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.