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

Chapter 6 - Section 6.3 - Exercise Set: 95 (Answer) For $3x^2 + bx - 5$, all positive values of ‘b’ are 2 and 14.
Chapter 6 - Section 6.3 - Exercise Set: 95 (Solution) Explanation : $3x^2 + bx - 5$ = $(3x + p)(x + q)$ Since $p\cdot q$ = -5, p and q have to be of opposite signs For (p + 3q) = b and positive ‘b’ required, If p = 1, q = -5 then b = -14 (Incorrect result) If p = 5, q = -1 then b = 2 (Correct) If p = -1, q = 5 then b = 14 (Correct) If p = -5, q = 1 then b = -2 (Incorrect result) So, all positive values of ‘b’ are 2 and 14.