Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 52



Work Step by Step

Using factoring techniques, then, \begin{array}{l} 9y^2+12y+4-x^2 \\= (9y^2+12y+4)-x^2 \\= (3y+2)^2-z^2 \\= [(3y+2)-z][(3y+2)+z] \\= (3y+2-z)(3y+2+z) .\end{array}
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