 Hi and welcome to the session. I am Asha and I am going to help you solve this problem which says determine which of the following polynomials has x plus 1 a factor. So we have given 4 polynomials and we have to check whether x plus 1 is a factor of each of the following or not. So before finding the solution let us first learn what does factor theorem say. This theorem is going to help us in finding the solution of this problem. So according to this theorem if px is any polynomial of degree greater than or equal to 1 and a is any real number first is x minus a is a factor of px a is equal to 0 and second is p is equal to 0 if x minus a is a factor of px. So this theorem is a key idea that we will be using in this problem to check whether x plus 1 is a factor of given polynomials or not. Let us now start with the solution and the first one is x cube plus x square plus x plus 1. So let us name this polynomial as px which is equal to x cube plus x square plus x plus 1 and we have to check whether x plus 1 is a factor of px that is x plus 1 is a factor of px that is we have to check whether px minus 1 is equal to 0 or not. If px minus 1 comes 0 this implies x plus 1 is a factor of px if px minus 1 do not come 0 this implies it is not a factor of px. So replacing the variable x by minus 1 we have minus 1 whole cube minus 1 whole square plus minus 1 plus 1 which is further equal to minus 1 plus 1 plus into minus is minus so again minus 1 plus 1 minus 1 cancels out with plus 1 again minus 1 cancels out with plus 1 and we are left with 0. So this implies px minus 1 is equal to 0 so this further implies x minus of minus 1 according to the factor theorem is a factor of px which is the given polynomial. This implies x plus 1 is a factor this configures the first part it is now proceed on to the second part and let us name the polynomial as qx, qx is equal to x raised to the power 4 plus x raised to the power 3 plus x raised to the power 2 plus x plus 1 and we have to check according to the question is x plus 1 the given linear polynomial a factor of qx or not. A key idea we know that if on substituting minus 1 in place of x we get the answer is 0 then x plus 1 is a factor of qx otherwise it is not a factor of qx. So given linear polynomial is x plus 1 which can be written as x minus of minus 1 so to find the solution let find the value of this polynomial when x is replaced by minus 1. So we have minus 1 raised to the power 4 plus minus 1 raised to the power 3 plus minus 1 raised to the power 2 plus minus 1 plus 1. Now minus 1 raised to the power 4 is 1 plus minus 1 raised to the power 3 is minus 1 minus 1 raised to the power 2 is again 1 plus into minus is minus 1 plus 1. So plus 1 answers out with minus 1 this plus with minus and we are left with 1. So this implies q at minus 1 is equal to 1 which is not equal to 0 and hence x plus 1 is not a factor of qx. So this completes the second part now proceeding on to the third part and let us name the polynomial as rx which is equal to x raised to the power 4 plus 3x cube plus 3x square plus x plus 1 and we have to check is x plus 1 a factor of rx. Now according to the key idea x plus 1 is a factor of rx rx minus 1 is equal to 0 since x plus 1 can be written as x minus of minus 1 and minus into minus is plus. So to find the answer of this problem what we will find is the value of polynomial rx when x is replaced by minus 1. So we have minus 1 raised to the power 4 plus 3 into minus 1 raised to the power 3 plus 3 into minus 1 raised to the power 2 plus minus 1 plus 1 which is equal to minus 1 raised to the power 4 gives 1 plus 3 into minus 1 plus 3 into 1 plus minus 1 plus 1 which is further equal to 1 minus 3 plus 3 minus 1 plus 1 and plus 1 cancels up with minus 1 minus 3 with plus 3 and we have 1. This implies the value of the polynomial rx when x is replaced by minus 1 is 1 which is not equal to 0 and hence x plus 1 is not a factor of the given polynomial rx. So this completes the third part and now proceeding on to the last one and let us denote this polynomial as sx. So sx is equal to x cube minus x square minus 2 plus root 2x plus root 2 and we have to take s plus 1 a factor of sx. According to the key idea x plus 1 is a factor of sx if s at minus 1 is equal to 0 since x plus 1 can be written as x minus of minus 1. Now let us find the value of the polynomial sx when x is replaced by minus 1 we have minus 1 whole cube minus of minus 1 whole square minus 2 plus root 2 into minus 1 plus root 2 which is equal to minus 1 whole cube is minus 1 minus 1 whole square is 1 minus 2 plus root 2 into minus 1 plus root 2. This order is equal to minus 1 minus 1 is minus 2 and multiplying this minus with minus we get plus 2 plus root 2 plus root 2 plus root 2 minus 2 cancels out and we are left with 2 root 2 and thus the value of the polynomial sx when x is replaced by minus 1 is 2 root 2 which is not equal to 0 and hence x plus 1 is not a factor of so this completes the fourth part the answer is x plus 1 is a factor of the first polynomial but not the factor of second third fourth one so this completes the solution and please remember the factor theorem while doing these types of problems so hope you enjoyed this session have a good day bye