 Hi and welcome to the session. I am Asha and I am going to help you solve this problem which says Find the value of k if x-1 is a factor of px in each of the following cases. So first let us learn the factor theorem with the help of which will solve this problem. So this theorem says if px is any polynomial of degree greater than or equal to 1 at a is any real number first is x-a is a factor of px p at a is equal to 0 and second is p is equal to 0 x-a is a factor of px. So this theorem is a key idea that will help us to determine the value of k. Let us now start with the solution and the first part is px is equal to x square plus x plus k and we have to find the value of k if x-1 is a factor of px. Now since x-a is a factor of px this implies p is equal to 0. Therefore in this problem we are given that x-1 is a factor of px. So we must have p at 1 is equal to 0 or replacing the variable x in px by 1 we have 1 square plus 1 plus k is equal to 0 or 1 square is 1 plus 1 plus k is equal to 0 or 2 plus k is equal to 0 or k is equal to minus 2. Value of k first part is minus 2. So this completes the first part. Now proceeding on to the next part where px is equal to 2x square plus kx plus root over 2 and we have to find the value of k if x-1 is a factor of px. A very clear idea. If x-1 is a factor of px then we must have p at 1 is equal to 0. This is by factorial. Now replacing the variable x by 1 in the polynomial px we have 2 1 raise to the power 2 plus k into 1 plus root over 2 is equal to 0. So we have 2 plus k plus root over 2 is equal to 0 or k is equal to minus 2 minus root over 2 which is already equal to taking minus sign common 2 plus root over 2. Hence the value of k the second part is 2 plus root over 2. This completes the second part. And now proceeding on to the third part where px is equal to kx square minus root over 2x plus 1. Again here we are given that x-1 is a factor of px. Therefore by factor theorem we must have p at 1 is equal to 0. Thus replacing the variable x by 1 in px we have k 2 1 square minus root over 2 into 1 plus 1 is equal to 0 or k minus root over 2 plus 1 is equal to 0 or k is equal to root over 2 minus 1. Thus the value of k in the third part is root over 2 minus 1. This completes the third part. And now proceeding on to the last part where px is equal to kx square minus 3x plus k. Here again x-1 is a factor of px and we are required to find the value of k. Therefore by factor theorem we must have p at 1 is equal to 0. So replacing the variable x by 1 in px we have k 1 square minus 3 into 1 plus k is equal to 0 or k minus 3 plus k is equal to 0 or k plus k is 2k 3 on taking on the right hand side or k is equal to 3 upon 2. Hence the value of k in the last part is 3 upon 2. So this completes the last part. So hope you enjoyed this session. Take care and have a good day.