 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that find a fourth degree polynomial function with real coefficients that has minus 3, 3 and iota as zeros such that f of 2 is equal to 150. We know that linear factor theorem states 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 with this key idea we shall move on to the solution. In this question we have to find a fourth degree polynomial function with real coefficients and we are given that minus 3, 3 and iota are its zeros we know that complex zeros always occur in conjugate pairs so if iota is the root then minus iota will also be the root hence the function has four roots or zeros and these are minus 3, 3 iota and minus iota. Now we use linear factor theorem given in the key idea we write the polynomial function as f of x is equal to a n into x minus 3 the whole into x minus of minus 3 the whole into x minus iota the whole into x minus of minus iota the whole this implies that f of x is equal to a n into x minus 3 the whole into x minus of minus 3 that is x plus 3 the whole into x minus iota the whole into x minus of minus iota that is x plus iota the whole which implies f of x is equal to a n into now x minus 3 the whole into x plus 3 the whole can be written as x square minus 3 square the whole by using the formula a minus b the whole into a plus b the whole is equal to a square minus b square into now using the same formula here we get x square minus iota square the whole since iota square is equal to minus 1 so we get f of x is equal to a n into x square minus 9 the whole into x square minus of minus 1 the whole which implies f of x is equal to a n into x square minus 9 the whole into x square plus 1 the whole now solving it further we get f of x is equal to a n into now x square into x square will be equal to x raised to the power 4 now x square into 1 is plus x square now minus 9 into x square is minus 9 x square minus 9 into 1 will be minus 9 the whole this gives f of x is equal to a n into x raised to the power 4 now plus x square minus 9 x square can be written as minus 8 x square minus 9 the whole now to determine this function we have to find the value of the leading coefficient a n now in the question it is also given that f of 2 is equal to 150 so we put x is equal to 2 in the function f of x and equate it to 150 so we get a n into 2 raised to the power 4 minus 8 into 2 square minus 9 is equal to 150 which implies a n into now 2 raised to the power 4 is 16 minus 8 into 2 square that is 4 minus 9 the whole is equal to 150 which implies that a n into 16 minus 32 minus 9 the whole is equal to 150 which implies a n into minus 25 is equal to 150 which further implies a n is equal to 150 upon minus 25 this implies a n is equal to minus 6 now substituting a n is equal to minus 6 in the function f of x is equal to a n into x raised to the power 4 minus 8 x square minus 9 the whole we get f of x is equal to minus 6 into x raised to the power 4 minus 8 x square minus 9 the whole which implies f of x is equal to minus 6 x raised to the power 4 plus 48 x square plus 54 thus the required polynomial function is f of x is equal to minus 6 x raised to the power 4 plus 48 x square plus 54 which is the required answer this completes our session hope you enjoyed this session