 Welcome again to another session on problem solving and we have been solving problems related to factor theorem and remainder theorem so the given question says find without actual Substitution the value of the given polynomial. So the polynomial looks very ugly. It's very big and The question says you have to find the value of this one Right at x equals to 15 now you can imagine if you put 15 Over here, then the calculation is going to be little tedious not little. It's two tedious, right? So if you see Value of a polynomial you have to find and how do we find out value of a polynomial? You simply put x equals to 15 in the given polynomial That is you have to calculate this number 19 times 15 to the power 5 plus 69 times 15 to the power of 4 this is multiplication minus 1 5 1 times 15 cubed plus 2 2 9 times 15 squared plus 1 6 6 times 15 and plus 26 now you could have imagined by now that It's really really tedious task and you know to calculate 15 to power 6 my god So there must be some alternative way and here is where the remainder theorem will come into play So how to do that? So basically we'll try to Take some factors out of this polynomial, let's say this polynomial is fx now from this fx We want to take out the you know Factors where in Or what factor what linear factor basically so I'll try to take out x minus 15 and then We'll see or basically will divide the entire fx by x minus 15 And try to find out the remainder in fact the remainder will be nothing but Whatever, you know, you just put the value x over there in the remainder and that will be the value of the function fx or polynomial fx why let me tell you how so let's say you could write fx as x minus 15 times Some gx you don't know the gx right now and finally our x Right, I can do that We have learned that dividend is equal to divisor into Potion plus remainder. So let's say if you divide the given polynomial fx by x minus 15 You would obtain some gx whatever it is and you'll also obtain rx. So hence What will be f 15 then so f of 15 if you see is nothing but If I put f of 15 here, it will become 15 minus 15 times gx Plus r 15, isn't it? Right, so that is what is you know, what we'll get this will become zero. This is zero, right? So hence f 15 is nothing but r 15, right? So r 15 if you divide the given polynomial by x minus 15 clearly, right? So let's now try to divide or basically try to extract x minus 15 or rather try to find out remainder Okay, let's try to find out a remainder of this. So how do we do it? So Let's try to factorize to an extent that we get x minus 15 So look at it carefully how I'm trying to factorize it. So x to the power 6 is there So if I need x minus 15 as a factor So don't you think I must have 15 x to the power 5 here? So the moment these two You know, so the terms are there in the after from the first two terms. I am extracting these two terms So clearly x minus 15 will be factor of this particular polynomial here small part of that polynomial, right? Now but in the given expression, there are 19 x to the power 5 minus 19 minus 15 I have extracted already. So how much is left? I am left with minus 4 x to the power 5, right? So x to the power 5 is taken care of Now for minus 4 x to the power 5, I would be needing plus 60 x to the power 4 Why? Because you can see again, you can take x to the power 4 common and you'll again get x minus 15 out of these two terms These two terms, let it be plus here Correct now, but there were 69 X to the power 4 so 9 is left. So now I will write 9 x to the power 4 So 69 x to the power 4 is taken care of for 9 x to the power 4 I need 15 times 9 that is 135 Is it it? I need 135 x to the power 3 So let me take 135, but there were 151 x cube. So how many left? around 16 left, right? So hence plus minus 16 x cube and for 16 x cube, I would be needing 16 times 50 that is 240. So plus 240 x square would be required Okay, now plus 240 x squared would be required, but 240 were not there Only 229 were there that means I have to subtract what? Minus 11 if I'm not wrong minus 11. Yes, so I will write minus 11 x square and for 11 I need 165. So I need 165 x Why I'm doing this after in the next step, it will become much clearer, right? So 165 x and what is left then guys? I'm left with one x only correct, so can I write that as X and then I need what minus 15 only but there was 26 so hence to compensate I will have to write 41 Correct. Now you could have imagined why did I do all of this manipulation? So I am purposefully trying to Get x minus 15 as factor. So now from the first two terms, I can get x to the power 5 common and within brackets x minus 15 That is what my intention was in the next case again I can take minus 4 x to the power 4 Common and then again, I will get x minus 15 In the third pair right that is 9 x 4 and minus 135 x cube you can again get 9 x cubed common To get x minus 15 over there And then again minus 16 x squared will be common and you'll get x minus 15 once again Followed by minus 11 common x minus 15 11 x common rather right and then plus x minus 15 Plus 41 correct. This is your fx and So you basically reduced the polynomial fx in terms of linear polynomial x minus 15 and some remainder now Clearly what will be f 15? So this is what was the objective. So f 15 is clearly x to the power 5 15 minus 15 or you know, so whatever it is Um, let's say x to the power 5 itself will be what? So I don't need to calculate all of these. I will just write for the sake of understanding So 15 to the power 4 times 15 minus 15 plus 9 times 15 to the power cube 15 to power 3 minus and multiplied by 15 minus 15 minus 16 15 square 15 minus 15 And then minus 11 times 15 Then 15 minus 15. So basically what I'm doing is wherever there is x I am deploying 15 over there because that is what is the value of polynomial fx at 15 x equals to 15 and then finally 41 here So if you notice, there is nothing much to discover here. So these are all zero. So all these terms are becoming zero And then my calculation becomes much simpler. So f 15 is simply 41, right? So the value of this big calculation here, it will reduce to simply 41 right now because you see f 15 everything gets cancelled out and only 41 is left. So f 15 is 41, so what is the learning learning is for such big polynomials if you could express the entire polynomial in terms of x minus 15 right then Or whatever the case maybe let's say tomorrow if you had to find out Let's say the same thing at x equals to let's say at x equals to 20 Then you will express the entire fx in terms of x minus 20 times some gx Plus rx you will do and then What will happen f of 20 will simply be nothing but the remainder When the entire polynomial is divided by x minus 20. So f 20 will be simply r 20 Okay, so this is what You know is the application of remainder theorem once again and you can use this theorem for such bigger calculations as well