#### Answer

Chapter 6 - Section 6.2 - Exercise Set: 89 (Answer)
All positive values of ‘c’ are 15, 28, 39, 48, 55, 60, 63 and 64.

#### Work Step by Step

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.