 Hello and welcome to the session. In this session we will discuss the fundamental theorem of algebra. Fundamental theorem of algebra states that if f of x is a polynomial of degree 1 or higher, then there is at least one complex root. We should keep in mind that complex numbers include real numbers. Real numbers are complex numbers where the imaginary part is 0. One of the important consequence of fundamental theorem is the linear factor theorem which states that if f of x has degree n greater than or equal to 1 with nonzero leading coefficient a n, then f of x has exactly n linear factors and can be written as f of x is equal to a n into x minus c 1 the whole into x minus c 2 the whole into and so on up to x minus c n the whole where c 1, c 2, c 3 and so on up to c n are real or complex zeros. Also some of the zeros and associated factors may be repeated. In other words we can say that a polynomial function of degree n where n is greater than equal to 1 can be factorized into n not necessarily distinct linear factors over the complex number field. The power on any repeated factors is known as its multiplicity factors that are not repeated have multiplicity is equal to 1. We should also note that complex zeros always occur in conjugate pairs that is if x plus iota y is a zero then its conjugate that is x minus iota y will also be the zero. For the timing you should know that complex numbers are those that can be written in the form a plus iota b where a and b both are real numbers and iota is the imaginary number. So complex numbers consist of real and imaginary parts. Let us consider an example write a polynomial function of third degree with real coefficients and leading coefficient as 1 its zeros are given as 3 and iota. Now let us start with the solution of the given problem. Here we are given that 3 and iota are the roots or zeros of the polynomial and we know that complex zeros always occur in conjugate pairs. So if iota is the root then minus iota will also be the root. So now we have three zeros of the given polynomial and these are three iota and minus iota. Now we use the three zeros and write f of x as product of three factors. Now the leading coefficient is also given as 1 so by linear factor theorem we get f of x is equal to 1 into x minus 3 the whole into x minus iota the whole into x minus of minus iota the whole which implies that f of x is equal to now 1 into x minus 3 the whole is x minus 3 into x minus iota the whole into x minus of minus iota the whole that is x plus iota the whole. This further implies that f of x is equal to x minus 3 the whole. Now we know that a plus b the whole into a minus b the whole can be written as a square minus b square. So using this formula here we can write x square minus iota square the whole and we know that the value of iota square is equal to minus 1 so here we get f of x is equal to x minus 3 the whole into x square minus of minus 1 the whole which implies that f of x is equal to x minus 3 the whole into x square plus 1 the whole which further implies that f of x is equal to now x into x square will be x cube x into 1 is x so we have x cube plus x now minus 3 into x square is minus 3 x square and minus 3 into 1 is minus 3 or it can be written as f of x is equal to x cube minus 3 x square plus x minus 3. Thus the required polynomial function is f of x is equal to x cube minus 3 x square plus x minus 3 this is the required answer this completes our session hope you enjoyed this session.