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: 89

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.