 Hello and welcome to another session on polynomials and today We are going to learn a very interesting method of multiplication of polynomials, so In the you know going forward you are going to use multiplication of polynomials a lot and hence We must have a very convenient method of multiplying So one of those method is called method of detached coefficients And it saves a lot of time in writing the you know variables multiple number of times while multiplying to polynomials So we will explain this process using an example so you can see there is a There is a product which has to be determined x square minus x plus 1 is to be multiplied by x square plus x plus 1 So first of all if the expressions or the polynomials are given you have to make it complete. What does that mean? You have to make sure that all the powers of the variable are there from the minimum to the maximum that degree of the polynomial so in this case if you see the first factor or the Multiplicant in this case The degree is 2 so hence powers of you know x should be 1 Sorry 0 1 and 2 so you can see it is completely Written like that so power of x is 0 then 1 then 2 similarly here also It is complete polynomial because all the powers of x are reflecting over there Now let's say if you had an expression like x cube plus 1 Okay, and one of the multiplicands would have been this Then you have to write it like x cube plus 0 times x squared Plus 0 times x and then finally 1 right. So this is the complete polynomial now once you have written the polynomial completely that means all the powers of the variables have been expressed Then you need to just detach the coefficients. So, you know all the coefficients which you can see So, let's take the first polynomial here x square minus x plus 1 this one the detached coefficients will be simply 1 Minus 1 and 1 so I'm writing it like that. Okay. Similarly in the second case it is 1 plus 1 and Plus 1 isn't it? So this is what it is Right now We have to just follow the normal process of multiplication, you know the way we do multiplication now first thing is Multiply this 1 by plus 1. So what is the result? You will get plus 1 Okay, then multiply this by this You will get minus 1 and then finally this by this. So what do you get you get plus 1 again? Now for the second line what you need to do is now you have to take this one and multiply with this one and Write just like you shift in your normal multiplication process. So you will write plus 1 here Then this one will get multiplied by this minus 1 here And the product is minus 1 just right in the left of it and then finally this one. So this will be plus 1 Okay, now you would have guessed it by now. So what you need to do next take? this one and multiply by right and write it here The next one, you know, so exactly underneath this one. So this is plus 1 and then Right, so multiply that you'll write minus 1 and this Here you write plus 1 here and then finally you sum them the way you Do it in normal multiplication. So this is one clearly This one is 0 This one is plus 1 again. This one is 0 and this one is plus 1 again and now you know how to Though what is the final result? So you have to just start putting the powers of or the x is here So x to power 0 x to power 1 x to power 2 x to power 3 and x to power 4 Right. So the product is simply x to the power 4 plus x square plus 1 you can check that this is the product Okay, let's take another example to make it more clear now, let's say We have been given to polynomials one is x to the power 5 plus x cube minus 2 x square Plus 3. Okay, this is polynomial number one and it has to be multiplied by this one x to the power 4 minus 3 x cube plus 4 x square minus 1 Okay, now if you can you see closely these two polynomials are not complete. There are few powers of x which are missing So let me, you know complete the expression. So it will be x to the power 5 plus x 0 times x to the power 4 plus x to the power 3 minus 2 times x squared plus 0 times x plus 3 Correct. And then this one is multiplied by 2 x to the power 4 minus 3 x cubed plus 4 x squared Plus 0 times x minus 1. Isn't it? Now, let's detach the coefficient. So detaching the coefficient you'll get Let me start from here 1 0 1 minus 2 0 3 Correct. 1 0 1 minus 2 0 3 and the next one will be simply 2 minus 3 4 0 minus 1 Okay, now, you know what to do simply multiply like that you can you know If you're very very Let's say you have practiced enough you don't need to write in this structure also you can do from Here itself if you if you see you know if you have practiced enough from here itself You know what to multiply when so if you now start multiplying so minus 1 goes to this So it becomes minus 3 then it becomes like that. So I will do for 1 It becomes 2 It becomes minus 1 It becomes 0 and this becomes minus Correct. This is the first one and next one. Anyways, you know, this is 0 all zeros. How many zero six times zeros? You have to just keep shifting one by one like that. Next is 4. So 4 3s are 12 Okay, so right full no carry and all so 4 3s are 12 4 0 4 times 0 0 So I'm what I'm doing. I hope you're understanding now 4 3 4 0 4 minus 2 4 1 4 0 and then finally 4 1 I will do so I'm writing the results now. So 12 0 minus 8 and 4 and 0 and 4 next one will be minus 9 then 0 minus plus 6 minus 3 0 and minus 3 Okay, and then finally with 2 it is 6 0 minus 4 Then 2 0 and 2 and then simply add Add all of them. You'll get minus 3 0 14 Then minus 10 then minus 2 then 9 Then minus 7 then 6 minus 3 and 2 Okay, so hence now you have to just put the Variable as in variable. So this is x to the power 0 x to power 1 x to power 2 x to power 3 x to power 4 x to power 5 6 7 8 9 Okay, so the answer is 2 x to the power 9 minus 3 x to the power 8 plus 6 x to the power 7 minus 7 x to power 6 plus 9 x 5 minus 2 x 4 minus 10 x cubed Plus 14 x squared 0 x n minus 3 This is the product. Okay, so you might be thinking that it is taking a little bit of more time while writing all of that but you don't need to Write in this format. You can do it verbally as well. Okay, so that will help in You know multiplying two polynomials faster. So this is called multiplication by detached Coefficient, I hope you understood this process try some more taking some example polynomials yourself and then try and then It will become easier to multiply bigger polynomials very very easily