Let the polynomial represented by f(x). ∴f(x)=(x−1)(x−2) (x−4)−90 . . . (i) In the options given factors are (x+14), (x−14),(x−6) and (x−7) We know that (x−a) is a factor of the polynomial f(x), if f(a)=0 Factor (x+14) Put x+14=0 or x=−14 into f(x), we get f(−14)=(−14−1)(14−2)(−14−4)−90 =(−15)(−16)(−18)−90 =−4320−90 =−4410≠0 Factor (x−14) : Put x−14=0 or x=14 into f(x), we get f(14)=(14−1)(14−2)(14−4)−90 =(13)(12)(10)−90=1560−90≠0 Factor (x−6): Put x−6=0 or x=6 into f(x), we get f(6)=(6−1)(6−2)(6−4)−90 =5×4×2−90=40−90≠0 Factor (x−7): Put x−7=0 or x=7 into f(x), we get f(7)=(7−1)(7−2)(7−4)−90 =6×5×3−90 =90−90=0 Hence, (x−7) is a factor of the given polynomial.