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

Chapter 6 - Section 6.2 - Exercise Set: 91 (Solution) Explanation : $y^2 + by + 20$ = $(y + p)(y + q)$ Since $p\cdot q$ = +20, 1) p and q have to be of the same signs For (p + q) = b and positive ‘b’ required, 2) both p and q cannot be of negative values, hence, only positive pairs are to be considered For p = 1, q = 20, b = 21 For p = 2, q = 10, b = 12 For p = 4, q = 5, b = 9 There is no need to consider both positive signs further as the sum of ‘b’ will be same and cyclic again So, all positive values of ‘b’ are 9, 12 and 21.