 Hello friends, so as discussed in the previous session We will now in this session study division of polynomials using division algorithm So one way of dividing two polynomials You already know and that is called long division method a long division method where we divide as per the division rules Right now in this session. We are going to discuss this division of polynomials using division algorithm Which we are not going to perform long division method at all so let us say the question is fx is x to the power 4 minus 3x squared plus 4x Plus 5 right mind you there is no x cube term in it. So hence degree fx is Nothing but 4 Right now I have to divide this by I'm taking the same example Which I took in the previous session x square minus x plus 1 in the last session We do did that long division this method this session. We are not going to do that We are going to use the division algorithm. Okay now degree of gx Clearly is 2 so by division algorithm I can say if I must get qx and rx such as fx is equal to gx times a quotient polynomial plus a remainder polynomial Okay, now Degree of fx is 4 right degree of fx is 4 that means and And degree of gx is 2 that means qx must have a degree of 2 as well Is it it you understand? See fx has a degree of 4 So this will be a polynomial of something like fx is anyways is x to the power 4 The first term is x to the power 4 minus all that other other stuff now gx has the first Gx is like this Okay, x square minus x now definitely I must have a power of x to here Then in in qx such that when I open this bracket by multiplication So I will get x square times x square as this x to the power 4 Correct if it was not x square Let us say if you counter-argue know why can't it be only x then In no in nowhere here the entire in the entire multiplication I will get a term which is x to the power 4 Understand see if it is x the highest power is x only So if I multiply highest power of this to highest power of this I will get x cube Maximum but then how do I find out x 4 if LHS is equal to RHS then there must be x to the power 4 term in RHS as well But you are saying the maximum is x cube only because all others will be lesser than one here, isn't it? So that means there must be a term containing x square here and Of and also it cannot be x cube. There cannot be any term like x cube here Why because if there is an x cube then when you multiply this with x square You will get a term of x to the power 5 which is not there in the LHS So hence with this logic I can say that this term must be having The second term this one must be having an x square in it and no more Less can be possible, right, but it cannot be More it cannot be x cube x to the power Cube but certainly x square has to be there and other lesser degree terms are also possible But you cannot have a polynomial here, which doesn't have x square only Okay, I hope you understood it. So hence if the if the Divider divisor had if the divisor had x square in it then The quotient must also have x square in it then only you can equate the two degrees on the left-hand side in the right-hand side Plus we have What do I have in the remainder? I have a remainder as Rx Now I know degree of Rx will be less than degree of this. That means dig Rx Rx is either a linear factor linear polynomial or it is a constant polynomial Constant polynomial. Why because we know that the remainder degree of Rx degree of Rx is less than degree of gx Isn't it so hence degree of gx is to which is quadratic So Rx is less left over the left over Rx will have only either linear or a constant polynomial, right Now then what is the conclusion conclusion is qx is a quadratic is degree to polynomial polynomial and Rx is Either linear or constant. Okay, so if qx is degree to polynomial, what kind of Degree to polynomial are like a standard form is a x square plus bx plus c Correct, let qx be this and let this guy Rx be let us say dx plus e Linear polynomial you will say what if it is constant So let's say eventually if we come to know that it is constant then D will come out to be 0 Right, if it is a constant polynomial, let us see how so What is the x so let them let me rewrite the entire expression So hence now I can rewrite x to the power 4 minus 3x square Plus 4x plus 5. This is the dividend What was the divisor x square minus x plus 1 What is the quotient? I am saying it is a x square plus bx plus c and what is the remainder? Let us say dx plus e Okay Now lhs is equal to rhs then the coefficient must also be equal So let me just simplify this on the right hand side. I will get what a x to the power 4 then A x to the power 4 then minus bx to the power 3 then I'm sorry plus bx to the power 3 bx to the power 3 and then cx square isn't it I'm now opening and multiplying one by one then Minus ax cube. So I'm writing it here minus ax cube then minus bx square then minus cx Right and then finally ax squared Right and then bx Plus c and then there is a dx and plus e Right, so eventually if you take all the like terms together, so it is ax to the power 4 plus b minus ax cube Plus c minus b plus ax squared Plus b plus dx Plus c plus e isn't it? Now lhs is equal to rhs, isn't it? So on hence all the coefficient must Also be same since x is not equal to zero See if x equals to zero then though there is no problem right a b c d can be anything But if x is not equal to zero Then what I have to This particular lhs will be equal to rhs only when one is equal to a why? Because here is the coefficient is one here. It is a so it must a must be equal to one Then the coefficient of x cube is zero Isn't it? There's no x cube here. So hence what will happen? Zero is equal to b minus a and a was one that means b also is equal to one Right then what the third is minus three and it is c minus b plus a Right now b is one a is one. So a minus b is zero. So hence we'll get c is equal to minus three Isn't it c is equal to minus three then what b plus d on the left hand side is four. So b plus d now b is one this implies d is equal to three Correct and c plus e so five this last term is the constant c plus e So c plus e now c is minus three this implies E is equal to five minus c which is eight Isn't it? So What do we get? We got all the a v c d value. So hence G sorry q x c without division itself We could find out q x now q x was a x square plus b x plus c Which is now equal to what? x square a is one b is one. So x Plus c minus three. So x square plus x minus three is the quotient and what is the remainder rx is Rx is dx plus e now Wait a minute four three plus d Oh, I'm sorry here. There is a mistake. The mistake is it is b plus d Minus c is missing. So minus cx is also there, right? So b plus d minus c. So please do a correction minus cx. So hence here you will get Minus c and hence d is equal to so this is four minus one Plus minus three which is equal to zero. Okay. So d is zero guys. So hence d is zero So hence, what will you get? You will get only e Which is equal to eight. So rx is equal to eight, right? So in this process, what do we get? q x the quotient is x square plus x minus three And rx is equal to eight. So without division we could find out what is q x and is very minor