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 45: 118



Work Step by Step

With $81y^2=(9y)^2$ and $\frac{1}{49}=(\frac{1}{7})^2$, the given binomial is equivalent to: $=(9y)^2-(\frac{1}{7})^2$ Factor using the formula $a^2-b^2=(a+b)(a-b)$ with $a=9y$ and $b=\frac{1}{7}$ to obtain: $=\color{blue}{(9y+\frac{1}{7})(9y-\frac{1}{7})}$
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