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.