Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 44: 64


$(6z + 9y^{2})(6z - 9y^{2})$

Work Step by Step

This polynomial is a difference of squares, so $a^{2} - b^{2} = (a-b)(a+b)$. In this case, $a^{2} = 36z^{2}$ so $a = 6z$ and $b^{2} = 81y^{4}$ so $b = 9y^{2}$. Therefore, $(a-b)(a+b) = (6z + 9y^{2})(6z - 9y^{2})$.
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