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: 117



Work Step by Step

With $49x^2=(7x)^2$ and $\frac{1}{25}=(\frac{1}{5})^2$, the given binomial is equivalent to: $=(7x)^2-(\frac{1}{5})^2$ Factor using the formula $a^2-b^2=(a+b)(a-b)$ with $a=7x$ and $b=\frac{1}{5}$ to obtain: $=\color{blue}{(7x+\frac{1}{5})(7x-\frac{1}{5})}$
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