The Chief Examiner reported that this question was attempted by majority of the candidates and they performed well in it. Candidates were able to show that since       f(-1) = 0, then (x + 1) is one of the factors. They also divided f(x) by (x + 1) to obtain f(x) = (x + 1)(6x2 + 7x – 5).  (6x2 + 7x – 5) was then factorized to obtain                (6x2 + 7x – 5) = (2x – 1)(3x + 5). Thus, f(x) = (x + 1)(2x – 1)(3x + 5).  i.e. factors of f(x) are (x + 1), (2x + 1) and (3x – 5). It was however observed that some candidates found the zeroes of f(x) instead of its factors. Candidates should note this difference.