 Hello friends welcome to another session on factorization and in this session we are going to discuss factorization of quadratic polynomial by splitting the middle term. Now this is a very famous technique and very commonly used technique as well to factorize quadratic polynomials as well as few in few cases polynomials which can be reduced in the form of quadratic polynomials and it is also used while solving quadratic equations so hence it's very important to master this concept. Secondly it will not you will not be able to learn this process overnight for any mathematical concept learning you have to solve a lot of problems and get exposed to variety of such problems so that you can understand where and how this particular technique can be used either either for factorizing a polynomial or solving a quadratic equation. So let us first try to understand the concept and then we'll take series of examples which you must go through to understand how this particular technique works. So first of all let's try to understand what the topic says factorization so we are understanding this particular we know what factorization means we have to convert a given polynomial into its factors and product of its factors we have to express it like expresses express it like and now you have quadratic polynomial so what is quadratic polynomial so any polynomial of the form so I'm saying p of x is a polynomial of the form ax squared plus bx plus c where a is not equal to 0 and a b and c are real numbers any such any any such polynomial will be called quadratic it is called quadratic quadratic comes from Latin word quadratus quadratus which meant square in Latin correct hence if you see the degree here is 2 so hence it is a square polynomial you know the degree is 2 so degree of px is 2 so hence we call it quadratic polynomial now how to factorize it so basically we have to express px into two factors so px is equal to ax square plus bx plus c I am saying after factorization let us say px will be of the form of px plus q and rx rx plus s these are the two factors right now any quadratic factor polynomial we have will have at max two linear factors why because if you have let's say any you know let's say three linear factors let's say tx plus u then what will happen if you see when you open this bracket all the three what will happen you will first multiply this term with this there will be a case when px will be multiplied to rx will be multiplied to tx right so you open up the first two bracket you will multiply px to rx and then when the second bracket these brackets are open then px would be multiplied to px times rx so there will be one term which would be like px times rx times px which will be nothing but p times r times t x cubed now in the LHS there is no cubic term no cube term so hence it cannot have more than two linear factors right so always remember if the degree is to the number maximum number of linear factors possible is n itself though if the degree is to then two linear factors if the degree is three three linear factors like that what is the linear factor so these are called linear factors right why linear factors because the power of x is here one similarly this is another linear factor where power of x is one okay so let us say that after factorization it becomes px plus q and rx plus s if you open these brackets what will you get you will get px times rx is so open this bracket so px times rx first term then px times s plus q times r times x plus qs so if you club them together it is pr x squared right and then plus you if you see this is nothing but ps plus qr times x plus qs isn't it now if you compare you know if you really want to if you if both the px are same then you can compare the two coefficients if you see here this coefficient a must be equal to pr so I can write a is equal to pr correct similarly b must be equal to ps plus qr correct so b is equal to ps plus qr and c is nothing but qs if you check the c is the constant term here qs isn't it now basically what you need to do is what you need to do is then how to find basically you have to find out p q r and s is it p q r and s if you get p q and r and s then you can you know the factorization know the factors so if you if you notice clear carefully if you multiply these two terms ac what is ac guys ac is pr qs correct now b the b has been divided into two parts pr so we have split b so imagine b to be split into two parts ps plus qr such that their product as well so if you multiply ps into qr you get ps qr which can be rewritten as pr qs and if you see this is same so if you see b has been split into two parts b1 and b2 whose product b1 b2 is nothing but a c this is the catch of splitting the technique so what you need to do in any such cases for again repeating it once more so ax square plus bx plus c equals 0 so what you need to do is multiply a and c okay so first find a and c so you got ac okay now split b now you'll say how to split it what is the rule so initially you have to use a lot of trial and error but there is a rule also to do it and I will take examples to understand it better but you have to keep this in mind that b has to be split into two parts b1 and b2 splitting means let's say that is b is equal to 10 so either you split like 5 plus 5 or 6 plus 4 or 7 plus 3 likewise whichever suits the case so you have to split it in such a way that that b1 times b2 must be equal to a times c right so keep this in mind you have to split b in such a way that the two parts when multiplied gives you a times c let's take an example and understand let us say we have a equation or polynomial px is equal to x square plus let's say 5x plus 4 okay now what is a here a is equal to 1 b is equal to 5 and c is equal to 4 multiply a and c right multiply a and c what will you get ac is equal to 4 now you have to split b what is b 5 split b into 2 b1 plus b2 such that b1 times b2 must be equal to ac which is equal to 4 so how can I split 5 5 can be split it as split as 5 sorry 0 cannot be used so 1 plus 4 2 plus 3 right 3 plus 2 and 4 plus 1 these are all the splits of 5 so if you see 1 times 4 gives you 4 right product of 1 into 4 so if you notice this is b1 and this is b2 now b1 and b2 is given to is found to be 4 b1 b2 is 4 which is matching with ac so we can stop here if you take 2 and 3 the product is 6 which is not equal to 4 not equal to ac so these are ruled out option again 4 and 1 and 1 and 4 are same so hence the split of 5 will be 5 should be written as 1 plus 4 let's do that so hence if you see x square then 5 is written as 1 plus 4 x plus 4 correct now you can expand it is nothing but x square plus x 1 times x is x plus 4 x plus 4 right now whatever techniques we have learned before group the terms and take comments okay so it is common from the first 2 x plus 1 I write x into x plus 1 and then plus 4 x plus 1 so hence it is x plus 1 times x plus 4 okay you see you got the two factors okay this is how the splitting the middle term works let's take another example another example is x square let's say minus 9x plus 14 okay this is px I have to factorize it what is a guys a is 1 b is clearly minus 9 and c is equal to 14 so what is a times c a times c is ac is equal to 14 now b is equal to minus 9 has to be broken down into b1 plus b2 such that b1 b2 is equal to 14 right now b1 b2 is 14 which is positive guys so hence b1 b2 either both are positive or both are negative then only the product will be positive isn't it now here b is minus 9 so when you split it you must have both negative so that the sum is negative isn't it so b1 plus b2 is minus b1 plus b2 is minus 9 what could be the possibilities possibilities could be 1 minus 1 minus 8 minus 2 minus 7 minus 3 minus 4 sorry minus 5 oh sorry minus 6 here then minus 4 and minus 5 and then it will repeat right you will see minus 5 minus 4 all are repeating so if you take these two product you'll get it if you take these two you'll get 14 you'll get 18 here you'll get 20 here right and what what do we need we need 14 see this is here 14 so hence if you break it down into minus 2 and minus 7 split it into minus 2 and minus 7 you'll get the desired result so let us do that so x square I can write minus 9x as minus 2x minus 7x plus 14 now again take commons x common x minus 2 minus 7 common x minus 2 again if you see you can even directly write this factor here and just try to match with the multiplication so that so if you write x minus 2 here what should be written over here so that the product gives you minus 7x so clearly minus 7 should be fitted in here right what I mean is let us say the moment you get the first factor x minus 2 so you write x minus 2 here directly now if x minus 2 is the factor then I have to get what minus 7 and 14 so what should I multiply here what should be written over here so that it gives you minus 7x and 14 so clearly it is minus 7 now you just take x minus 2 common and this is x minus 7 correct so this is what is the factorization now problem arises when you know it is you know of the form x square plus 3 root 3x now radicals are being deployed and this is px I'm saying px is equal to x square plus 3 root 3x plus 6 now again the same thing happens here so what is ac if you see a into c is 1 into 6 that is 6 now I have to split b is 3 root 3 so 3 root 3 has to be split into two parts such that the product is 6 right I'm saying b1 plus b2 is 3 root 3 and b1 b2 is 6 correct now b1 plus b2 is 3 root 3 now 3 root 3 is an irrational number guys so b1 and b2 yeah both have both have to be irrational right because because if one of them is rational and one of them is irrational then the sum will not be pure ration irrational number you understand so 2 plus root 3 will never be pure irrational it is rational plus irrational correct to get 3 root 3 we have to split into 2 irrational number and that to carrying root 3 in them for example what I'm trying what I mean to say is 3 root 3 can be expressed as let's take some examples 1 root 3 plus 2 root 3 this will be 3 root 3 is it and this is the only possibility of splitting right because the other possibility is 2 root 3 plus root 3 which is same as the previous one right now we have to just check whether the multiplication the product of this leads to 6 or not it indeed leads to 6 why 2 root 3 into root 3 is clearly 6 so hence we got our splits so it is x square plus root 3 x plus 2 root 3 x plus 6 now you can interchange these two the order will not create any difference now so hence now what to do x I know what is common between the first two terms x so x plus root 3 the moment you get that you just write x plus root 3 directly no issues now what you have to find something here okay now what is look at this what should I multiply this x here to get 2 root 3 x obviously 2 root 3 so you got the other factor as well 2 root 3 here now you just check 2 root 3 times root 3 must be equal to 6 which is actually true so hence I got the factors x plus root 3 and x plus 2 root 3 this is how factorization has to be done using splitting the middle term now we will be taking up more examples in the subsequent sessions