 Hi and welcome to the session. I am Asha and I am going to help you with the following question which says find P0, P1 and P2 for each of the following polynomials. Starting with the first one which is Py is equal to y square minus y plus 1 and we have to find P0, P1 and P2. Now to find the values of P0, P1 and P2 we will replace the variable y in the polynomial Py by 0, 1 and 2 respectively. So first let us find the value of P0 which we will get on replacing this y by 0. So first we have 0 square minus 0 plus 1 which is equal to 0 minus 0 plus 1 which is further equal to 1. So this implies P at 0 is equal to 1. Now we have to find the value of the polynomial Py at y is equal to 1. So P1 is replacing this variable y by 1, we have 1 square minus 1 plus 1 which is equal to 1 minus 1 plus 1 and minus 1 plus 1 cancels out we are left with 1. Thus the value of the polynomial Py at y is equal to 1 is also 1. And lastly we have to find the value of the polynomial Py at y is equal to 2. So replacing the variable y by 2 we have 2 square minus 2 plus 1 which is equal to 2 square is 4 minus 2 plus 1. Now 4 plus 1 is 5 and 5 minus 2 is 3. Thus we have 3 and hence the polynomial Py at y is equal to 2 is equal to 3. Thus we have P0 is equal to 1, P1 is equal to 1 and P2 is equal to 3. So this completes the first part and now proceeding on to the next one where the polynomial is Pt is equal to 2 plus t plus 2t square minus d cube and we have to find the values of the polynomial Pt at 0 at t is equal to 1 and t is equal to 2. So first let us find the value of the polynomial Pt at t is equal to 0. So we have on substituting t as 0 in the polynomial we get 2 plus 0. Now 0 square is 0 on multiplying it with 2 we get 0 and 0 cube is again 0. So the value of the polynomial Pt at t is equal to 0 is 2. Now let us find the value of the polynomial Pt at t is equal to 1. So we have 2 plus 1 plus 2 into 1 square minus 1 cube which is equal to 2. 2 plus 1 is 3 and 1 square is 1 on multiplying it with 2 it gives 2 and 1 cube is 1. Thus 3 plus 2 is 5 and 5 minus 1 is 4 and hence the value of the polynomial Pt at t is equal to 1 is 4 and lastly let us find the value of the polynomial Pt at t is equal to 2 and on substituting t as 2 we have 2. In place of t we will put 2 and we have 2 plus t square t is 2 whole square minus t cube is 2 whole cube. Now 2 plus 2 is 4 and we have plus 2, 2 square is again 4 and 2 cube is 2 into 2 into 2 that is 8 and 2 into 4 is 8. So plus 8 minus 8 cancels out and we have 4. Thus the value of the polynomial Pt at t is equal to 2 is 4. Hence we have P0 is equal to 2, P1 is equal to 4 and P2 is equal to 4. So this completes the second part. Let us now proceed on to the third part where Px is equal to x cube and we are required to find P0, P1 and P2. First let us find the value of P0 which we get on replacing the variable x by 0 that is we have 0 whole cube which is equal to 0. Thus P at 0 is equal to 0. Now let us find the value of the polynomial Px at x is equal to 1 which we get on replacing 1 x by 1 which is equal to 1 and thus P at 1 is equal to 1. And lastly on finding the value of the polynomial Px at x is equal to 2 we have 2 whole cube which is equal to 8 and thus the value of the polynomial at x is equal to 2 is 8 and thus we have P at 0 is equal to 0, P at 1 is equal to 1 and P at 2 is equal to 8. So it completes the third part and now proceeding on to the last part which is Px is equal to x minus 1 into x plus 1. And we are required here to find the values of the polynomial Px at x is equal to 0, x is equal to 1 and x is equal to 2. So first let us find the value of the polynomial Px at x is equal to 0. We have 0 minus 1 and 0 plus 1, 0 minus 1 is minus 1 and 0 plus 1 is 1 and on multiplying minus 1 with 1 we get minus 1 and thus the value of the polynomial Px at x is equal to 0 is minus 1. Now let us find the value of the polynomial Px at x is equal to 1, thus we have 1 minus 1 into 1 plus 1 and 1 minus 1 is equal to 0 and 1 plus 1 is equal to 2 and on multiplying 0 with 2 we get 0, thus the value of the polynomial P1 is equal to 0. And now let us find the value of the polynomial Px at x is equal to 2, thus replacing the variable x by 2 we have 2 minus 1 into 2 plus 1 and 2 minus 1 is 1 and 2 plus 1 is 3 and on multiplying 0 with 1 we get 3, thus the value of the polynomial Px at x is equal to 2 is equal to 3 and thus P at 0 is equal to minus 1, P at 1 is equal to 0 and P at 2 is equal to 3. So this completes the solution, so hope you enjoyed it, just be careful in adding, subtracting and multiplying according to the sign and replace the variable by the given values to find the value of the given polynomial. So take care and have a good day.