be. The constant term, so
we are looking now for two

numbers that multiply together
to give us minus 6 and add

together to give us 5.

Well. 3 times by two.

Well, three times by two would
give us plus 6.

Minus three times by minus two
would also give us plus six, so

that's not good. Six and one.

Well, if we could have 6
times by minus one, that

would give us minus six and
six AD minus one would give

us 5. So this looks like the
combination that we want.

So we take three X squared.

Plus 6X minus X

minus 2. Let's have a
look for a common factor here.

Well, there's a three X squared
and a 6X, so there's obviously a

tree is a factor, and also an X.
So let's take out three X leaves

me X +2.

3X times my X gives us the three
X squared 3X times by two gives

us 6X and now want to common
factor for these two terms minus

X minus two. We don't seem to
share anything in common.

I've got a common factor and
it's minus 1 - 1 times minus one

times by something has to give
me minus X, so that must be X

and minus one times by something
has to give me minus two. Well,

that's got to be +2.

So now I've got these two lumps
of algebra again, this one and

this one, and each lump has the
same factor in it. This common

factor of X +2, so I'll take
that one out X +2 and I've got

X +2. Multiplying 3X and X +2,
multiplying minus one.

And so there's the factorization
of the expression that we began

with. Let's take another
one, two X squared.

Plus 5X
minus 7.

So we're looking for two numbers
that will multiply together to

give us 2 times by minus Seven,
so they must multiply together

to give us minus 14. Just write
down again at the side that the

minus 14 comes from minus Seven
times by two.

And then these two numbers,
whatever they are, I've got to

add together to give us the
coefficient of X +5.

So what are these two numbers?
Well, Seven and two seem

reasonable factors of 14, and
they are factors of 14 which

have a difference. If you like a
five, so they seem good options

7 and 2. Seven and two.

But we've got to make a balance
here. We need +5 and we need

minus 4T. So one of these is got
to be negative, and it looks

like it's going to have to be
negative two in order that 7

plus negative two should give us

the five there. So now we
can write this down as two

X squared.

Breaking up that plus 5X as
plus 7X minus 2X and then

minus Seven at the end.

What have we got here as a
common factor? Well, we've got

an X in each term, so we can
take that out, giving us 2X plus

Seven. And here again, what have
we got for a common factor? Or

the only thing that's in common
is one and there's a minus sign

with each one, so it's minus 1 *
2 X plus 7 - 1 times by two. X

gives us minus two X minus one.

Plus Seven gives us minus Seven
close the bracket.

Two lumps of algebra. Again,
this one, and this one. In each

one. There's this common factor
of 2X plus Seven, so we take

that out 2X plus 7 and that's
multiplying X and it's

multiplying minus one, and so we
have got this factorization of

the expression that we began

with. Take

another example.
Six X squared.

Minus 5X minus four. Now what's
different here is that this is

not a prime number. We've had a
two and we've had a 3, but this

is a 6.

You might have been able to do
the other two by inspection, but

this one is more difficult to do
by inspection, and really, we

perhaps are going to have to
depend upon the method we just

learned, so we're looking again
for two numbers that will

multiply together to give us 6
times by minus four. In other

words, minus 24.

OK, I'll just write that down so
we can see where it's come from.

Minus 24 is 6 times by minus
four and we want these two

numbers. Whatever they are.
They've also got to add together

to give us the coefficient of X,
so they must add together to

give us minus 5.

So 2 numbers that might multiply
to give us 24, eight, and three

good options, and eight and
three do have a difference of

five, so they look options we
can use. Now let's juggle the

signs we need to have minus
five, so that would suggest that

the 8's got to be the negative
one. So let's have minus 8 times

by three. That will give us
minus 24 and minus 8.

Plus three that will give us
minus five, so we've got six X

squared. Minus 8X plus
3X breaking down that

5X. Minus 4.

Common factor here.

Well, the six and the three
share a common factor of three.

And of course we've X squared
and X common factor of X, so we

can take out three X.

And that will leave us 2X
plus one.

These two terms, what do we got
for a common factor? Well, they

clearly share a common factor of
four and also a minus sign. So

we take minus four times by.

Now, minus four times by
something has to give us minus

8X, so that's going to be 2X and
then minus four times by

something has to give us minus
four, so that's got to be plus

one again to lump sum algebra
sharing. This common factor of

2X plus one. So we'll take that

out. And then we have two
X plus one multiplying 3X.

And two X plus one multiplying

minus 4. Will take
1 final example of

this kind. So
I've got 15 X

squared. Minus three X

minus 80. Now in all the
others, the thing that we

haven't checked at the
beginning, and perhaps we should

have done is do the
coefficients. The numbers that

multiply the X squared.

That multiply the X
and the constant term

share a common factor.

And in this case they do as a
common factor of 3.

And where there is a common
factor, we need to take it

out to begin with, so will
take the three out.

3 times by something as to give
us 15 X squared so three times

by 5 X squared will do that.

3 times by something has to give
us minus 3X so three times by

minus X will do that.

3 times by something has to give
us minus 18 and so minus six

will do that.

Now we're looking at Factorizing
this xpression Here in the

bracket and we're looking for
two numbers that were multiplied

together to give us five times
by minus six, which is minus 30.

And again, I'll just write down
where that came from. Minus 30

was five times by minus six, and
I'm looking for two numbers that

will add together to give Maine.
Now here the number that's

multiplying the X is.

Minus one. So I want two
numbers to multiply together to

give me minus 30 and add
together to give me minus one,

well, five and six seem like
obvious choices 'cause they've

got a difference of one and they
multiply together to give 30. So

how can I juggle the signs with
the five and the six?

Well, 5 + 6 has to give me minus
one. It looks like the six is

going to have to carry the minus
sign, so 5 plus minus six does

give me minus one and five times
by minus six does give me minus

30, so I'm going to have three.

Brackets five X squared plus 5X
minus six X, so we broken down

this minus X into 5X, minus 6X
and then the final term on the

end minus six and close the

bracket. Keep the three outside.
Let's look at the front two

terms here. There's a common
factor of 5X. Let's take that

out. Five X.

X plus One 5X times by X gives
me the five X squared 5X times

by one gives me the 5X.

Common factor here is minus six.
Each of these terms shares A6

and the minus sign, so will take
out the factor minus six, and

then we need minus six times by
has to give us minus six X, so

that's times by X minus six
times by something has to give

us minus six, so that's minus
six times by one, and then I

need to make sure I close the
whole bracket with that big one

there. Two lumps of algebra,
each sharing this common factor

of X plus one. Let's take that

out three. Bracket X plus
one times by the X

Plus One Times 5X.

The X plus one also times
minus six, and so we've

completed that factorization.

OK, we've looked at a series of

examples. An we've developed.

A way of handling these that
enables us to factorize these

quadratics. Hasn't really been
anything special about them, but

I want to do now is have a look
at three particular special

cases. Let's begin with the
first special case. By having a

look at X squared minus 9.

Now it's obviously different
about this one.

From the previous examples is
that there's no external, just

says X squared minus 9.

So let's do it in the way that
we would do we look for two

numbers that would multiply
together to give us minus 9 and

add together to give us the X
coefficient, but there is no X

coefficient. That means the
coefficient has to be 0.

0 times by XOX is.

So I've got to find 2 numbers
that multiply together to give

minus 9 and add together to give

0. Obviously they've got to be
the same size but different

sign, so minus three and
three fit the bill perfectly,

so we have X squared
minus 3X plus 3X minus

9. Look at the front two terms.
There's a common factor of X.

Leaving me with X minus three.
The back two terms as a common

factor of 3 leaving X minus
three and now two lumps of

algebra each sharing this common
factor of X minus three X minus

three, multiplies the X and
multiplies the three.

Well, we compare this

with this. Not only is
this X squared that we get X

times by X, but this 9.

Forgetting the minus sign for a
moment is 3 squared.

3 times by three.

So in fact this expression could
be rewritten as X squared minus

3 squared. In other words, it's
the difference of two squares.

So let's do this again.

But more generally, In other
words, instead of minus nine,

which is minus 3 squared, let's
write minus a squared. So we

have X squared minus a squared.

We want to factorize it, so
we're looking for two numbers

that multiply together to give
minus a squared and add together

to give 0. Because there are
no access, so minus A and

a fit the bill. So we're
going to have X squared minus

8X Plus 8X minus.

A squared. Common factor
of X here X minus

a. Under common factor
of a here, X minus a

again 2 lumps of algebra, each
one sharing this common factor

of X minus a.

X minus a multiplies
X, an multiplies a.

So we now have.

What is, in effect a standard
result we have what's called the

difference of two squares and

its factorization. So let's just
write that down again. X squared

minus a squared is always equal
to X minus a X plus

A. So that if we can identify
this number that appears, here

is a square number.

We can use this
factorization immediately.

So what if we had something say
like X squared minus 25?

Well, we recognize 25 as
being 5 squared, so

immediately we can write this
down as X minus five X +5.

What if we had something like
2 X squared minus 32?

Doesn't really look like that,
does it? But there is a common

factor of 2, so as we've said
before, take the common factor

out to begin with.

Leaving us with X
squared minus 16 and

of course 16 is 4
squared, so this is

2X minus four X +4.

Files and we had nine X
squared minus 16.

What about this one?

Again, look at this term here
9 X squared. It is a

complete square. It's 3X times
by three X.

So instead of just working with
an X, why can't we just work

with a 3X?

And of course, that's what we
are going to do. This must be 3X

and 3X and the 16 is 4 squared,
so 3X minus four, 3X plus four.

And again, this is still the
difference of two squares, so

that's one. The first special
case, and we really do have to

learn that one. And remember it
'cause it's a very, very

important factorization. Let's
have a look now.

Another factorization

special case. Having
just on the difference of two

squares looking at this
quadratic expression, we've got

a square front X squared and the
square at the end 5 squared.

But we've got 10X in the middle.

OK, we know how to handle it, so
let's not worry too much. We

want two numbers that multiply
together to give 25 and two

numbers that add together to
give us 10.

The obvious choice for that is 5

and five. Five times by 5 is 20
five 5 + 5 is 10.

So we break that middle term
down X squared plus 5X Plus

5X plus 25.

Look at the front. Two terms are
common factor of X, leaving us

with X +5.

Look at the back to terms are
common factor of +5 leaving us

with X +5.

Two lumps of algebra sharing a
common factor of X +5.

X +5
multiplies, X&X,

+5, multiplies

+5. These two
are the same, so this is

X +5 all squared.

In other words, what we've got
here X squared plus 10X plus 25

is a complete square X +5 all
squared. We call that a complete

square. Take
another

example, X
squared minus

8X plus 16.

We recognize this is a square
number X squared and this is a

square number. 16 is 4 squared.

And of course.

Minus 4 plus minus four would
give us 8.

As indeed minus four times, my
minus four would give us plus

16. So we immediately.

Recognizing this as this
complete square X minus

four all squared.

One more. 25
X squared minus

20 X +4.

25 X squared is a complete
square. It's a square number.

It's the result of multiplying
5X by itself.

4 is a square
number, it's 2

times by two.

Minus 20

X. Well, if I say
this is a complete square,

then I've got to have minus
two in their minus two times,

Y minus two would give me 4 -
2 times by the five. X would

give me minus 10X and I'm
going to have two of them

minus 20X. So again I
recognize this as a complete

square.

Might have found that a little
bit confusing and a little bit

quick. That doesn't matter
be'cause. If you don't recognize

it, you can still use the
previous method on it.

Let's just check that 25 X
squared minus 20X plus four. So

if we didn't recognize this as a
complete square, we would be

looking for two numbers that
would multiply together to give

us 100. The 100 is 4 times
by 25. Just write this at the

side four times by 25 and two
numbers that would.

Add together to give us minus
20. The obvious choices are 10

and 10, well minus 10 and minus
10 because we want minus 10

times by minus 10 to give plus
100 and minus 10 plus minus 10

to give us minus 20.

So we take this expression
25 X squared.

And we breakdown the minus 20X
minus 10X minus 10X.

And we have the last terms plus

4. We look at the front two
terms for a common factor and

clearly there is 5X.

Gives me 5X minus two. Now I
look for a common factor in the

last two terms there is a common
factor of 2 here, because 2 * 5

gives us 10 and 2 * 2 gives us
4. But there is this minus sign,

so perhaps we better take minus
two is our common factor.

Which will give us minus two
times by something has to give

us minus 10X. That's going to be
5X and minus two times by

something has to give us plus
four which is going to have to

be minus 2.

Two lumps of algebra. The common
factor is 5X minus 2.

5X minus two
is multiplying 5X.

And it's also multiplying minus

2. So we have got this again
as a complete square, but not

by inspection, but using our
standard method.

So that deals with the 2nd

special case. Let's now have a
look at the 3rd and final

special case, and this is when
we don't have a constant term

and we've got something like 3 X
squared minus 8X. What do we do

with that? Well, clearly there
is a common factor of X, so we

must take that out to begin
with. So we take out X.

We've got three X squared, so X
times by three X must give us

the three X squared.

And then X times by something to
give us the minus 8X. Well it's

got to be minus 8, and that's
all that we need to do. We've

broken it down into two brackets
X times by three X minus 8.

What if say we had 10

X squared? Plus

5X. Again, we look
for a common factor.

Obviously there's an ex as a
common factor again, but

there's more this time
because there's ten and five

which share a common factor
of 5, so we need to pull out

the whole of that as 5X.

Five times by two will give us
the 10, so that's 5X times by

two. X will give us the 10 X
squared plus 5X times by

something to give us 5X that
must just be 1.

So, so long as we remember
to inspect the quadratic

expression 1st and check
for common factors, this

particular one shouldn't
cause us any difficulties.