 Hello friends, welcome to another problem-solving session on polynomials and we are dealing with remainder theorem right application of remainder theorem Now the question says if the polynomials fx equals ax cube plus 4x square plus 3x minus 4 and Gx equals x cube minus 4x plus a leave the same remainder When divided by x minus 3 so this is the divisor and they leave the same remainder when both of them are divided by x minus 3 you have to find out the value of a this a here, right? What is the value of a so what we have done is this is fx fx is equal to this much We have written that down and divisor is x minus 3 Now from remainder theorem, what do we see that if fx is divided by ax plus b then remainder will be simply f of Minus b by a correct remainder will be the value of the polynomial If you replace x or the variable by minus b by a so clearly in this case the first case The what is the value of b and a's in terms of the divisor? I'm talking about divisor don't get confused by the a which is here It's a little misleading but anyways General term I'm talking about where you have to divide by a linear ax plus b You can see this is our divisor in this case. What is this? ax plus b form and I'm comparing this with x minus 3 so clearly a is one now this a is not this one So let me do one thing. I am you know replacing it with capital a and Yeah, so that will make it much easier. Okay, so this a here. This divisor is of the form of ax plus b Okay, so ax plus b x minus 3 so a is 1 and b is minus 3 so minus b by a is 3 So you have to simply deploy 3 find the value of the polynomial at x equals to 3 So that's what we have done We have found this out and as I have removed or let's say this a is I Am converting it to capital a so let me convert it to capital a everywhere so that there is no confusion so this a Okay, 27 a plus 41 right so after calculation you get first remainder is 27 a plus 41 Now let's take the another point the second polynomial right second polynomial is x cube minus 4x plus a now I'm again remove replacing this a by capital a so that you guys are not Disturbed or confused Okay, so let me just remove that. Yeah So this is this a Okay, so we are finding capital a in both cases. Okay Now in the second case, this is our polynomial. This polynomial is given. So let me again write it as capital a Okay, and again, we'll apply remainder theorem. So hence g3 will give you the remainder so g3 remainder when gx is divided by x minus 3 will be simply g3 G of 3 so let me write that here and I have written that and you have to just replace this by capital a everywhere So capital a capital a Capital a right. So if you deploy 3 in the gx, you will get the remainder second remainder r2, which is 15 plus a Okay, now it's given that both of them leave the same remainder That means if you divide this expression by x minus 3 and this expression or polynomial by x minus 3 They leave the same remainder. So that means the remainders can be equated both of them are same So I have equated our one is going to art and let me just replace these by capital a everywhere so that you are not confused So this one is also 27 capital a so we are talking in terms of capital a only Okay, so this one is capital a so if you solve this expression 27 a plus 41 was the remainder obtained by First division 15 plus a is a remainder obtained by second division You will quit both of them and you solve you will get a is equal to minus one This is the solution so if capital a is minus one then both these polynomials leave the same remainder when divided by this particular Devise I hope you understood the application of this theorem in this question