## Algebra: A Combined Approach (4th Edition)

Chapter 6 - Section 6.2 - Exercise Set: 89 (Solution) Explanation : $n^2 - 16n + c$ = $(n + p)(n + q)$ Since c = $p\cdot q$, for all positive ‘c’, 1) p and q have to be of the same signs For (p + q) = -16 2) both p and q cannot be of positive values, hence, only negative pairs are to be considered For p = -1, q = -15, c = 15 For p = -2, q = -14, c = 28 For p = -3, q = -13, c = 39 For p = -4, q = -12, c = 48 For p = -5, q = -11, c = 55 For p = -6, q = -10, c = 60 For p = -7, q = -9, c = 63 For p = -8, q = -8, c = 64 There is no need to consider both negative signs further as the product of ‘c’ will be same and cyclic again So, all positive values of ‘c’ are 15, 28, 39, 48, 55, 60, 63 and 64.