Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.2 - Factoring Trinomials of the Form x2+bx+c - Exercise Set: 91

Answer

Chapter 6 - Section 6.2 - Exercise Set: 91 (Answer) All positive values of ‘b’ are 9, 12 and 21.

Work Step by Step

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.
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